A mathematical account of the electromagnetic structure of photonic crystals
can be found in [83]. We shall start by arguing that the propagation of TE
polarised light, with $E=(E_{1},E_{2},0)$ and $H=H_{3}\hat{e}_{3}$, where
$\hat{e}_{i}$ is the $i$-th unit vector, in two dimensional photonic crystals
can be reduced to a periodic eigenvalue problem with Helmholtz operator
$\mathcal{L}=\nabla\cdot D_{e}\nabla$

$\mathcal{L}\psi=\omega^{2}\psi$ (125)

with $\psi=H_{3}.$ Here, $D_{e}=\frac{J^{T}\kappa J}{\mu_{3}}$ is a periodic
function, with $J=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix},$ that only depends
on the permittivity and permeability of the material and will be specified
below. We start from the Maxwell equations

$\begin{split}\nabla\times E&=-\partial_{t}B,\ \nabla\times
H=\partial_{t}D,\\\ \nabla\cdot D&=0,\ \nabla\cdot B=0.\end{split}$

We then find that Maxwell’s equation can be written as a Schrödinger equation
with

$i\partial_{t}\Psi=M\Psi$

with $\Psi=(E,H)^{t}$ and

$M=R^{-1}\begin{pmatrix}0&i\nabla\times\\\ -i\nabla\times&0\end{pmatrix}.$

Here, $R=\begin{pmatrix}\varepsilon&\zeta\\\ \zeta^{*}&\mu\end{pmatrix}$ with
permittivity tensor $\varepsilon$, permeability tensor $\mu$, and bianisotropy
tensor $\zeta.$

For photonic analogues of 2D material, we shall only consider matrix entries
of $R$ that vary in the two-dimensional plane. In addition, we assume that the
coupling between the longitudinal and transversal direction is zero.

This way, we may decouple the $6\times 6$ Maxwell equations into two coupled
equations with matrices

$R_{\perp}=\begin{pmatrix}\varepsilon_{\perp}&\xi_{\perp}\\\
\xi_{\perp}&\mu_{\perp}\end{pmatrix}\text{ and
}R_{||}=\operatorname{diag}(\varepsilon_{||},\mu_{||})$

that are the form

$\begin{split}\partial_{t}(E_{\perp},H_{\perp})&=R_{\perp}^{-1}(-J\nabla_{\perp}H_{3},J\nabla_{\perp}E_{3})\text{
and }\\\ \partial_{t}(E_{3},H_{3})&=R_{||}^{-1}(-J\nabla_{\perp}\cdot
H_{3},J\nabla_{\perp}\cdot E_{3}).\end{split}$ (126)

Here $\nabla_{\perp}=(\partial_{x_{1}},\partial_{x_{2}}).$ Eliminating
$(E_{\perp},H_{\perp})$ and using
$R_{\perp}^{-1}=:\begin{pmatrix}\kappa&\zeta\\\ \zeta^{*}&\iota\end{pmatrix},$
we find under the assumption of pure TE waves, that is $E_{3}=0$, the wave
equation

$\partial_{t}^{2}H_{3}(x_{1},x_{2},t)=-\mathcal{L}H_{3}(x_{1},x_{2},t).$

Observe that by (126), $(E_{\perp},H_{\perp})$ are directly determined from
$H_{3}.$ By looking at planar waves
$H_{3}(x_{1},x_{2},t)=H_{3}(x_{1},x_{2})e^{\pm i\omega t},$ we reduce the wave
equation to (125).

In magneto-optic materials, one has

$R_{\perp}=\operatorname{diag}(\varepsilon\sigma_{0}-\gamma\sigma_{2},\mu\sigma_{0}),$

where $\gamma$, with $|\gamma|<\varepsilon,$ is the strength of the Faraday-
rotation. This way,
$\kappa=\frac{\varepsilon}{\varepsilon^{2}-\gamma^{2}}\sigma_{0}+\frac{\mu}{\varepsilon^{2}-\gamma^{2}}\sigma_{2}.$

####  IX-A2 Bloch-Floquet theory and band structure of photonic crystals

Photonic crystals are periodic with respect to some lattice $\Gamma.$ This is
reflected in the periodicity of the operator $\mathcal{L}$. We can thus apply
the standard Gelfand transform and find that that $\mathcal{L}$ is unitarily
equivalent to the direct integral operator

$\int_{\mathbb{C}/\Gamma}^{\oplus}\mathcal{L}(k)\frac{dk}{|\mathbb{C}/\Gamma|}\text{
with }\mathcal{L}(k)=-(\nabla+ik)\cdot D_{e}(\nabla+ik)$

for which the eigenvalue problem (125) converts into

$\mathcal{L}(k)H_{3}(x,k)=\omega^{2}(k)H_{3}(x,k),\quad
H_{3}(x+\Gamma,k)=H_{3}(x,k)$

for $k\in\mathbb{C}/\Gamma^{*}$, the Brillouin zone. By standard arguments,
the set of admissible $\omega$ for given $k$ is discrete
$(\omega_{n}(k)^{2})_{n\in\mathbb{N}}$ with associated fields
$(H_{3,n}(x,k)).$

The function $\omega(k)^{2}$ is also called the dispersion relation and
$\partial_{k}\omega(k)$ is the speed of propagation.

Thus, any eigenfunction of $\mathcal{L}(k)$ has a Fourier expansion

$\phi(x)=\sum_{m\in\mathbb{Z}^{2}}\phi_{m}e^{i(m_{1}k_{1}+m_{2}k_{2})\cdot
x},$

where $\Gamma^{*}=k_{1}\mathbb{Z}+k_{2}\mathbb{Z}.$

####  IX-A3 The quantisation of the EM field

In order to study quantum effects such as spontaneous emission, it is
essential to use a field theoretic description of the electromagnetic field
inside a photonic crystal[81, 82].

The filling of bands in photonic crystals by photons can be described by
quantizing the magnetic field and consider field operators

$\begin{split}\widehat{H_{3}}(x,k,t)&=\sum_{n\in\mathbb{N}}h_{n}(k)\Big{(}H_{3,n}(x,k)a_{n}(k)e^{-i\omega_{n}(k)t}\\\
&+\overline{H_{3,n}(x,k)}a_{n}(k)^{*}e^{i\omega_{n}(k)t}\Big{)}\end{split},$

where $h_{n}(k)=\frac{\hbar\omega_{n}(k)}{2\mu|\mathbb{C}/\Gamma|}$ and
$a_{n}(k),a_{n}(k)^{*}$ are annihilation and creation operators of photons
with wavevector $k$ in band $n$. The macroscopic limit is then obtained by
taking the expectation value of the field vector
$\langle\widehat{H_{3}}(x,k,t)\rangle$ with respect to the photonic state.

####  IX-A4 Dirac points

As a simple example, we shall consider the case that $D_{e}$, appearing in
(125), has honeycomb-lattice symmetries, i.e. $\Gamma$ is an equilateral
triangular lattice with $2\pi/3$-rotational symmetry In this case, it is well-
known, see e.g. [83], that the spectrum of $\mathcal{L}(k)$ at suitable
energies can be effectively described, close to some positive energy
$E_{D}>0$, by a two-dimensional Dirac operator

$H(k)=\begin{pmatrix}0&2D_{\bar{z}}+k\\\ 2D_{z}+\bar{k}&0\end{pmatrix}\text{
with }z\in\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$

and $k\in\mathbb{C}$ with periodic boundary conditions where
$D_{z}=\frac{1}{i}\partial_{z}.$ This operator can be diagonalised with Bloch
functions $\left(\begin{array}[]{c}1\\\
\pm\frac{\bar{k}}{|k|}\end{array}\right)$ and eigenvalues $\pm|k|.$ This means
that the original frequencies satisfy for $k$ small enough
$\omega(k)^{2}=E_{D}\pm|k|,$ where $E_{D}$ is the energy level of the Dirac
cones in the spectrum of $\mathcal{L}(k).$

####  IX-A5 Strain-induced pseudomagnetic fields

Since photons do not directly interact with electromagnetic fields, the effect
of electromagnetic fields on the electronic band structure cannot be modelled
directly using external fields. However, it has been observed that physical
strain can be used to imitate the effect of electromagnetic fields, see e.g.
[84]. Indeed, consider the displacement field $T(z)=z+u(z)$ with $u$ a
displacement vector. In the case of a honeycomb-lattice symmetries, the
effective strain-induced magnetic potential is then given as

$A(z)\propto\operatorname{Tr}(U(z)\sigma_{3})-i\operatorname{Tr}(U(z)\sigma_{1})$

where $U(z)=\frac{1}{2}(Du+Du^{t}),$ with Jacobi matrices $Du.$ This way, by
applying suitable strain, we obtain a pseudo-magnetic potential
$A(z)=\frac{iBz}{2}$ associated with a constant magnetic field $B>0$ and the
effective Hamiltonian is the magnetic Dirac operator

$H=\begin{pmatrix}0&2D_{\bar{z}}-\frac{1}{2}iBz\\\
2D_{z}+\frac{1}{2}iB\bar{z}&0\end{pmatrix},$ (129)

with constant magnetic field $B>0.$

An example of, up to a change of gauge, pseudomagnetic field inducing strain
is

$T(z)=z+\kappa^{2}\Im(z)^{2}.$

Figure 6: Different strain profiles leading to effective constant magnetic
fields. Increasing field strenghts from left to right.

We then have the conjugation relation

$e^{-B|z|^{2}/4}(2D_{\bar{z}})e^{B|z|^{2}/4}=2D_{\bar{z}}-\frac{i}{2}Bz.$

The infinitely degenerate ground state satisfies then

$u(z)=f(z)e^{-\frac{B|z|^{2}}{4}},f\in\mathscr{O}(\mathbb{C}),\int_{\mathbb{C}}|f(z)|^{2}e^{-B|z|^{2}/2}dz<\infty,$

where $\mathscr{O}(\mathbb{C})$ is the space of entire functions. The
infinitely degenerate ground state can be equivalently interpreted as a flat
band, see e.g. [85].

Let $a=2D_{z}-\overline{A(z)},$ one then has that $[a^{*},a]=2B$ which means
that to find the zero energy band for the Hamiltonian (129), we have to find
zero modes to $a^{*}$ satisfying Bloch-Floquet boundary conditions.

Let $E_{D}>0$ be the energy of the Dirac point, then the lowest band is just
$\omega(k)^{2}=E_{D}$ for all $k$. Correspondingly, by the commutation
relation of the ladder operators, we find that the other (also flat) bands are
of the form $\omega(k)^{2}=E_{D}^{2}\pm\sqrt{2nB}$ with $n\in\\{0,...,N\\}$
for some large enough $N$ which is in fact independent of $k$, since the band
is flat.

As a simple example, we may choose photons in the strained crystal with state

$|\psi\rangle=\prod_{i=1}^{n}|0\rangle_{k_{i}}$

with harmonic oscillator frequencies $\omega(k_{i})\equiv E_{D}.$ In Figure 7
we see that high-energy states are in general harder to reconstruct as their
characteristic function is more extended in phase space.

