# Quantum simulation of Ising spins on Platonic graphs

Andrew Byun, Minhyuk Kim, and Jaewook Ahn  Department of Physics, KAIST,
Daejeon 34141, Korea

###### Abstract

We present quantum simulation experiments of Ising-like spins on Platonic
graphs, which are performed with two-dimensional arrays of Rydberg atoms and
quantum-wire couplings. The quantum wires are used to couple otherwise
uncoupled long-distance atoms, enabling topology-preserving transformtions of
the three-dimensional graphs to the two-dimensional plane. We implement three
Platonic graphs, tetrahedron, cube, and octahedron of Platonic solids, and
successfully probe their ground many-body spin configurations before and after
the quasi-adiabatic control of the system Hamiltonians from the paramagnetic
phase to anti-ferromagnetic-like phases. Our small-scale quantum simulations
of using less than 22 atoms are limited by experimental imperfections, which
can be easily improved by the state-of-the-art Rydberg-atom technologies for
more than 1000-atom scales. Our quantum-wire approach is expected to pave a
new route towards large-scale quantum simulations.

##  I Introduction

Breakthroughs in artificial quantum many-body systems have brought unforeseen
opportunities to explore problems in physics, chemistry, material science, and
medicine [1, 2, 3, 4, 5]. In recent years, Rydberg-atom systems in particular
have advanced rapidly, demonstrating quantum-mechanical simulations of the
nature of complex quantum materials [6, 7, 8, 9, 10]. There are many
advantages to use Rydberg atoms in quantum simulations. Interactions of
Rydberg atoms are strong and short-ranged enough to feature the mesoscopic
nature of lattice Hamiltonians such as Ising, XY, and XXZ models [11, 12, 13,
14, 15]. Rydberg-atom arrays are easily scalable [16, 17, 18] to generate
entanglements of a huge number of atoms [19], to investigate many-body
dynamics near quantum phase transitions [20, 21, 22, 23, 24], and to study
topological effects [25]. Three-dimensional arrangements [26, 27] of Rydberg
atoms are used to probe the physical properties of tree lattices [28] and are
expected to investigate many-body physics of more complex arrangements [27].
In reported experiments [29, 30], the number of Rydberg atoms starts to exceed
a few hundred, boosting the expectation towards computational advantages of
Rydberg-atom systems in solving combinatorial optimization problems [31, 32,
33, 34].

Rydberg-atom technology uses neutral atoms captured and arranged by focused
optical beams and coupled with each other through Rydberg-atom dipole blockade
interactions [6]. So the inter-atomic distances of coupled Rydberg atoms are
lower and upper bounded by the Rayleigh range and Rydberg blockade distance,
respectively, which strongly limits geometries and topologies of the qubit
couplings of a Rydberg-atom array [27]. One of the latest Rydberg-atom
technologies is quantum wiring, the use of auxiliary atoms to couple remote
atoms, which enables otherwise impossible, complex qubit coupling networks
[35, 33]. In this paper, we use quantum wires to demonstrate topology-
preserving transformations of a Rydberg-atom system from a 3D surface to a
plane. Three Platonic graphs, the tetrahedron, cube, and octahedron graphs,
are experimentally implemented and their many-body ground states are probed
via quantum adiabatic controls [36, 20].

The rest of the paper is organized as follows. The model Hamiltonian of
Rydberg-atom arrays is defined in Sec. II along with the usages of the quantum
wires in our experiments. After we describe the phase diagrams of the Rydberg-
atom arrays for the Platonic graphs in Sec. III and the experimental procedure
in Sec. IV, the resulting quantum simulation data of the Platonic graphs are
reported in Sec. V. We conclude by discussing the scaling issues and outlooks
in Sec. VI.