Figure 7:  Reconstruction error for single particle characteristic function
with $n=5$ particles for $E_{D}=0.5$ and $E_{D}=5$ with $N=150$. High energies
are harder to reconstruct, as they are more

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##  Appendix A Proof of (41)

The purpose of this appendix is to provide a self-contained proof of the
representation (41) of the Gaussian white noise channel. We start by observing
that it suffices to test (41) on all projectors $\ket{s}\\!\\!\bra{s}$ on the
coherent states, where $s\in\mathds{R}^{2m}$. This is because the linear span
of such projectors is dense in the space of Hilbert–Schmidt operators. To
prove this latter claim, we can take an arbitrary Hilbert–Schmidt operator $X$
and assume that it is orthogonal to
$\operatorname{span}\\{\ket{s}\\!\\!\bra{s}\\}_{s\in\mathds{R}^{2m}}$, in
formula
$\operatorname{Tr}\left[X\ket{s}\\!\\!\bra{s}\right]=\braket{s}{X}{s}=0$ for
all $s\in\mathds{R}^{2m}$. We can express this in words by saying that the
Husimi Q-function $Q_{X}(s)\coloneqq\frac{1}{(2\pi)^{m}}\braket{s}{X}{s}$ of
$X$ vanishes everywhere. Since the characteristic function (28) of $X$ is a
point-wise multiple of the Fourier transform of $Q_{X}$, in formula [66, §
4.5]

$\displaystyle\chi_{X}(t)=e^{\|t\|^{2}/4}\int
d^{2m}s\,Q_{X}(s)\,e^{is^{\intercal}\Omega t}\,,$ (130)

we have that also $\chi_{X}\equiv 0$ identically. Since the correspondence
between Hilbert–Schmid operators and characteristic functions is an isometry
and in particular injective [67, Theorem 5.3.3], we conclude that $X=0$. Since
$X\in\operatorname{span}\\{\ket{s}\\!\\!\bra{s}\\}_{s\in\mathds{R}^{2m}}^{\perp}$
was arbitrary, this entails that
$\operatorname{span}\\{\ket{s}\\!\\!\bra{s}\\}_{s\in\mathds{R}^{2m}}$ is
dense.

Therefore, let us verify (41) by letting the right-hand side act on an
arbitrary $\ket{s}\\!\\!\bra{s}$. We obtain that

$\displaystyle\int\\!\\!\frac{d^{2m}x}{(2\pi\lambda)^{m}}\,\pazocal{D}(x)\,\tau_{\\!\frac{1}{2\lambda}-\frac{1}{2}}^{\otimes
m}\pazocal{D}(x)^{\dagger}\ket{s}\\!\\!\bra{s}\pazocal{D}(x)\,\tau_{\\!\frac{1}{2\lambda}-\frac{1}{2}}^{\otimes
m}\pazocal{D}(x)^{\dagger}$ $\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(i)}}}}{{=}}\int\\!\\!\frac{d^{2m}x}{(2\pi\lambda)^{m}}\,\pazocal{D}(x)\,\tau_{\\!\frac{1}{2\lambda}-\frac{1}{2}}^{\otimes
m}\ket{s-x}\\!\\!\bra{s-x}\,\tau_{\\!\frac{1}{2\lambda}-\frac{1}{2}}^{\otimes
m}\pazocal{D}(x)^{\dagger}$ $\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(ii)}}}}{{=}}\left(\\!\frac{2\lambda}{1\\!+\\!\lambda}\\!\right)^{\\!\\!2m}\\!\\!\\!\int\\!\\!\frac{d^{2m}x}{(2\pi\lambda)^{m}}\,e^{-\frac{2\lambda}{(1+\lambda)^{2}}\,\|s-x\|^{2}}$
$\displaystyle\hskip
64.58313pt\cdot\pazocal{D}(x)\,\Ket{\tfrac{1\\!-\\!\lambda}{1\\!+\\!\lambda}(s\\!-\\!x)}\\!\\!\Bra{\tfrac{1\\!-\\!\lambda}{1\\!+\\!\lambda}(s\\!-\\!x)}\,\pazocal{D}(x)^{\dagger}$
$\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iii)}}}}{{=}}\left(\\!\frac{2\lambda}{1\\!+\\!\lambda}\\!\right)^{\\!\\!2m}\\!\\!\\!\int\\!\\!\frac{d^{2m}y}{(2\pi\lambda)^{m}}\,e^{-\frac{2\lambda}{(1+\lambda)^{2}}\|y\|^{2}}$
$\displaystyle\hskip
64.58313pt\cdot\pazocal{D}(s-y)\,\Ket{\tfrac{1\\!-\\!\lambda}{1\\!+\\!\lambda}y}\\!\\!\Bra{\tfrac{1\\!-\\!\lambda}{1\\!+\\!\lambda}y}\,\pazocal{D}(s-y)^{\dagger}$
$\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iv)}}}}{{=}}\left(\\!\frac{2\lambda}{1\\!+\\!\lambda}\\!\right)^{\\!\\!2m}\\!\\!\\!\int\\!\\!\frac{d^{2m}y}{(2\pi\lambda)^{m}}\,e^{-\frac{2\lambda}{(1+\lambda)^{2}}\|y\|^{2}}\Ket{s\\!-\\!\tfrac{2\lambda}{1\\!+\\!\lambda}y}\\!\\!\Bra{s\\!-\\!\tfrac{2\lambda}{1\\!+\\!\lambda}y}$
$\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(v)}}}}{{=}}\int\\!\\!\frac{d^{2m}z}{(2\pi\lambda)^{m}}\,e^{-\frac{\|z\|^{2}}{2\lambda}}\Ket{s\\!+\\!z}\\!\\!\Bra{s\\!+\\!z}$
$\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(vi)}}}}{{=}}\int\\!\\!\frac{d^{2m}z}{(2\pi\lambda)^{m}}\,e^{-\frac{\|z\|^{2}}{2\lambda}}\pazocal{D}(z)\ket{s}\\!\\!\bra{s}\pazocal{D}(z)^{\dagger}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(vii)}}}}{{=}}\mathcal{N}_{\lambda}(\ket{s}\\!\\!\bra{s})\,.$

Here, (i), (iv), and (vi) follow from (26) and (25); in (ii) we employed the
identity

$\displaystyle\tau_{\nu}\ket{t}=\frac{1}{\nu+1}\,e^{-\frac{2\nu+1}{4(\nu+1)^{2}}\,\|t\|^{2}}\Ket{\tfrac{\nu}{\nu+1}\,t},$
(131)

which is readily verified in Fock basis by combining (27) and (42); in (iii)
we performed the change of variables $y\coloneqq s-x$; in (v) we set
$z\coloneqq-\frac{2\lambda}{1+\lambda}y$; finally, (vii) is simply (39). The
proof of (41) is complete.

##  Appendix B Fourier representation for characteristic function

Throughout this appendix, we will prove rigorously that the function
$f_{\mu,T}$ defined by (57) has a well-defined Fourier transform, which
moreover coincides with itself. This will establish (64). We formalise these
facts as follows:

######  Lemma 13.

For every measure $\mu$ on the symplectic group and for every symplectic $T$,
the function $f_{\mu,T}$ defined in (57) is in $L^{1}(\mathds{R}^{2m})\cap
L^{2}(\mathds{R}^{2m})$; more precisely,

$\displaystyle\int d^{2m}x\,\left|f_{\mu,T}(x)\right|=(2\pi)^{m}\,,\qquad\int
d^{2m}x\,f_{\mu,T}^{2}(x)\leq\pi^{m}\,.$ (132)
$\displaystyle{\hbox{Moreover,}}\quad\widetilde{f}_{\mu,T}(x)=f_{\mu,T}(x).$

###### Proof.

We start with the first identity, which follows by writing

$\displaystyle\int d^{2m}x\,\left|f_{\mu,T}(x)\right|$ $\displaystyle=\int
d\mu(S)\int
d^{2m}x\,e^{-\frac{1}{2}\,x^{\intercal}(TS)^{{}^{\intercal}}(TS)\,x}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(ii)}}}}{{=}}(2\pi)^{m}\,.$

Here in (ii) we combined the Gaussian integral formula

$\displaystyle\int
d^{N}x\,e^{-\frac{1}{2}x^{\intercal}Ax+it^{\intercal}x}=\sqrt{\frac{(2\pi)^{N}}{\det
A}}\,e^{-\frac{1}{2}t^{\intercal}A^{-1}t}\,,$ (133)

valid for $N\times N$ positive definite matrices $A>0$, with the observation
that $\det(TS)=1$ because both $T$ and $S$ — and hence $TS$, too — are
symplectic.

Since $f_{\mu,T}$ is in $L^{1}(\mathbb{R}^{2m})$, positive, and bounded by
$1$, it is clear that it must be also in $L^{2}(\mathbb{R}^{2m})$. However, it
requires little effort to prove directly the second inequality, which is
anyway tighter than what one would obtain simply by the above observation. It
suffices to compute

$\displaystyle\int d^{2m}x\,f_{\mu,T}^{2}(x)$ (134) $\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iii)}}}}{{=}}\int\\!\\!d\mu(S)\,d\mu(S^{\prime})\int\\!\\!d^{2m}x\,e^{-\frac{1}{2}x^{\\!\intercal}\left((TS)^{\\!{}^{\intercal}}\\!(TS)+(TS^{\prime})^{\\!{}^{\intercal}}\\!(TS^{\prime})\right)x}$
$\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iv)}}}}{{=}}\int
d\mu(S)\,d\mu(S^{\prime})\,\frac{(2\pi)^{m}}{\sqrt{\det\left((TS)^{\\!{}^{\intercal}}\\!(TS)+(TS^{\prime})^{\\!{}^{\intercal}}\\!(TS^{\prime})\right)}}$
$\displaystyle\
\stackrel{{\scriptstyle\text{(v)}}}{{\leq}}\int\\!\\!d\mu(S)\,d\mu(S^{\prime})\,(2\pi)^{m}\frac{1}{2^{m}}=\pi^{m}\,.$

Here, (iii) is just Tonelli’s theorem, (iv) is again an application of (133),
and (v) follows from the Minkowski determinant inequality, which states that
the function $A\mapsto(\det A)^{1/N}$ is concave on the set of positive semi-
definite $N\times N$ matrices [86, Eq. (4.21)], combined with the fact that
$\det(TS)=1$.

Now that we know that $f_{\mu,T}$ is highly regular, we can manipulate the
integrals in its Fourier transform more safely:

$\displaystyle\widetilde{f}_{\mu,T}(x)$
$\displaystyle=\int\frac{d^{2m}u}{(2\pi)^{m}}\,f_{\mu,T}(u)\,e^{ix^{\intercal}\Omega
u}$ $\displaystyle=\int
d\mu(S)\int\frac{d^{2m}u}{(2\pi)^{m}}\,e^{-\frac{1}{2}x^{\intercal}(TS)^{\intercal}(TS)x+ix^{\intercal}\Omega
u}$ $\displaystyle=\int
d\mu(S)\,e^{-\frac{1}{2}x^{\intercal}\Omega(TS)^{-1}(TS)^{-\intercal}\Omega^{\intercal}x}$
$\displaystyle=\int
d\mu(S)\,e^{-\frac{1}{2}x^{\intercal}(TS)^{\intercal}(TS)x}=f_{\mu,T}(x)\,.$

In the above derivation, we have used (133) together with the fundamental
identity $W\Omega=\Omega W^{-\intercal}$, valid for any symplectic matrix $W$.
∎

##  Appendix C The double truncation lemma

######  Lemma 7.

For all $\alpha>0$, all non-negative integers $m$ (number of modes) and $M$
(Fock truncation number), and all $\eta\geq 0$, it holds that

$\displaystyle\left\|\mathcal{L}_{(I+N_{m})^{\alpha}}\mathcal{R}_{(I+N_{m})^{\alpha}}\left(\mathcal{P}_{M}-\widetilde{\mathcal{P}}_{{M}}\right)\right\|_{2\to
1}$ (135)
$\displaystyle\qquad\leq(mM+1)^{2\alpha}(M+1)^{m}\,3^{mM}\left(\frac{\Gamma\left(2M+1,\tfrac{\eta^{2}}{2}\right)}{(2M)!}\right)^{m/2}$
$\displaystyle\qquad=(mM+1)^{2\alpha}(M+1)^{m}\,3^{mM}e^{-\frac{m}{4}\eta^{2}}\left(\sum_{p=0}^{2M}\frac{\eta^{2p}}{2^{p}p!}\right)^{m/2},$

where in the second line we introduced the incomplete Gamma function, given by
[87, § 6.5]

$\displaystyle\Gamma(N,x)\coloneqq\int_{x}^{\infty}ds\,e^{-s}s^{N-1}=(N-1)!\,e^{-x}\sum_{p=0}^{N-1}\frac{x^{p}}{p!}$
(136)

for $N>0$ and $x\geq 0$.

###### Proof.

Let $T$ be an arbitrary Hilbert–Schmidt operator acting on
${\pazocal{H}}_{m}$, the Hilbert space of an $m$-mode system. In what follows,
we denote by $[M]^{m}\coloneqq\\{0,\ldots,M\\}^{m}$ the set of possible Fock
number vectors with all entries bounded by $M$, and for $\mathbf{n}\in[M]^{m}$
we will use the notation $|\mathbf{n}|\coloneqq\sum_{j=0}^{m}\mathbf{n}(j)$ to
denote the corresponding total photon number. We start by writing

$\displaystyle\left\|\mathcal{L}_{(I+N_{m})^{\alpha}}\mathcal{R}_{(I+N_{m})^{\alpha}}\left(\mathcal{P}_{M}(T)-\widetilde{\mathcal{P}}_{{M}}(T)\right)\right\|_{1}$
(137) $\displaystyle\
\leq\sum_{\mathbf{n}_{1},\mathbf{n}_{2}\in[M]^{m}}(1\\!+\\!|\mathbf{n}_{1}|)^{\alpha}(1\\!+\\!|\mathbf{n}_{2}|)^{\alpha}\left|\operatorname{Tr}\left[\left(\widetilde{Z}_{\mathbf{n}_{1}\mathbf{n}_{2}}-\ket{\mathbf{n}_{1}}\\!\\!\bra{\mathbf{n}_{2}}\right)T\right]\right|$
$\displaystyle\
\leq\sum_{\mathbf{n}_{1},\mathbf{n}_{2}\in[M]^{m}}(1\\!+\\!|\mathbf{n}_{1}|)^{\alpha}(1\\!+\\!|\mathbf{n}_{2}|)^{\alpha}\left\|\widetilde{Z}_{\mathbf{n}_{1}\mathbf{n}_{2}}-\ket{\mathbf{n}_{1}}\\!\\!\bra{\mathbf{n}_{2}}\right\|_{2}\|T\|_{2}$
$\displaystyle\
\leq(mM\\!+\\!1)^{2\alpha}(M\\!+\\!1)^{m}\\!\\!\max_{\mathbf{n}_{1},\mathbf{n}_{2}\in[M]^{m}}\left\|\widetilde{Z}_{\mathbf{n}_{1}\mathbf{n}_{2}}-\ket{\mathbf{n}_{1}}\\!\\!\bra{\mathbf{n}_{2}}\right\|_{2}\|T\|_{2}$

We can then continue by estimating

$\displaystyle\left\|\widetilde{Z}_{\mathbf{n}_{1}\mathbf{n}_{2}}-\ket{\mathbf{n}_{1}}\\!\\!\bra{\mathbf{n}_{2}}\right\|^{2}_{2}$
(138) $\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(i)}}}}{{=}}\prod_{j=1}^{m}\int\frac{d^{2}u_{j}}{2\pi}\left|\chi_{\widetilde{Z}_{\mathbf{n}_{1}(j)\mathbf{n}_{2}(j)}}(u_{j})-\chi_{\ket{\mathbf{n}_{1}(j)}\\!\bra{\mathbf{n}_{2}(j)}}\big{(}u^{(j)}\big{)}\right|^{2}$
$\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(ii)}}}}{{=}}\prod_{j=1}^{m}\int\frac{d^{2}u_{j}}{2\pi}\left|\chi_{\ket{\mathbf{n}_{1}(j)}\\!\bra{\mathbf{n}_{2}(j)}}(u_{j})\right|^{2}\left(1-\xi_{\eta,R}(u_{j})\right)^{2}$
$\displaystyle\
\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iii)}}}}{{\leq}}\prod_{j=1}^{m}\int_{\|u_{j}\|\geq\eta}\frac{d^{2}u_{j}}{2\pi}\left|\chi_{\ket{\mathbf{n}_{1}(j)}\\!\bra{\mathbf{n}_{2}(j)}}(u_{j})\right|^{2}$

Here, (i) comes from (34), in (ii) we remembered (76), and in (iii) we
observed that $1-\xi_{\eta,R}(u_{j})=0$ if $\|u_{j}\|\leq\eta$, and estimated
$1-\xi_{\eta,R}(u_{j})\leq 1$ otherwise. In order to continue, we should upper
bound $\chi_{\ket{n}\\!\bra{n^{\prime}}}(u)$ for vectors $u\in\mathds{R}^{2}$
of sufficiently large modulus, and for arbitrary integers $n,n^{\prime}\leq
M$. To this end, let us write

$\displaystyle\left|\chi_{\ket{n}\\!\bra{n^{\prime}}}(u)\right|^{2}$
$\displaystyle=\left|\braket{n^{\prime}}{\pazocal{D}(u)}{n}\right|^{2}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(iv)}}}}{{=}}e^{-\frac{1}{2}\|u\|^{2}}\left|\braket{n^{\prime}}{e^{\alpha(u)a^{\dagger}}e^{-\alpha(u)^{*}a}}{n}\right|^{2}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(v)}}}}{{\leq}}e^{-\frac{1}{2}\|u\|^{2}}\left\|e^{-\alpha(u)^{*}a}\ket{n}\right\|^{2}\left\|e^{\alpha(u)^{*}a}\ket{n^{\prime}}\right\|^{2}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(vi)}}}}{{=}}e^{-\frac{1}{2}\|u\|^{2}}\left\|\sum_{k=0}^{n}\frac{(-\alpha(u))^{k}}{k!}\sqrt{\frac{n!}{(n-k)!}}\,\ket{n-k}\right\|^{2}$
$\displaystyle\quad\cdot\left\|\sum_{k=0}^{n^{\prime}}\frac{(\alpha(u)^{*})^{k}}{k!}\sqrt{\frac{n^{\prime}!}{(n^{\prime}-k)!}}\,\ket{n^{\prime}-k}\right\|^{2}$
$\displaystyle=e^{-\frac{1}{2}\|u\|^{2}}\left(\sum_{k=0}^{n}\frac{|\alpha(u)|^{2k}}{(k!)^{2}}\frac{n!}{(n-k)!}\right)$
$\displaystyle\quad\cdot\left(\sum_{k=0}^{n^{\prime}}\frac{|\alpha(u)|^{2k}}{(k!)^{2}}\frac{n^{\prime}!}{(n^{\prime}-k)!}\right)$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(vii)}}}}{{=}}e^{-\frac{1}{2}\|u\|^{2}}L_{n}\big{(}-|\alpha(u)|^{2}\big{)}\,L_{n^{\prime}}\big{(}-|\alpha(u)|^{2}\big{)}$
$\displaystyle=e^{-\frac{1}{2}\|u\|^{2}}L_{n}\big{(}-\tfrac{1}{2}\|u\|^{2}\big{)}\,L_{n^{\prime}}\big{(}-\tfrac{1}{2}\|u\|^{2}\big{)}$
$\displaystyle\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(viii)}}}}{{\leq}}e^{-\frac{1}{2}\|u\|^{2}}L_{M}\big{(}-\tfrac{1}{2}\|u\|^{2}\big{)}^{2}.$

The steps of the above derivation can be justified as follows: (iv) is an
application of (23), and we recall that
$\alpha(u)\coloneqq\frac{1}{\sqrt{2}}(u_{1}+u_{2})$; (v) is simply the
Cauchy–Schwarz inequality; (vi) can be verified by a repeated application of
(18); in (vii) we introduced the Laguerre polynomials

$\displaystyle
L_{n}(x)\coloneqq\sum_{k=0}^{n}\binom{n}{k}\frac{(-x)^{k}}{k!}\,;$ (139)

and in (viii) we noted that $L_{n}(-x)$ is monotonically non-decreasing in the
integer $n$ for $x\geq 0$, essentially because $\binom{n}{k}$ is monotonically
non-decreasing in $n$ for fixed $k$. Continuing, we deduce that

$\displaystyle\int_{\|u\|\geq\eta}\frac{d^{2}u}{2\pi}\left|\chi_{\ket{n}\\!\bra{n^{\prime}}}(u)\right|^{2}$
$\displaystyle\quad\leq\
\int_{\|u\|\geq\eta}\frac{d^{2}u}{2\pi}\,e^{-\frac{1}{2}\|u\|^{2}}L_{M}\big{(}-\tfrac{1}{2}\|u\|^{2}\big{)}^{2}$
$\displaystyle\quad\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(ix)}}}}{{=}}\
\int_{s\geq\eta^{2}/2}ds\,e^{-s}L_{M}\big{(}-s\big{)}^{2}$
$\displaystyle\quad\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(x)}}}}{{=}}\
\sum_{h,k=0}^{M}\binom{M}{h}\binom{M}{k}\frac{1}{h!k!}\int_{s\geq\eta^{2}/2}ds\,e^{-s}s^{h+k}$
$\displaystyle\quad\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(xi)}}}}{{=}}\
\sum_{h,k=0}^{M}\binom{M}{h}\binom{M}{k}\frac{1}{h!k!}\,\Gamma\left(h+k+1,\tfrac{\eta^{2}}{2}\right)$
$\displaystyle\quad=\
\sum_{h,k=0}^{M}\binom{M}{h}\binom{M}{k}\binom{h+k}{h}\frac{\Gamma\left(h+k+1,\tfrac{\eta^{2}}{2}\right)}{(h+k)!}$
$\displaystyle\quad\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(xii)}}}}{{\leq}}\
\sum_{h,k=0}^{M}\binom{M}{h}\binom{M}{k}\,2^{h+k}\frac{\Gamma\left(2M+1,\tfrac{\eta^{2}}{2}\right)}{(2M)!}$
$\displaystyle\quad\stackrel{{\scriptstyle\mathclap{\scriptsize\mbox{(xiii)}}}}{{=}}\
3^{2M}\,\frac{\Gamma\left(2M+1,\tfrac{\eta^{2}}{2}\right)}{(2M)!}$

Here: in (ix) we introduced the new variable
$s\coloneqq\tfrac{1}{2}\|u\|^{2}$; in (x) we expanded the square thanks to the
expression (139) for the Laguerre polynomials; in (xi) we introduced the
incomplete Gamma function (136); in (xii), besides using that
$\binom{N}{k}\leq 2^{N}$ for all non-negative integers $N,k$, we also observed
that by virtue of the expansion in (136) it is easy to verify that
$\tfrac{1}{(N-1)!}\Gamma(N,x)$ is a monotonically non-decreasing function of
the integer $N\geq 1$ for all fixed $x>0$; finally, (xiii) is just the
binomial theorem, applied twice to $3^{M}=(1+2)^{M}$.