##  II Quantum wires for Platonic graphs

We consider a mathematical graph, $G(V,E)$, to model the interactions of an
atom array, in which the vertices, $V$, and edges, $E$, respectively,
represent the atoms and pairwise atom-atom interactions. In the Rydberg-atom
blockade regime [6] of adjacent atoms, the Hamiltonian $\hat{H}_{G}$ of the
atom array in $G$ is given by

$\hat{H}_{G}=\frac{\hbar}{2}\sum_{i\in
V}\left(\Omega\hat{\sigma}_{x,i}-\Delta\hat{\sigma}_{z,i}\right)+\sum_{(i,j)\in
E}U\hat{n}_{i}\hat{n}_{j},$ (1)

where $\Omega$ and $\Delta$ are the Rabi frequency and detuning of the optical
excitation of the atoms from the ground state, denoted by
$\ket{\downarrow}=\ket{0}$, to a Rydberg state, $\ket{\uparrow}=\ket{1}$, and
$U=C_{6}/d^{6}$ is the van der Waals interaction, of coefficient $C_{6}$,
between the adjacent atoms. The distance of all adjacent atom pairs is kept to
be $d$, which is smaller than the blockade distance,
$d_{R}=(C_{6}/\hbar\Omega)^{1/6}$, i.e., $d<d_{R}$. $\hat{\sigma}_{x,z}$ and
$\hat{n}=(\hat{\sigma}_{z}+1)/2$ are the operators for Pauli $x,z$ and Rydberg
excitation number. This Hamiltonian for quantum Ising spins [21, 20] has been
considered for maximum independent set problems (MIS) of $G$ [31, 32, 33, 34].

Figure 1: Platonic graphs and their two-dimensional arragnements: (a) The
tetragonal graph $K_{4}$, (b) the cube graph $Q_{3}$, and (c) the octahedron
graph $K_{2,2,2}$, in which the yellow edges are to be replaced by quantum
wires. (d) The quantum-wired tetragonal graph, $K_{4}^{\prime}$, uses one
quantum wire, $W$, of two wire atoms, to replace the four yellow edges of
$K_{4}$, so there are six atoms and five equal-length edges in
$K_{4}^{\prime}$. (e) The quantum-wired cube graph, $Q_{3}^{\prime}$ uses four
quantum wires, $W_{1,2,3,4}$, each of two wire atoms, to replace the four
yellow edges of $Q_{3}$, so there are 16 atoms and 20 edges in
$Q_{3}^{\prime}$. (f) The quantum-wired octahedron graph, $K_{2,2,2}^{\prime}$
uses three quantum wires, $W_{1,2,3}$, each of four wire atoms, to replace the
three yellow edges of $K_{2,2,2}$ so there are 18 atoms and 24 edges in
$K_{2,2,2}^{\prime}$. The lengths are $d^{\prime}$ for the edges involved with
quantum wires and $d$ for all other edges.

Figure 1 shows, in the first row, three Platonic graphs to be constructed in
this work. They are, in the graph theory notations, $K_{4}$, the tetrahedron
graph ($f=4$), $Q_{3}$, the cube graph ($f=6$), and $K_{2,2,2}$, the
octahedron graph ($f=8$), as shown in Figs. 1(a,b,c), respectively, where $f$
denotes the facet number, i.e., $f=2-||V||+||E||$ in Euler’s polyhedron
formula. While the given platonic solids are three-dimensional, their graphs
are planar, so they are in principle implementable on the plane. However,
planar arrangements of atoms for these planar graphs require non-adjacent
atoms to be coupled, as if they were adjacent, so we choose to use quantum
wires [33].

Quantum wires [35, 33] can be substituted for chosen edges of a graph $G$, in
cases when we determine ground-state spin configurations and energies of
$H_{G}$. For the MIS phase which is defined by the parameter conditions,
$\Omega=0$ and $0<\Delta<U$, in Eq. (1), the edge of $G$ can be replaced by a
quantum wire, which is simply a chain of an even number of wire atoms in an
anti-ferromagnetic (AF) spin state. In the MIS phase, in which no adjacent
atoms are simultaneously excited to the Rydberg state, the state of the
quantum wire, say one with two atoms, must be in either $\ket{00}_{W}$,
$\ket{01}_{W}$, or $\ket{10}_{W}$, so the quantum-wire introduces no
additional energies [33].