Plugging the above estimate into (138), we find that

$\displaystyle\left\|\widetilde{Z}_{\mathbf{n}_{1}\mathbf{n}_{2}}-\ket{\mathbf{n}_{1}}\\!\\!\bra{\mathbf{n}_{2}}\right\|_{2}\leq
3^{mM}\,\left(\frac{\Gamma\left(2M+1,\tfrac{\eta^{2}}{2}\right)}{(2M)!}\right)^{m/2}$
(140)

for all $\mathbf{n}_{1},\mathbf{n}_{2}\in[M]^{m}$. From (137) we then deduce
that

$\displaystyle\left\|\mathcal{L}_{(I+N_{m})^{\alpha}}\mathcal{R}_{(I+N_{m})^{\alpha}}\left(\mathcal{P}_{M}(T)-\widetilde{\mathcal{P}}_{{M}}(T)\right)\right\|_{1}$
(141)
$\displaystyle\quad\leq(mM\\!+\\!1)^{2\alpha}(M\\!+\\!1)^{m}\,3^{mM}\left(\frac{\Gamma\left(2M\\!+\\!1,\tfrac{\eta^{2}}{2}\right)}{(2M)!}\right)^{m/2}\\!\\!\|T\|_{2}\,,$

concluding the proof. ∎


# End Behavior of Ramanujan’s Taxicab Numbers

Brennan Benfield, Oliver Lippard, and Arindam Roy  Brennan Benfield:
Department of Mathematics and Statistics, University of North Carolina at
Charlotte, 9201 University City Blvd., Charlotte, NC 28223, USA  Oliver
Lippard: Department of Mathematics and Statistics, University of North
Carolina at Charlotte, 9201 University City Blvd., Charlotte, NC 28223, USA
Arindam Roy: Department of Mathematics and Statistics, University of North
Carolina at Charlotte, 9201 University City Blvd., Charlotte, NC 28223, USA

###### Abstract.

Generalized taxicab numbers are the smallest positive integers that are the
sum of exactly $j$, positive $k$th powers in exactly $m$ distinct ways. This
paper is considers for which values of $m$ does a smallest such integer exist
as $j$ gets large. There appear to be only two possible outcomes, leading to
curious results like there is no positive integer that can be expressed as the
sum of exactly $10$ positive squares in exactly $3$ ways. This paper resolves
a number of conjectures found in the OEIS by considering generalized Taxicab
numbers in the setting of the theory of partitions.

The authors extend their sincerest gratitude to Michael De Vlieger.

Mathematics Subject Classifications. 11P05, 11D09, 11D25, 11P81, 11Y50
Keywords. Waring’s problem, quadratic and cubic Diophantine equations,
partitions

##  1\. Introduction

In his biographical lectures titled _Ramanujan_ [Hardy], G. H. Hardy reflects
on the famous exchange that occurred some years earlier:

> He could remember the idiosyncrasies of numbers in an almost uncanny way. It
> was Littlewood who said that every positive integer was one of Ramanujan’s
> personal friends. I remember going to see him once when he was lying ill in
> Putney. I had ridden in taxi-cab No. 1729, and remarked that the number
> seemed to me rather a dull one, and that I hoped that it was not an
> unfavourable omen. “No,” he replied, “it is a very interesting number; it is
> the smallest number expressible as a sum of two cubes in two different
> ways.”

That must have startled Hardy; in fact, Ramanujan had previously studied
Euler’s Diophantine equation: $A^{3}+B^{3}=C^{3}+D^{3}$ and had written in his
famous Lost Notebook [Bernt] some remarkable parametric solutions.

As Ramanujan and Hardy’s conversation has been retold, $Ta(m)$ has come to be
known as the $m$th taxicab number, defined as the least integer which can be
expressed as a sum of two positive cubes in $m$ distinct ways, up to order of
summands. Trivially, $Ta(1)=2$ since $2=1^{3}+1^{3}$. Ramanujan made famous
$Ta(2)=1729$, but it was known as early as 1657, mentioned in a correspondence
between Wallice and Frenicle [Frenicle]. With computer assistance, $Ta(3)$ was
found in 1957 by Leech [Leech], followed by $Ta(4)$ in 1989 by Dardis &
Rosenstiel [Dardis], $Ta(5)$ in 1999 by Wilson [Wilson], and $Ta(6)$ in 2008
by Hollerbach [Hollerbach], who confirmed a conjecture made five years earlier
by Calude, Calude & Dinneen [Claude]. No further taxicab numbers are known,
although upper bounds were given in 2006 by Boyer [Boyer] for $Ta(7)$ through
$Ta(12)$. Despite knowing so few taxicab numbers, Fermat proved [Boyer] that
for all $m$, there exists a positive integer that can be expressed as the sum
of two cubes in $m$ distinct ways.

$2=1^{3}+1^{3}$ | $48988659276962496$
---|---
| $38787^{3}+365757^{3}=$
$1729=12^{3}+1^{3}$ | $107839^{3}+362753^{3}=$
${\color[rgb]{1,1,1}1729}=9^{3}+10^{3}$ | $205292^{3}+342952^{3}=$
| $221424^{3}+336588^{3}=$
$87539319=1673^{3}+4363^{3}$ | $231518^{3}+331954^{3}=$
${\color[rgb]{1,1,1}87539319}=2283^{3}+4233^{3}$ |
${\color[rgb]{1,1,1}87539319}=2553^{3}+4143^{3}$ | $24153319581254312065344$
| $582162^{3}+28906206^{3}=$
$6963472309248=24213^{3}+190833^{3}$ | $3064173^{3}+28894803^{3}=$
${\color[rgb]{1,1,1}6963472309248}=54363^{3}+189483^{3}$ | $8519281^{3}+28657487^{3}=$
${\color[rgb]{1,1,1}6963472309248}=102003^{3}+180723^{3}$ | $16218068^{3}+27093208^{3}=$
${\color[rgb]{1,1,1}6963472309248}=133223^{3}+166303^{3}$ | $17492496^{3}+26590452^{3}=$
| $18289922^{3}+26224366^{3}=$
Table 1. The six known taxicab numbers.

Numbers that can be expressed as the sum of $j$, $k$th positive powers in $m$
distinct ways have come to be known as generalized taxicab numbers.

In 2019, Dinitz, Games, & Roth [Dinitz] denote by $T(k,j,n)$ the smallest
number that can be expressed as the sum of $j$, positive $k$th powers in at
least $m$ ways. They prove that for all $k$ and $j$ there exists a positive
integer that can be expressed as the sum of $j$, positive $k$th powers in at
least $m$ ways:

######  Lemma 1.1 (Dinitz, Games, & Roth).

For all $j,m\geq 1$, there exists a smallest positive integer $s_{0}$ such
that $T(s_{0}+\lambda,j,m)=T(s_{0},j,m)+\lambda$ for every $\lambda\geq 0$.

Let $\text{Taxicab}(k,j,n)$ denote the smallest positive integer that can be
expressed as the sum of $j$, positive $k$th powers in exactly $m$ ways. For
example, $\text{Taxicab}(1,2,2)=4$, $\text{Taxicab}(2,2,2)=50$,
$\text{Taxicab}(3,2,2)=1729$, and $\text{Taxicab}(4,2,2)=635318657$, but
$\text{Taxicab}(5,2,2)$ is unknown [Claude]. It could be that no such number
exists - this paper gives evidence that for many values of $m$,
$\text{Taxicab}(k,j,m)$ has no solution, and proves nonexistence for several
particular cases.

######  Theorem 1.2.

There is no positive integer that can be expressed as the sum of 10 or more
positive squares in exactly 3 ways.

######  Theorem 1.3.

For $n>686362371661$, there is no positive integer that can be expressed as
the sum of 98 or more positive cubes in exactly 90 ways.

######  Conjecture 1.4.

There is no positive integer that can be expressed as the sum of 98 or more
positive cubes in exactly 90 ways.

In other words, $\text{Taxicab}(2,10,3)=\emptyset$ and
$\text{Taxicab}(3,90,98)=\emptyset$. It is not obvious for which values of $m$
there always exists a value for $\text{Taxicab}(k,j,m)$. This paper also
establishes an upper bound $N^{k}(m,j)$ on how large $n$ needs to be to
determine if $\text{Taxicab}(k,j,m)$ exists.

######  Theorem 1.5.

For $k=2$, define

$N^{2}(m,j)=(mj+j+14)^{2}$

For each positive integer $j>6$ and $m>1$, if no solution to
$\text{Taxicab}(2,j,m)$ exists for $n<N_{m}^{j}$, then no solution exists,
else $\text{Taxicab}(2,j+1,m)=\text{Taxicab}(2,j,m)+1$ for all $n>N_{m}^{j}$.

######  Conjecture 1.6.

For $k=3$, define

$N^{3}(m,j)=(mj+j+150)^{3}$

For each positive integer $j>14$ and $m>1$, if no solution to
$\text{Taxicab}(3,j,m)$ exists for $n<N_{m}^{j}$, then no solution exists,
else $\text{Taxicab}(3,j+1,m)=\text{Taxicab}(3,j,m)+1$ for all $n>N_{m}^{j}$.

######  Conjecture 1.7.

For all $m\geq 2$ and for a fixed $k\geq 2$, there exists a $J$ such that for
all $j\geq J$, either there exists a positive integer that satisfies
$\text{Taxicab}(k,j+1,m)=\text{Taxicab}(k,j,m)+1$, or $\text{Taxicab}(k,j,m)$
does not exist for all $j\geq J$.

##  2\. Partitions

Another way to frame Ramanujan’s Taxicab numbers is within the setting of
integer partitions. The partition function $p(n)$ enumerates the number of
partitions of a positive integer $n$ where the partitions are positive integer
sequences $\lambda=(\lambda_{1},\lambda_{2},...)$ with
$\lambda_{1}\geq\lambda_{2}\geq~{}\dots>0$ and $\sum_{j\geq
1}{\lambda_{j}}=n$. For example, $p(4)=5$ since the only ways to partition $4$
are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. The sequence given by $p(n)$
has a generating function [Abramowitz, Andrews] given by:

$\sum_{n}p(n)x^{n}=\prod_{i=1}^{\infty}\frac{1}{1-x^{i}}=\frac{1}{(1-x)(1-x^{2})(1-x^{3})\cdots}.$

It has become increasingly popular to consider restricted partition functions,
typically denoted $p_{A}(n)$ where $A$ is some constraint on $\lambda$. For
instance, the integer $4$ could be partitioned using only prime numbers, only
positive squares, only powers of $2$, only odd numbers, etc:

$\displaystyle 4=2+2$ $\displaystyle\ \ \qquad 4$ $\displaystyle=2^{2}$
$\displaystyle\ \ \qquad 4$ $\displaystyle=2^{2}$ $\displaystyle\ \ \qquad 4$
$\displaystyle=3+1$ $\displaystyle=1^{2}+1^{2}+1^{2}+1^{2}$
$\displaystyle=2^{1}+2^{1}$ $\displaystyle=1+1+1+1$
$\displaystyle=2^{0}+2^{0}+2^{0}+2^{0}$

There are two restricted functions that are of particular interest regarding
generalized taxicab numbers: The $k$th power partition function and the
partitions into exactly $j$ parts function.

The $k$-th power partition function (see [Benfield, [Gafni], [Vaughan],
[Wright]]), denoted $p^{k}(n)$, counts the number of ways that a positive
integer $n$ can be written as the sum of positive $k$-th powers. Note that
$p^{k}(n)=p(n)$ when $k=1$, and for $k=2$, $p^{2}(n)$ restricts $\lambda$ to
positive squares. For example, $p^{2}(4)=2$ since
$4=2^{2}=1^{2}+1^{2}+1^{2}+1^{2}$ and $p^{k}(4)=1$ for all $k\geq 3$. The
generating function for the sequence given by $p^{k}(n)$:

$\sum_{n}p^{k}(n)x^{n}=\prod_{i=1}^{\infty}\frac{1}{1-x^{i^{k}}}=\frac{1}{(1-x^{k})(1-x^{2^{k}})(1-x^{3^{k}})\cdots}.$

The partitions into exactly $j$ parts function (see [ahlgren, kim]), denoted
$p(n,j)$ counts the number of ways that a positive integer $n$ can be written
as a sum of exactly $j$ integers. For example, $p(4,2)=2$ because there are
exactly two ways to write $4$ as a sum of $2$ integers: $4=3+1=2+2$. The
sequence given by $p(n,j)$ has a generating function [Andrews] given by:

$\sum_{n,j}p(n,j)x^{n}y^{j}=\prod_{i=1}^{j}\frac{1}{1-yx^{i}}=\frac{1}{(1-yx)(1-yx^{2})(1-yx^{3})\cdots}.$

Synthesizing the combinatorial properties of these two restricted partition
functions creates the $k$th power partition into exactly $j$ parts function,
denoted $p^{k}(n,j)$.

######  Theorem 2.1.

The generating function for $p^{k}(n,j)$ is given by

$\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}p^{k}(n,j)x^{n}y^{j}=\prod_{i=1}^{\infty}\frac{1}{1-x^{i^{k}}y}=\frac{1}{(1-yx^{k})(1-yx^{2^{k}})(1-yx^{3^{k}})\cdots}.$

###### Proof.

The generating function for the unrestricted partition function counts all
partitions of $n$ with weight $x^{n}.$ It is possible to build a generating
function for $p^{k}(n,j)$ in two variables by counting partitions of $n$ into
$j$ parts (in this case, restricted to $k$-th powers) with weight
$x^{n}y^{j}.$

The power of $x$ counts the sum of the partition, and the power of $y$ counts
the number of parts. Thus, if $i^{k}$ appears in the partition, then there
should be a term of $x^{i^{k}}y,$ because it adds $i^{k}$ to the sum and $1$
to the number of parts. Hence, the term contributed to the generating function
is $\frac{1}{1-x^{i^{k}}y}.$ Any term that can be expressed as $i^{k}$ can be
inserted into the partition, so the generating function becomes

$\prod_{i=1}^{\infty}\frac{1}{1-x^{i^{k}}y},$

as desired. ∎

Given a positive integer $n$, there is a relation between this restricted
partition function and generalized taxicab numbers:

$\text{Taxicab}(k,j,m)=n\qquad\qquad p^{k}(n,j)=m.$

$\text{Taxicab}(k,j,m)$ is simply the smallest $n$ satisfying $p^{k}(n,j)=m$.
Theorem 1.2 and Conjecture 1.4 can be restated: there is no positive integer
$n$ such that $p^{2}(n,10)=3$ or $p^{3}(n,90)=98$.

This paper also establishes an identity for $p^{k}(n,j)$ relating successive
values of $n$ and $k$.

######  Theorem 2.2.

For all $n,j\geq 2$,

$p^{k}(n,j)\leq p^{k}(n+1,j+1)$

with equality whenever $n<2^{k}j$.

###### Proof.

Suppose $n$ is the sum of $j$, positive $k$th powers in exactly $m$ ways. The
smallest possible representation of $n$ without a $1^{2}$ term is

$n=\underbrace{2^{k}+2^{k}+\ldots+2^{k}}_{j\ \text{times}}=2^{k}j.$

For all $n<2^{k}j$, every representation of $n$ as a sum of $j$ positive $k$th
powers contains a $1^{2}$ term which can be subtracted off to obtain the
identity

$p^{k}(n,j)=p^{k}(n-1,j-1).$

And if $n>2^{k}j$, then adding $1^{2}$ to each representation of $n$ as a sum
of $j$ positive $k$th powers yields $m$ representations of $n+1$ as the sum of
$j+1$, positive $k+1$th powers. But there may be other representations of
$n+1$ that contain no $1^{2}$ term. Hence,

$p^{k}(n,j)\leq p^{k}(n+1,j+1).$

∎

Further restricting the summands of $p^{k}(n,j)$, denote by $p^{k}(n,j,\mu)$
the number of partitions of $n$ into $j$ positive $k$th powers with a maximum
part $\mu$. For example, $p^{2}(50,2,25)=1$ since this function counts
$50=5^{2}+5^{2}$ but not $50=1^{2}+7^{2}$. Note that $p^{k}(n,j)=p^{k}(n,j,n)$
and for $n\geq\mu$, the following identity holds.

######  Theorem 2.3.

For all positive integers $n,k,j,\mu$,

$p^{k}(n,j,\mu)=\sum_{\begin{subarray}{c}i=1\\\
i^{k}\leq\mu\end{subarray}}p^{k}(n-i^{k},j-1,i^{k}).$

###### Proof.

The largest possible part of each partition given by $p^{k}(n,j,\mu)$ is
determined by the sequence $i^{k}$ for
$i=1,2,\ldots,\lfloor\sqrt[k]{\mu}\rfloor$. Let $S_{k}$ be the set of
partitions with largest part $i^{k}.$ Then, the term $i^{k}$ can be removed
from any partition in $S_{k}$ to obtain a partition of $n-i^{k}$ into $j-1$
parts. Since $i^{k}$ is the largest part in the partition before its removal,
no parts in the resulting partition can be greater than $i^{k}.$ Thus,
$|S_{k}|=p^{k}(n-i^{k},j-1,i^{k}).$ Summing the number of partitions for each
possible $k$ yields the identity. ∎

######  Corollary 2.4.

The generating function for $p^{k}(n,j,\mu)$ is given by

$\sum_{\begin{subarray}{c}i=1\\\
i^{k}\leq\mu\end{subarray}}p^{k}(n-i^{k},j-1,i^{k})x^{n}y^{j}=\prod_{\begin{subarray}{c}i=1\\\
i^{k}\leq\mu\end{subarray}}\frac{1}{1-x^{i^{k}}y}.$

###### Proof.

This follows naturally from Theorem 2.2, with the generating function
truncated at $\mu$ because no larger parts are allowed. ∎

Starting the product for the generating function of $p^{k}(n,j)$ (or
$p^{k}(n,j,\mu)$) at $i=0$ instead of $i=1$ generates the sequence of
partitions of $n$ into $j$ non-negative $k$th powers (with maximum part
$\mu$). In other words, starting at $i=0$ allows $0^{k}$ as a summand. This is
also equal to the sequence of partitions of $n$ into at most $j$ positive
$k$-th powers.

##  3\. Positive Squares

Consider first when $k=2$, for which values of $m$ does there exist a positive
integer that can be written as the sum of $j$ positive squares in exactly $m$
ways? For a fixed $m$, as $j$ increases one of two things will happen:
increasing the value from $j$ to $j+1$ is accomplished by adding $1^{2}$ to
each representation of $\text{Taxicab}(2,j,m)$, or no such number exists for
all larger values of $j$; these are the only two possible cases.

### Case 1

Consider the sequence given by $\text{Taxicab}(2,j,2)$, the smallest number
that can be expressed as the sum of $j$ positive squares in exactly 2 ways as
$j$ increases. The smallest integer that can be expressed as the sum of two
squares in exactly two ways is
$50=1^{2}+7^{2}=5^{2}+5^{2}=\text{Taxicab}(2,2,2)$. The sequence begins:

Table 2. $\text{Taxicab}(2,j,2)$ for $j\leq 10$. | $j=1$ | $j=2$ | $j=3$ | $j=4$ | $j=5$ | $j=6$ | $j=7$ | $j=8$ | $j=9$ | $j=10$
---|---|---|---|---|---|---|---|---|---|---
$\text{Taxicab}(2,j,2)$ | 2 | 50 | 27 | 31 | 20 | 21 | 22 | 23 | 24 | 25

Notice the pattern emerging at $j=5$: $\text{Taxicab}(2,j+1,2)$ is obtained by
simply adding $1^{2}$ to every representation of $\text{Taxicab}(2,j,2)$.

$\displaystyle 20$ $\displaystyle=2^{2}+2^{2}+2^{2}+2^{2}+2^{2}$
$\displaystyle=4^{2}+1^{1}+1^{2}+1^{2}+1^{2}$ $\displaystyle 21$
$\displaystyle=2^{2}+2^{2}+2^{2}+2^{2}+2^{2}\mathbf{{\color[rgb]{0,0,1}+1^{2}}}$
$\displaystyle=4^{2}+1^{1}+1^{2}+1^{2}+1^{2}\mathbf{{\color[rgb]{0,0,1}+1^{2}}}$
$\displaystyle 22$
$\displaystyle=2^{2}+2^{2}+2^{2}+2^{2}+2^{2}+1^{2}\mathbf{{\color[rgb]{0,0,1}+1^{2}}}$
$\displaystyle=4^{2}+1^{1}+1^{2}+1^{2}+1^{2}+1^{2}\mathbf{{\color[rgb]{0,0,1}+1^{2}}}$
$\displaystyle\vdots$

This is one possible end behavior for a fixed $m$ as $j$ increases. There is
only one other thing that can happen for a fixed $m$ as $j$ gets large.

### Case 2

The other possible end behavior for a fixed $m$ is that there exists a $J$
such that for all $j\geq J$, there does not exist a positive integer that
satisfies $\text{Taxicab}(2,j,m)$. The smallest example occurs when $m=3$ and
$j=10$; there does not exist an integer that can be expressed as the sum of
$10$ or more squares in exactly $3$ ways. Consider the first few cases for
$\text{Taxicab}(2,j,3)$.