Figure 2: Phase diagrams of Platonic graphs: (a) The tetrahedron graph
$K_{4}$, (b) the cube graph $Q_{3}$, and (c) the octahedron graph $K_{2,2,2}$.
There are common paramagnetic phases, PM↓ and PM↑ in the regions,
$\Delta_{f}<0$ and $\Delta_{f}>3U$, respectively, and graph-dependent AF-like
phases between them. The first AF-like phases above PM↓ are the MIS phases of
the graphs, in which all adjacent atoms that are denoted by the edges of the
graphs are Rydberg blockaded. (a) The MIS phase of $K_{4}$ allows one Rydberg
atom, i.e., the ground many-body state is the superposition of four singly-
excited configurations in Eq. (2). The other AF-like phases of $K_{4}$ are the
superpositions of doubly-excited and triply-excited configurations,
respectively located in $U<\Delta_{f}<2U$ and $2U<\Delta_{f}<3U$. (b) The MIS
phase of $Q_{3}$ is the superposition of two quadruply-excited configurations
as in Eq. (5). (c) The MIS phase of $K_{2,2,2}$ is the superposition of three
doubly-excited configurations, as in Eq. (6), and the other AF-like phase is
the superposition of three quadruply-excited configurations, located in
$2U<\Delta<3U$. Experiments in Sec. IV are performed by adiabatically evolving
the atom arrays from the PM↓ to the MIS phases.

As an example, let us consider the ground-state spin configurations of the
tetrahedron graph, $K_{4}$, and compare them with the ones of a quantum-wired
graph, $K_{4}^{\prime}$. As shown in Fig. 1(d), our choice of $K_{4}^{\prime}$
has two wire atoms, $a$,$b$, which are used to couple the four atoms, 1,2,3,4
of $K_{4}$. The MIS ground state of the tetrahedron graph, $K_{4}$, is given
by

$\ket{{\rm MIS}(K_{4})}=\frac{\ket{1000}+\ket{0100}+\ket{0010}+\ket{0001}}{2}$
(2)

which is the four-qubit $W$ state. The MIS ground state of the quantum-wired
graph, $K_{4}^{\prime}$, is given by

$\displaystyle\ket{{\rm MIS}(K_{4}^{\prime})}$ $\displaystyle=$
$\displaystyle\sqrt{\frac{3}{32}}\ket{10}_{W}\otimes\left(\ket{0001}+\ket{0010}\right)+\sqrt{\frac{3}{32}}\ket{01}_{W}\otimes\left(\ket{0100}+\ket{1000}\right),$
(3) $\displaystyle+$
$\displaystyle\sqrt{\frac{5}{32}}\ket{00}_{W}\otimes\left(\ket{0101}+\ket{0110}+\ket{1001}+\ket{1010}\right)$

where the first two terms satisfy the AF condition of the quantum wire but the
last term fails. Therefore, if we simply discard the last term, by performing
a conditional measurement of $\ket{{\rm MIS}(K_{4}^{\prime})}$, given the
condition of the quantum-wire’s AF configurations, i.e., $\ket{10}_{W}$ and
$\ket{01}_{W}$ only, we can obtain the probability distribution of $\ket{{\rm
MIS}(K_{4})}$. With the a priori information of the symmetric phase relation
of the graph, we obtain $\left(\bra{01}_{W}+\bra{10}_{W}\right)\ket{{\rm
MIS}(K_{4}^{\prime})}\rightarrow\ket{{\rm MIS}(K_{4})}$.