Table 3. Values for $\text{Taxicab}(2,j,3)$ for $j\leq 10$. | $j=1$ | $j=2$ | $j=3$ | $j=4$ | $j=5$ | $j=6$ | $j=7$ | $j=8$ | $j=9$ | $j=10$
---|---|---|---|---|---|---|---|---|---|---
$\text{Taxicab}(2,j,3)$ | 3 | 325 | 54 | 28 | 29 | 30 | 31 | 35 | 49 | ${\emptyset}$

But why not simply add $1^{2}$ to each representation of $49$ and obtain
$T(2,10,3)=50$? In fact, $50$ has four representations as the sum of $10$
squares:

$\displaystyle T(2,9,3)=49$
$\displaystyle=5^{2}+3^{2}+3^{2}+1^{2}+1^{2}+1^{2}+1^{2}+1^{2}+1^{2}$
$\displaystyle=3^{2}+3^{2}+3^{2}+3^{2}+3^{2}+1^{2}+1^{2}+1^{2}+1^{2}$
$\displaystyle=4^{2}+3^{2}+3^{2}+2^{2}+2^{2}+2^{2}+1^{2}+1^{2}+1^{2}$
$\displaystyle 50$
$\displaystyle=5^{2}+3^{2}+3^{2}+1^{2}+1^{2}+1^{2}+1^{2}+1^{2}+1^{2}\mathbf{\color[rgb]{0,0,1}{+1^{2}}}$
$\displaystyle=3^{2}+3^{2}+3^{2}+3^{2}+3^{2}+1^{2}+1^{2}+1^{2}+1^{2}\mathbf{\color[rgb]{0,0,1}{+1^{2}}}$
$\displaystyle=4^{2}+3^{2}+3^{2}+2^{2}+2^{2}+2^{2}+1^{2}+1^{2}+1^{2}\mathbf{\color[rgb]{0,0,1}{+1^{2}}}$
$\displaystyle\mathbf{\color[rgb]{1,0,0}{\
=3^{2}+3^{2}+2^{2}+2^{2}+2^{2}+2^{2}+2^{2}+2^{2}+2^{2}+2^{2}}}$

Adding $1^{2}$ to the representations of $\text{Taxicab}(2,9,3)$ does not
yield $\text{Taxicab}(2,10,3)$. This is the only other possible end behavior
of $\text{Taxicab}(2,j,m)$ for a fixed $m$ as $j$ gets large.

##  4\. Numerical Data for Taxicab(2,j,m)

Figure 1 is a bit map of $\text{Taxicab}(2,j,m)$ for $2\leq j\leq 75$ running
from left to right along the horizontal axis and $1\leq m\leq 500$ running top
to bottom along the vertical axis. This map shows a white pixel if
$\text{Taxicab}(2,j,m)$ exists and a black pixel if not.

Figure 1. $\text{Taxicab}(2,j,m)$ for $j\leq 75$ and $m\leq 500$.

The end behavior of $\text{Taxicab}(2,j,m)$ splits the positive integers into
two sets: numbers for which there always exists a solution for
$\text{Taxicab}(2,j,m)$ for sufficiently large $j$, and the complement,
numbers for which there does not exist a solution for $\text{Taxicab}(2,j,m)$
for sufficiently large $j$. Denote by $\\{m_{i}^{k}\\}$ the sequence of $m$
such that $\text{Taxicab}(k,j,m)$ has a solution for a fixed $k$ and for
sufficiently large $j$, and denote by $\\{\mathbb{N}\setminus m_{i}^{k}\\}$
the sequence of $m$ such that no solution to $\text{Taxicab}(k,j,m)$ exists
for sufficiently large $j$. Then $m_{i}^{2}$ and its complement sequence
$\mathbb{N}\setminus m_{i}^{2}$ begin:

$\displaystyle m_{i}^{2}$
$\displaystyle=1,2,4,5,6,7,8,9,10,12,16,14,15,16,18,19,20,21,22,24,25,26,27,28,29,30,\ldots$
$\displaystyle\mathbb{N}\setminus m_{i}^{2}$
$\displaystyle=3,11,17,23,32,34,35,36,38,41,43,45,46,47,49,54,55,57,58,60,61,63,64,65,\ldots$

######  Conjecture 4.1.

The sequences $\\{m_{i}^{2}\\}$ and $\\{\mathbb{N}\setminus m_{i}^{2}\\}$ each
contain infinitely many elements.

######  Question 4.2.

For the sequence $\\{m_{i}^{2}\\}$, is it possible to find an explicit or an
asymptotic formula?

It appears that $\\{m_{i}^{2}\\}$ is growing exponentially, indicating that
values of $m$ for which $\text{Taxicab}(2,j,m)$ exists become increasingly
rare as $m$ gets large. In other words, it appears that ’most’ values of $m$
do not allow a solution for $\text{Taxicab}(2,j,m)$. Utilizing Mathematica, a
reasonable approximation for $\\{m_{i}^{2}\\}$ is $c(x)\approx
17.5603e^{-0.0327825x}$. Figure 2 maps $\\{m_{i}^{2}\\}$ and $c(x)$ for $m\leq
500$.

Figure 2. $c(x)$ and $\\{m_{i}^{2}\\}$ for $m\leq 500$.

##  5\. Proof of Theorems 1.2 and 1.5

To prove that there is no positive integer $n$ that can be written as the sum
of 10 squares in exactly 3 ways, consider one of the 3 representations of $n$:

$\displaystyle n$
$\displaystyle=\underbrace{x_{1}}_{a_{1}}+\underbrace{x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}+x_{8}+x_{9}+x_{10}}_{b_{1}}$

Let $n=a+b$ where $a=x_{1}^{2}$ and
$b=n-a=x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}+x_{7}^{2}+x_{8}^{2}+x_{9}^{2}+x_{10}^{2}$.
Then, whenever there are $3\binom{10}{1}+1=31$ ways to choose $a$ such that
$a$ is a perfect square and $b$ is the sum of $9$ positive squares, there must
be at least $4$ ways to express $n$ as the sum of $10$ positive squares. In
1911, Dubouis [Dubouis] (and independently in 1933, Pall [Pall]) determined
exactly which positive integers are not the sum of exactly $9$ positive
squares. Grosswald [Grosswald] summarizes the result elegantly: let
$\mathbf{B}^{2}=\\{1,2,4,5,7,10,13\\}$ and denote by $\mathbf{B}_{j}^{k}$ the
set of positive integers that are not the sum of exactly $j$ positive $k$th
powers.

######  Lemma 5.1 (Dubouis & Pall).

Every positive integer $n$ can be expressed as the sum of $j\geq 6$ positive
squares provided $n>j-1$ and $n\neq j+b$ for $\beta\in\mathbf{B}^{2}$. For
$j=5$, the result holds for $\mathbf{B}^{2}\cup\\{28\\}$.

Lemma 5.1 enumerates which positive integers cannot be expressed as the sum of
$9$ squares: $1,2,3,4,5,6,7,8$, and $10,11,13,14,16,19,22$. Let $a\geq 961$;
there are at least $\lfloor\sqrt{961}\rfloor=31$ positive squares less than
$a$. For all $n\geq 961+22+1=984$, note that $n=a+b$ where $a$ is a perfect
square and $b$ is a sum of $9$ squares, and there are at least $4$ ways to
express $n$ as the sum of $10$ positive squares. A computer search verifies
that no $n<961$ can be expressed as the sum of $10$ positive squares in
exactly $3$ ways.

In fact, $p^{2}(n,10)\geq 4$ for all $n\geq 48$. Note that $1^{2}$ can be
added $c$ times to each representation of $n$, hence $p^{2}(n,j)\leq
p^{2}(n+c,j+c)$ for all $j\geq 10$ and $n\geq 48$. In particular, for all $n$
and $j$, $p^{2}(n,j)\leq p^{2}(n+1,j+1)$. Similarly, if every representation
of $n$ as the sum of $j$ squares in exactly $m$ ways has $1^{2}$ as a term,
then $p^{2}(n,j)=p^{2}(n-1,j-1)$ by removing the $1^{2}$ term from each
representation. The minimum that a representation with no $1^{2}$ term can be
is $4j$ since the smallest such representation is simply $2^{2}$, added
exactly $j$ times:

$n=\underbrace{2^{2}+2^{2}+2^{2}+\ldots+2^{2}}_{j\ \text{times}}=4j$

Thus, for $n<4j$, the identity $p^{2}(n,j)=p^{2}(n-1,j-1)$ holds. The same
process outlined above for can be repeated for $j=11$ and $j=12$. A computer
search up to $(11\cdot 3+11+14)^{2}=3364$ ensures that
$\text{Taxicab}(2,11,3)$ does not exist and a search up to $(12\cdot
3+12+14)^{2}=3844$ ensures that $\text{Taxicab}(2,12,3)$ does not exist. In
other words, there is no $n$ such that $p^{2}(n,11)=3$ nor $p^{2}(n,12)=3$.
Let $j>12$ and suppose $n<38+j<4j$, then by Theorem 2.2,

$p^{2}(n,j)=p^{2}(n-1,j-1)=\ldots=p^{2}(n-j+12,12)\neq 3.$

Otherwise, if $n\geq 38+j$, then $n-j+12\geq 50$, and

$p^{2}(n,j)\geq p^{2}(n-1,j-1)\geq\ldots\geq p^{2}(n-j+12,12)\geq 4.\qed$

Proof of Theorem 1.2 relied on knowing which numbers are not a sum of $9$ (or,
in general, $j-1)$ squares. By Lemma 5.1, the largest number that cannot be
written as a sum of $j-1$ squares is $j-1+13$. Then, there are
$\binom{j}{1}=j$ ways to decompose a sum of $j$ squares into a square and a
sum of $j-1$ squares. If there are at least $mj+1$ representations of $n$ as a
perfect square plus a number that is the sum of $j-1$ squares, then there are
at least $m+1$ representations of $n$ as a sum of $j$ squares by the
Pigeonhole principle. This gives an upper bound on how high a computer needs
to check to determine for a given $m$ if $\text{Taxicab}(2,j,m)$ exists for
$j\geq 7$. Denote this upper bound by $N^{2}(m,j)$:

$N^{2}(m,j)=(mj+1+(j+13))^{2}=(mj+j+14)^{2}$

In other words, to determine if there exists an $n$ that is the sum of $j$
squares in exactly $m$ ways, a computer need only search up to $N_{m}^{j}$. By
Lemma 1.1, for $n>N_{j}^{k}$, if $\text{Taxicab}(2,j,m)$ exists, then
$\text{Taxicab}(2,j+1,m)=\text{Taxicab}(2,j,m)+1$; else, no such integer
exists. This proves Theorem 1.5. ∎

A reliable bound is obtained from Theorem 1.5, but a tighter bound exists.
Visible in Figure 1 is a curved boundary separating an unorganized region of
small values of $j$ and a regularly organized region of large values of $j$.
Theorem 1.5 holds for sufficiently large $n$, but it also holds for many
smaller values of $n$. Let $J^{2}(m)$ map the smallest $(m,J)$ such that
Theorem 1.5 holds for all $j\geq J$. Utilizing Mathematica, a close fit for
$J^{2}(m)$ appears to be $g_{2}(x)=45.06\sqrt[8]{x}-44.7873$. Figure 3 plots
$J(m)$ and $b(x)$ for $j\leq 75$ and $m\leq 500$.