In general, the MIS ground states of a general graph $G$ and its quantum-wired
graph $G^{\prime}$ are related as

$\bra{\rm AF}_{W}\ket{{\rm MIS}(G^{\prime})}\rightarrow\ket{{\rm MIS}(G)},$
(4)

where $\ket{\rm AF}_{W}$ denotes AF configurations of the quantum wire. In
experiments below, we use quantum-wired graphs, $K_{4}^{\prime}$ of one
quantum wire, $Q_{3}^{\prime}$ of four quantum wires, and $K_{2,2,2}^{\prime}$
of three quantum wires, as shown respectively in Figs. 2(d,e,f), to obtain the
MIS ground states of the Platonic graphs, $K_{4}$, $Q_{3}$, and $K_{2,2,2}$.

##  III Phase diagrams

The phase diagrams, or the ground-state spin configurations, of the
Hamiltonian $H_{G}(U,\Delta_{f},\Omega=0)$ are shown in Fig. 2 for the chosen
Platonic graphs. The Hamiltonian $H_{G}$ in Eq. (1) has two competing energy
terms: the laser detuning term which favors Rydberg atoms, or the up spins,
and the interaction term which favors no adjacent double excitations. So,
there are two paramagnetic phases, PM↓ in the $\Delta<0$ region and PM↑ in
$\Delta>3U$, and AF-like phases between them. Among the AF-like phases, we are
interested in the MIS phase, which has no adjacent spin pairs at all. The MIS
phase regions differ by graphs. Below we consider the MIS phases of the
tetrahedron, cube, and octahedron graphs, respectively.

Tetrahedron: The tetrahedron graph, $K_{4}$, has four vertices and each vertex
is edged to all others. In the MIS phase of no adjacent Rydberg atoms, only
one atom is allowed to be excited to the Rydberg state. So, the MIS ground
state is $\ket{\rm{MIS}(K_{4})}$ in Eq. (2), which is the super-atom[38]
superposition state. The MIS phase region, MIS($K_{4}$), is located in
$0<\Delta<U$, as shown in Fig. 2(a). Between MIS($K_{4}$) and PM↑, there are
two additional AF-like phases in regions $U<\Delta<2U$ and $2U<\Delta<3U$,
which have one and two adjacent Rydberg atom pairs, respectively.

Cube: The cube graph, $Q_{3}$, has eight vertices, each edged to three other
vertices. In the MIS phase, MIS($Q_{3}$), when one atom is in the Rydberg
state, its face-diagonal, three vertices are allowed to become Rydberg atoms,
so the MIS ground state is

$\ket{{\rm MIS}(Q_{3})}=\frac{\ket{01010101}+\ket{10101010}}{\sqrt{2}}.$ (5)

Any other states, having less or more Rydberg atoms, are not ground states in
any region in $0<\Delta<3U$, because Rydberg deexcitation or excitation of
$\ket{\rm{MIS}(Q_{3})}$ increases the energy by $\Delta>0$ or $3U-\Delta$,
respectively. So, the MIS phase is the only AF-like phase of the cube graph,
as shown in Fig. 2(b).

Octahedron: The octahedron graph, $K_{2,2,2}$, has six vertices, each edged to
four other vertices. There are two AF-like phases, the MIS phase of two
Rydberg atoms, i.e.,

$\ket{{\rm
MIS}(K_{2,2,2})}=\frac{\ket{100100}+\ket{010010}+\ket{001001}}{\sqrt{3}}$ (6)

and its inversion of four Ryberg atoms. Their equal energy boundary is
$\Delta=2U$, so the MIS phase of the octahedron graph is located in
$0<\Delta<2U$ and the inverted MIS phase is in $2U<\Delta<3U$, as shown in
Fig. 2(c).

##  IV Experimental procedure

The MIS phases of the Platonic graphs, $K_{4}$, $Q_{3}$, and $K_{2,2,2}$, are
probed with quantum adiabatic control of Hamiltonians, $H_{G}^{\prime}$, of
the quantum-wired graphs, $G^{\prime}=K_{4}^{\prime}$, $Q_{3}^{\prime}$, and
$K_{2,2,2}^{\prime}$. The Hamiltonians are changed from their paramagnetic
phase region, PM↓, to the MIS phase region, respectively. Quantum simulation
experiments of this kind were previously performed with various Rydberg-atom
arrays [12, 29, 21, 28, 33].

We used rubidium (87Rb) atoms, which were cooled in a magneto-optical trap and
optically pumped to the ground state
$\ket{0}=\ket{\downarrow}=\ket{5S_{1/2},F=2,m_{F}=2}$. The atoms were then
captured by far-off resonance dipole traps (optical tweezers) and arranged to
positions for $G^{\prime}$ [37]. Table 1 lists the atom positions used in our
experiments for $G^{\prime}=K_{4}^{\prime}$, $Q_{3}^{\prime}$, and
$K^{\prime}_{2,2,2}$. The nearest-neighbor distances were the same $d=8.0\pm
0.3$ $\mu$m and the blockade distance was $d_{B}=10.55$ $\mu$m at
$\Omega_{0}/2\pi=0.74$ MHz. The Rydberg state was
$\ket{1}=\ket{\uparrow}=\ket{71S_{1/2},J=1/2,m_{J}=1/2}$, excited by the two-
photon off-resonant transition, $\ket{0}\rightarrow\ket{m}\rightarrow\ket{1}$,
via the intermediate state
$\ket{m}=\ket{5P_{3/2},F^{\prime}=3,m_{F^{\prime}}=3}$. We used a 780-nm laser
field (Toptical DL Pro 780) of $\Omega_{780}=(2\pi)112$ MHz for
$\ket{0}\rightarrow\ket{m}$ transition and a 480-nm laser field (Toptica TA-
SHG Pro) of $\Omega_{480}=(2\pi)7.4$ MHz for $\ket{m}\rightarrow\ket{1}$, and
the intermediate detuning was $\Delta_{m}=(2\pi)560$ MHz. The effective Rabi
frequency $\Omega$ was varied up to the max
$\Omega_{0}=\Omega_{780}\Omega_{480}/2\Delta_{m}=(2\pi)0.74$ MHz.

Table 1: Atom positions of the experimental graphs Graphs | Atom positions $(x,y)$ [$\mu$m]
---|---
Tetrahedron $K^{\prime}_{4}$ | 1: | (-10.9, 4.0) | 2: | (-10.9, -4.0)
3: | (10.9, -4.0) | 4: | (10.9, 4.0)
$W$: | ($\pm$ 4.0, 0.0) |  |
Cube $Q^{\prime}_{3}$ | 1: | (-4.0, 4.0) | 2: | (-4.0, -4.0)
3: | (4.0, -4.0) | 4: | (4.0, 4.0)
5: | (9.7, 9.7) | 6: | (-9.7, 9.7)
7: | (-9.7, -9.7) | 8: | (9.7, -9.7)
$W_{1}$: | ($\pm$ 4.0, 15.3) |  $W_{2}$: | (-15.3, $\pm$ 4.0)
$W_{3}$: | (-4.0, $\pm$ 15.3) |  $W_{4}$: | (15.3, $\pm$ 4.0)
Octahedron $K_{2,2,2}^{\prime}$ | 1: | (0.0, 9.8) | 2: | (-4.0, 2.9)
3: | (-8.0, -4.0) | 4: | (0.0, -4.0)
5: | (8.0, -4.0) | 6: | (4.0, 2.9)
|  $W_{1}$: |  | (-6.9, 13.8) | (-14.9, 13.8)
---|---
(-18.9, 6.9) | (-14.9, 0.0)
|  $W_{2}$: |  | (-8.0, -12.0) | (-4.0, -18.9)
---|---
(4.0, -18.9) | (8.0, -12.0)
|  $W_{3}$: |  | (14.9, 0.0) | (18.9, 6.9)
---|---
(14.9, 13.8) | (6.9, 13.8)