Figure 3. $J^{2}(m)$ (black) and $g_{2}(x)$ (red)

##  6\. Conjectures from the OEIS

There are many entries in the Online Encyclopedia of Integer Sequences (OEIS)
[OEIS] that record various sequences as $k$, $j$, and $m$ are manipulated.
This section resolves several conjectures found in the OEIS.

### Entry A080673 [A080673]

Define the sequence where $A080673(n)$ as the “largest number with exactly n
representations as sum of five positive squares (or 0 if no number with
exactly n representations exists)”. There is a comment from Johnson [A080673]
that $A080673(0)=33$ (i.e. $33$ is the largest positive integer with $0$
representations as a sum of $5$ positive squares) and that there is no number
less than $10^{6}$ that can be expressed as the sum of five positive squares
in exactly 188 ways. Hasler asks if any of the undefined $0$’s in the sequence
have been proven. The fact that $A080673(0)=33$ follows immediately from Lemma
5.1. Using a similar technique as in the proof of Theorem 1.2, it is possible
to prove that in fact no number can be expressed as the sum of $5$ positive
squares in exactly $188$ ways.

######  Theorem 6.1.

$\text{Taxicab}(2,5,188)$ does not exist.

Lagrange [Lagrange] famously proved in 1770 that every number can be expressed
as the sum of four squares, but allowed $0^{4}$ as a term in the
representation - not every integer has a representation as the sum of four
positive squares. The sequence of positive integers that cannot be expressed
as the sum of four squares begins:

$\text{OEIS A000534
\cite[cite]{[\@@bibref{}{A000534}{}{}]}}:1,2,3,5,6,8,9,11,14,17,24,29,32,41,56,96,128,224,384,512,896,\ldots$

The complete list of such positive integers was conjectured in 1638 by
Descartes [Descartes] and proven in 1911 by Dubouis [Dubouis]. Recall that
$\mathbf{B}^{2}=\\{1,2,4,5,7,10,13\\}$.

######  Lemma 6.2 (Dubouis).

Every positive integer $n$ can be expressed as the sum of $4$ positive squares
provided $n>3$ and $n\neq 4+b$ for $\beta\in\mathbf{B}^{2}\cup\\{25,37\\}$ and
$n\neq 2*4^{\alpha},6*4^{\alpha},14*4^{\alpha}$ for $\alpha=0,1,2,\ldots$.

In other words, $\mathbf{B}_{4}^{2}=\\{1,2,3,5,6,8,9,11,14,17,29,41,2\cdot
4^{\alpha},6\cdot 4^{\alpha},14\cdot 4^{\alpha}\mid\alpha=0,1,2,\ldots\\}$.

###### Proof.

Consider an arbitrary representation of a positive integer $n$ as the sum of
$5$ positive squares. Let $n=a+b$ where $a=x_{1}^{2}$ and
$b=x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}$. By Lemma 6.2, the only positive
integers greater than $41$ that cannot be expressed as a sum of exactly four
squares are $2\cdot 4^{\alpha},6\cdot 4^{\alpha},$ and $14\cdot 4^{\alpha}$
for $\alpha=1,2,\ldots$ For all $n\geq 921681$, there are at most $21$
integers between $n-919681$ and $n$ that cannot be expressed as a sum of four
squares.

Let $\gamma(n)$ be the number of positive integers between $n$ and $n-919681$
such that $n$ cannot be expressed as a sum of exactly four squares. Since
$n\geq 921681,n-919681\geq 2000.$ Then, the smallest value of $\alpha$ that
yields an integer that cannot be expressed as a sum of four squares is
$\lceil\log_{4}(\frac{n-919681}{2})\rceil$, and the largest value of $\alpha$
that can be expressed as a sum of four squares is
$\lfloor\log_{4}(\frac{n}{14})\rfloor.$ Note that for each value of $\alpha$,
there are at most $3$ positive integers that cannot be expressed as the sum of
four positive squares. It follows that $\gamma(n)\leq
3(\log_{4}(\frac{n}{2})-\log_{4}(\frac{n-919681}{14})),$ which is a decreasing
function. Evaluation at $n=921681$ shows $\gamma(n)\leq 18$ for all $n\geq
921681.$ Therefore, there are at least $\lfloor\sqrt{919681}-18\rfloor=941$
combinations of a square and a sum of four squares that sum to $n$. At most
$\binom{5}{4}=5$ of these sums give the same combination of $5$ squares.
Hence, for all $n>921681$, there are at least $\lceil\frac{941}{5}\rceil=189$
ways to represent $n$ as a sum of $5$ positive squares. Johnson [A080673] has
already checked that $\text{Taxicab}(2,5,188)$ does not exist for $n<10^{6}$.
∎

Hasler [A080673] asks if any of the other listed elements in the sequence
where $A080673(n)=0$ are proven. This paper establishes nonexistence of
$\text{Taxicab}(2,5,m)$ for $m\leq 500$.

######  Theorem 6.3.

$\text{Taxicab}(2,5,m)$ does not exist for $m=259,304,308,372,394,483,497$.

### Entry A295702 [A295702]

Define the sequence $A295702(n)$ as the “largest number with exactly n
representations as a sum of $6$ positive squares (or 0 if no number with
exactly n representations exists)”. This entry contains the conjecture “It
appears that $A295702(36)$ does not exist.”

######  Theorem 6.4.

$\text{Taxicab}(2,6,36)$ does not exist.

###### Proof.

A computer search verifies that no $n\leq N^{2}(36,6)=(6\cdot
36+6+14)^{2}=55696$ can be expressed as the sum of $6$ positive squares in
exactly $36$ ways. By Theorem 1.5, $\text{Taxicab}(2,6,36)$ does not exist. ∎

### Entry A295795 [A295795]

Define the sequence $A295795(n)$ as the “largest number with exactly n
representations as sum of $7$ positive squares (or 0 if no number with exactly
n representations exists)”. This entry contains the conjecture “It appears
that $A295795(44)$ does not exist.”

######  Theorem 6.5.

$\text{Taxicab}(2,7,44)$ does not exist.

###### Proof.

A computer search verifies that no $n\leq N^{2}(44,7)=(7\cdot
44+7+14)^{2}=107584$ can be expressed as the sum of $7$ positive squares in
exactly $44$ ways. By Theorem 1.5, $\text{Taxicab}(2,7,44)$ does not exist. ∎

### Entry A025416 [A025416]

Define the sequence $A025416(n)$ as the sequence given by
$\text{Taxicab}(2,4,m)$ for $m=1,2,\ldots$ Perry [A025416] conjectures that
$A025416(n)$ never becomes monotone.

Figure 4. $\text{Taxicab}(2,4,m)$ for $m\leq 2861$.

$\text{Taxicab}(2,4,m):4,31,28,52,82,90,135,130,162,198,202,252,234,210,346,306,\ldots$

for $m=1,2,\ldots$ Perry’s conjecture appears to hold for all $j$ and is
generalized in the following conjecture.

######  Conjecture 6.6 (Perry).

Given a fixed $j>2$, as $m$ increases, the sequence given by
$\text{Taxicab}(2,j,m)$ never becomes monotone.

There is good evidence to believe Conjecture 6.6. Note that these sequences
are essentially the columns of Figure 1, and most values of $m$ and $j$ lie in
the highly regular region on the right hand side. On the OEIS, values are $0$
whenever $\text{Taxicab}(2,j,m)$ does not exist, and some positive integer
whenever $\text{Taxicab}(2,j,m)$ does exist. By that alone, it appears that
Perry’s conjecture holds, supposing there are infinitely many elements of both
$m_{i}$ and $\mathbb{N}\setminus m_{i}$. But Perry’s conjecture made for $m=4$
contains no values of $0$, that is, $\text{Taxicab}(2,4,m)$ always exists (at
least as far as OEIS A025416 has entries). Define $m_{i,j}^{2}$ as the
sequence of $m$ such that $\text{Taxicab}(2,j,m)$ exists. Then, Perry’s
conjecture may be generalized by the following:

######  Conjecture 6.7.

For each $m$ and $j$, the sequence $m_{i,j}^{2}$ never becomes monotone.

##  7\. Positive Cubes

For which values of $m$ does there exist a positive integer that can be
written as the sum of $j$ positive cubes in exactly $m$ ways? An analogous
result to Lemma 5.1 for cubic equations does not appear to have been
established. Define the set

$\displaystyle\mathbf{B}^{3}=\\{$ $\displaystyle
1,2,3,4,5,6,8,9,10,11,12,13,15,16,17,18,19,20,22,23,24,25,27,$ $\displaystyle
29,30,31,32,34,36,37,38,39,41,43,44,45,46,48,50,51,53,55,57,$ $\displaystyle
58,60,62,64,65,67,69,71,72,74,76,79,81,83,86,88,90,93,95,97,$ $\displaystyle
100,102,107,109,114,116,121,123,128,135,142,149\\}$

######  Conjecture 7.1.

For all $j\geq 4$, there are only finitely many positive integers that cannot
be expressed as the sum of exactly $j$ positive cubes.

######  Conjecture 7.2.

Every positive integer $n$ can be expressed as the sum of $j\geq 14$ positive
cubes provided $n>j-1$ and $n\neq j+\beta$ for $\beta\in\mathbf{B}^{3}$.

For $j=1,2,3$, there are infinitely many positive integers that cannot be
written as the sum of exactly $j$ positive cubes; it was shown by Davenport
[Davenport] that no positive integer congruent to $\pm 4$ modulo $9$ can be
expressed as the sum of $3$ cubes (it is an open question if this list is
exhaustive), and Euler is credited [Macys] with the proof of the particular
case of Fermat’s Last Theorem - that no positive cube can be expressed as the
sum of $2$ positive cubes.

There are $j$ ways to decompose a sum of exactly $j$ cubes into the sum of one
cube and the sum of $j-1$ cubes. Supposing Conjecture 7.2 holds, the largest
number that cannot be written as the sum of $j-1$ positive cubes is $j-1+149$.
Then a computer search need not exceed

$n\leq N^{3}(m,j)=(mj+(j-1+150)+1)^{3}=(mj+j+150)^{3}.$

If $\text{Taxicab}(3,j,m)$ does not exist for $n\leq N^{3}(m,j)$, then it
doesn’t exist. This, along with Lemma 1.1, proves Conjecture 1.6, supposing
Conjecture 7.2 holds. Analogous to the end behavior in the setting of positive
squares, it appears that for a fixed $m$, as $j$ increases, one of two things
occurs for $\text{Taxicab}(3,j,m)$.

######  Conjecture 7.3.

For all $m\geq 2$ and for a fixed $k\geq 2$, there exists a $J$ such that for
all $j\geq J$, either there exists a positive integer that satisfies
$\text{Taxicab}(3,j+1,m)=\text{Taxicab}(3,j,m)+1$, or $\text{Taxicab}(3,j,m)$
does not exist for all $j\geq J$.