The atoms were initially in the PM↓ phase, i.e.,
$\ket{\Psi(t=0)}=\ket{0}^{N^{\prime}}$, where $N^{\prime}$ the number of atoms
in $G^{\prime}$, and then adiabatically driven to their MIS phase,
$\ket{\Psi(t=t_{f})}=\ket{{\rm MIS}(G^{\prime})}$. The control parameters of
$H_{G^{\prime}}$ in Eq. (1) were changed in three stages. In the first stage,
the Rabi frequency $\Omega(t)$ was linearly tuned on from $\Omega(t=0)=0$ to
$\Omega(t_{1})=\Omega_{0}$ and the detuning $\Delta(t)$ was maintained at
$\Delta(t=0-t_{1})=-3$ MHz. In the second stage, we maintain the Rabi
frequency $\Omega(t=t_{1}-t_{2})=\Omega_{0}$ and linearly change the detuning
to $\Delta(t_{2})=\Delta_{f}$. In the final stage, the Rabi frequency was
linearly changed to zero, i.e., $\Omega(t_{f})=0$, and the detuning was
maintained at $\Delta(t=t_{2}-t_{f})=\Delta_{f}$. We used $t_{f}=4.0$ $\mu$s,
$t_{1}=t_{f}/10$, $t_{2}=t_{f}-t_{1}$, and $\Delta_{f}=2.0$ MHz for
tetrahedron experiment and 3.0 MHz for cube and octahedron experiments.
$\Omega(t)$ and $\Delta(t)$ were changed with a radio-frequency programmable
synthesizer (Moglabs XRF) and acousto-optic modulators (AOM) [28]. The final
Hamiltonian parameters are $(U/\Omega_{0},\Delta_{f}/\Omega_{0})=(5.27,2.70)$
for the tetragonal $K_{4}$ and $(5.27,4.05)$ for the cube $Q_{3}$ and
octahedron $K_{2,2,2}$ experiments, which are located in their respective MIS
phases in Fig. 3.

After the atoms were driven to the MIS phase, their final states were measured
by collecting the fluorescence images of the cyclic transition,
$\ket{5S_{1/2},F=2}\leftrightarrow\ket{5P_{3/2},F^{\prime}=3}$ in an electron
multiplying charge coupled device (Andor iXon Ultra 897). We repeated the
above measurement by 927 times for $K_{4}$, 3292 for $Q_{3}$, and 1574 for
$K_{2,2,2}$ to obtain their probabilities of all spin configurations.

##  V Results

Before we perform the quantum simulation of the platonic graphs, we first test
the working principle of the quantum wire by using an example of
$K_{4}^{\prime}$ and varying the edge distance of the quantum-wired graph. We
use two kinds of edges, $d_{12}=d_{34}=d$ and
$d_{1a}=d_{2a}=d_{ab}=d_{b3}=d_{b4}=d^{\prime}$, which are black- and yellow-
colored edges, respectively, in Fig. 1(d), where $a$,$b$ denote the two wire
atoms and $d^{\prime}$ is the length of the edges involved with the wire
atoms. We use six different atom-arrays of $d^{\prime}$ varied from $0.8d$ to
$1.3d$ as in Fig. 3(a) (from the left to the right). It is noted that, as
$d^{\prime}/d$ increases from zero to $\infty$, the six-atom system changes
from a super-atom to a pair of isolated dimers, and the $d^{\prime}=d$ case
corresponds to the $K_{4}^{\prime}$ graph in the third column in Fig. 3.