Conjecture 7.3 is supported by Figure 5, a bitmap of $\text{Taxicab}(3,j,m)$,
where values of $j$ for $7\leq j\leq 200$ run along the horizontal axis and
values of $m$ for $1\leq m\leq 500$ run down the vertical axis. Again, a white
pixel indicates that a solution to $\text{Taxicab}(3,j,m)$ exists and a black
pixel indicates that no such solution exists. A familiar pattern emerges in
Figure 5.

There is a chaotic region on the left corresponding to small values of $j$ and
a highly regular region on the right corresponding to large values of $j$.
These two regions of the graph are again separated by a boundary curve with a
familiar shape. Let $J^{3}(m)$ denote the smallest pair $(m,J)$ such that
Conjecture 7.3 holds for all $j\geq J$. Figure 5 graphs $J^{3}(m)$ for $m\leq
500$. Utilizing Mathematica, a close fit for $J^{3}(m)$ appears to be
$g_{3}(x)=15.8278\sqrt[3]{x}+12.0661$

Values of $m$ for which $\text{Taxicab}(3,j,m)$ exists creates the sequence
$\\{m_{i}^{3}\\}$ and its complement sequence $\\{\mathbb{N}\setminus
m_{i}^{3}\\}$. Figure 5 also plots $J^{3}(m)$ for $m\leq 500$.

$\displaystyle m_{i}^{3}$ $\displaystyle=1,2,3\ldots
89,91,92,93,95,96,97,98,99,100,101,102,104,105,107,108,109,\ldots$
$\displaystyle\mathbb{N}\setminus m_{i}^{3}$
$\displaystyle=90,94,103,106,113,118,138,146,149,156,157,160,164,165,168,173,176,180,\ldots$

Figure 5. (Left) $\text{Taxicab}(3,j,m)$, (Right Top) $m_{i}^{3}$, and (Right
Bottom) $J^{3}(m)$ and $g_{3}(m)$ for $m\leq 500$ and $7\leq j\leq 200$

Regarding Conjecture 1.4, $\text{Taxicab}(3,98,90)$ is the smallest instance
in the setting of cubes that there is (conjectured) no positive integer that
can be expressed as the sum of $98$ positive cubes in exactly $j\geq 90$ ways.
In Figure 5, this corresponds to the top most solid black line in
$\text{Taxicab}(3,j,m)$. It is computationally expensive to check small
values, but supposing Conjecture 1.6 holds, a computer need only search up to

$N^{3}(90,98)=(90\cdot 98+98+150)^{3}=745649162432.$

If no such value satisfies $p^{3}(n,98)=90$ for $n\leq 745649162432$, then
Conjecture 1.4 holds, and $\text{Taxicab}(3,98,91)$ does not exist. But there
are smaller values of $j$ and $m$ such that $\text{Taxicab}(3,j,m)$ does not
exist. In an effort to minimize the range of a computer search using this
method, consider the upper left-most black dot in Figure 5.

######  Conjecture 7.4.

No positive integer is the sum of exactly $47$ positive cubes in exactly $41$
ways.

Supposing Conjecture 1.6 holds, a computer need only search up to

$N^{3}(41,47)=(41\cdot 47+47+150)^{3}=9582162624.$

If no such value satisfies $p^{3}(n,47)=41$ for $n\leq 9582162624$, then
Conjecture 7.4 holds, and $\text{Taxicab}(3,47,41)$ does not exist.

##  8\. Higher Positive Powers

After hearing Ramanujan’s reply regarding $1729$, Hardy recalls [Hardy]

> I asked him, naturally, whether he could tell me the solution of the
> corresponding problem for fourth powers; and he replied, after a moment’s
> thought, that he knew no obvious example, and supposed that the first such
> number must be very large.

The answer to Hardy’s question is large indeed - the discovery that
$635318657=59^{4}+158^{4}=133^{4}+134^{4}$ is credited to Euler [Euler] but it
wasn’t until 1957 that Leech [Leech] proved that $635318657$ is in fact
$\text{Taxicab}(4,2,2)$. The existence of positive integers that are the sum
of $2$ positive $k$th powers in exactly $2$ ways is still open. In terms of
Diophantine equations, it is unknown [Ekl] if the equation

$A^{k}+B^{k}=C^{k}+D^{k}$

has solutions for $k>4$.

######  Conjecture 8.1.

For all $m\geq 2$ and for a fixed $k\geq 2$, there exists a $J$ such that for
all $j\geq J$, either there exists a positive integer that satisfies
$\text{Taxicab}(4,j+1,m)=\text{Taxicab}(4,j,m)+1$, or $\text{Taxicab}(4,j,m)$
does not exist for all $j\geq J$.

Conjecture 8.1 is supported by Figure 6, a bitmap of $\text{Taxicab}(4,j,m)$,
where values of $j$ for $j\leq 500$ run along the horizontal axis and values
of $m$ for $1\leq m\leq 500$ run down the vertical axis. Again, a white pixel
indicates that a solution to $\text{Taxicab}(4,j,m)$ exists and a black pixel
indicates that no such solution exists. A familiar pattern emerges in Figure
6.

Values of $m$ for which $\text{Taxicab}(4,j,m)$ exists creates the sequence
$\\{m_{i}^{4}\\}$ and its complement sequence $\\{\mathbb{N}\setminus
m_{i}^{4}\\}$. Figure 6 also plots $J^{4}(m)$ for $m\leq 500$.

$\displaystyle m_{i}^{4}$ $\displaystyle=1,2,\ldots
32,34,35\ldots,56,58,60,61,\ldots,94,96,97,\ldots$
$\displaystyle\mathbb{N}\setminus m_{i}^{4}$
$\displaystyle=33,57,59,95,113,116,118,120,140,141,143,144,148,152,160,162,172,\ldots$

There is a chaotic region on the left corresponding to small values of $j$ and
a highly regular region on the right corresponding to large values of $j$.
These two regions of the graph are again separated by a curve. Let $J^{4}(m)$
denote the smallest pair $(m,J)$ such that Conjecture 8.1 holds for all $j\geq
J$. Figure 6 graphs $J^{4}(m)$ for $m\leq 500$. Utilizing Mathematica, a close
fit for $J^{3}(m)$ appears to be $g_{4}(x)=43.106\sqrt[4]{x}-28.3045$.

Figure 6. (Left) $\text{Taxicab}(4,j,m)$, (Right Top) $m_{i}^{4}$, and (Right
Bottom) $J^{4}(m)$ and $g_{4}(m)$ for $m\leq 500$ and $7\leq j\leq 200$

For powers higher than $k=4$, there is every reason to suppose that something
similar occurs for every $m$ as $j$ gets large. Heuristically, for very large
$n$, there are many representations of $n$ as the sum of $j$ positive $k$th
powers. To determine if $\text{Taxicab}(k,j,m)$ exists for a particular $m$
and $j$, the only possibilities are relatively small values of $n$. Because
there are so few possibilities, there are many values of
$\text{Taxicab}(k,j,m)$ that do not exist.

##  9\. Final Thoughts

There are seemingly endless variations of partition functions and Taxicab
numbers obtained by manipulating $k$, $j$, and $m$. What is the smallest
number that can be expressed as the sum of $j$, integer $k$th powers in
exactly $m$ ways (without the restriction that the $k$th powers are positive)?
For $k=3$ and $j=2$, these numbers are called Cabtaxi numbers, denoted
$\text{Cabtaxi}(3,j,m)$. Note that $0^{k}$ is allowed as a summand in this
definition. For example, Viète [Vi`ete] discovered in 1591 that
$\text{Cabtaxi}(3,2,2)=91=3^{3}+4^{3}=6^{2}-5^{2}$. Escott [Escott] computed
$\text{Cabtaxi}(3,2,3)$ in 1902, followed by Rathburn [Rathbun] in 1992 who
computed $\text{Cabtaxi}(3,2,4),\ldots,\text{Cabtaxi}(3,2,7)$, then Bernstein
[Bernstein] in 1998 computed $\text{Cabtaxi}(3,2,8)$, Moore [Moore] in 2005
computed $\text{Cabtaxi}(3,2,9)$, and in 2006, Boyer [Boyer] conjectured
$\text{Cabtaxi}(3,2,10)$, which was proven in 2008 by Hollarbach [Hollerbach].
Even with modern computing power, only $10$ Cabtaxi numbers are currently
known.

$91=3^{3}+4^{3}$ | $728=8^{3}+6^{3}$ | $2741256=108^{3}+114^{3}$ | $6017193=166^{3}+113^{3}$
---|---|---|---
${\color[rgb]{1,1,1}91}=6^{3}-5^{3}$ | ${\color[rgb]{1,1,1}728}=9^{3}-1^{3}$ | ${\color[rgb]{1,1,1}2741256}=140^{3}-14^{3}$ | ${\color[rgb]{1,1,1}6017193}=180^{3}+57^{3}$
| ${\color[rgb]{1,1,1}72888}=12^{3}-10^{3}$ | ${\color[rgb]{1,1,1}2741256}=168^{3}-126^{3}$ | ${\color[rgb]{1,1,1}6017193}=185^{3}-68^{3}$
|  | ${\color[rgb]{1,1,1}2741256}=207^{3}-183^{3}$ | ${\color[rgb]{1,1,1}6017193}=209^{3}-146^{3}$
|  |  | ${\color[rgb]{1,1,1}6017193}=246^{3}-207^{3}$
$\text{Cabtaxi}(3,2,2)$ | $\text{Cabtaxi}(3,2,3)$ | $\text{Cabtaxi}(3,2,4)$ | $\text{Cabtaxi}(3,2,5)$

Define generalized Cabtaxi numbers as the smallest positive integer $n$ such
that $n=x_{1}^{k}\pm x_{2}^{k}\pm\ldots\pm x_{j}^{k}$ in exactly $m$ ways. The
relationship with the partition function $p^{k}(n,j)$ no longer holds because
partitions are strictly defined as decreasing sequences of positive integers
that sum to $n$. Without the restriction to addition, perhaps something
analogous occurs with Cabtaxi numbers, or perhaps a completely different
pattern will emerge.

######  Question 9.1.

For each $m$, what is the end behavior of $\text{Cabtaxi}(k,j,m)$ for a fixed
$k$ as $j$ gets large?

There are also popular iterations of Taxicab numbers that involve further
restrictions on the summands of $n$. What is the smallest number that can be
expressed as the sum of $j$ prime (or distinct, or coprime, etc.) positive
$k$th powers in exactly $m$ ways? The smallest $\text{Taxicab}(3,2,2)$ and
$\text{Cabtaxi}(3,2,2)$ where the summands are prime are:

$\displaystyle 6058655748$ $\displaystyle=613^{3}+18233^{3}$ $\displaystyle
62540982$ $\displaystyle=3973^{3}-313^{3}$ $\displaystyle=10493^{3}+16993^{3}$
$\displaystyle=18673^{3}-18613^{3}.$

######  Question 9.2.

For each $m$, what is the end behavior of $\text{Taxicab}(k,j,m)$ and
$\text{Cabtaxi}(k,j,m)$ for a fixed $k$ as $j$ gets large and the summands of
$n$ are restricted to prime (or distinct, or coprime, etc.) positive $k$th
powers?

##  10\. Statements and Declarations

No support was received from any organization for the submitted work nor was
any funding received by any author.