Figure 3: $K_{4}^{\prime}$-like graphs of two types of edges: (a) Six
different graphs are constructed with $d^{\prime}/d=0.8$, 0.9, 1.0, 1.1, 1.2,
and 1.3 (from left to right), among which the $d^{\prime}=d$ graph corresponds
to $K_{4}^{\prime}$. (b) Fluorescence images of the atom arrays driven to
$H(\Omega_{f}=0,\Delta_{f}=2.0\rm{MHz},U=3.9\rm{MHz})$ and sorted based on
their spin configurations. Six characteristic spin configurations are shown,
(from top to bottom) $\ket{00}_{W}\ket{1010}$, $\ket{00}_{W}\ket{1001}$,
$\ket{01}_{W}\ket{1000}$, $\ket{11}_{W}\ket{1010}$, and
$\ket{11}_{W}\ket{1001}$. The scale bar represents the relative probabilities
of the configurations per graph. The numbers of collected events are 1057,
971, 927, 906, 2866, and 918, respectively, for the graphs.

In Fig. 3(b), we show configuration-dependent fluorescence images of the atoms
adiabatically driven to the MIS-phase condition, i.e.,
$0<\Delta_{f}/U=0.51<1$. There are six characteristic spin-configurations,
which are, from the top to the bottom, $\ket{00}_{W}\ket{1010}$,
$\ket{00}_{W}\ket{1001}$, $\ket{01}_{W}\ket{1000}$, $\ket{11}_{W}\ket{1010}$,
and $\ket{11}_{W}\ket{1001}$. Other spin-configurations of the same rotational
or reflection symmetries are observed similarly. The contrasts of the images
in Fig. 2(b) are normalized per column to compare their relative occurrences
among spin-configurations and graphs. In the first column for
$d^{\prime}/d=0.8$, the most-strongly interacting six atoms, the most
frequently observed configuration is $\ket{00}_{W}\ket{1010}$, which has both
the wire atoms in the ground state. In the last column for $d^{\prime}/d=1.3$,
the pair of isolated dimers, $\ket{11}_{W}\ket{1001}$ is the most frequently
observed configuration, in which both the wire atoms are Rydberg atoms. Both
of these cases, $\ket{00}_{W}\ket{1010}$ and $\ket{11}_{W}\ket{1001}$ fail the
AF quantum-wire condition. On the other hand, in the third column, which
corresponds $K_{4}^{\prime}$ ($d^{\prime}=d$), the highlighted
$\ket{01}_{W}\ket{1000}$ configuration is significantly observed, which
satisfies the AF quantum-wire condition. The other two significantly observed
configurations are $\ket{00}_{W}\ket{1010}$ and $\ket{00}_{W}\ket{1001}$,
agreeing well with the expected ground many-body state of $K_{4}^{\prime}$ in
Eq. (3).

Figure 4: Probability distributions, $|\langle n\ket{G^{\prime}}|^{2}$, of
quantum-wired Platonic graphs driven to their MIS phases: (a)
$G^{\prime}=K_{4}^{\prime}$, (b) $Q_{3}^{\prime}$, and (c)
$K_{2,2,2}^{\prime}$. The numbers of collected experimental events are 927,
3292, and 1574, respectively. (d-f) After the AF conditions of the quantum
wires are imposed, there remain 469, 820, and 72 events, respectively, which
are used to reconstuct the probability distributions, $|\langle
n\ket{G}|^{2}$, of the Platonic graphs, (d) $G=K_{4}$, (e) $Q_{3}$, and (f)
$K_{2,2,2}$. In (d), a numerical simulation of the evolved dynamics of (a) is
shown in comparison.

Main results of the quantum simulation experiments of the platonic graphs are
summarized in Fig. 4. The atom arrays of the quantum-wired graphs,
$G^{\prime}=K_{4}^{\prime}$, $Q^{\prime}_{3}$, and $K^{\prime}_{2,2,2}$, are
adiabatically driven from the PM↓ phase to their MIS phases and their
probabilities are plotted per spin configurations, before and after the AF
quantum-wire conditions are imposed, in Figs. 4(a-c) and Figs. 4(d-e),
respectively. The experimental probability distributions of the platonic
graphs, $K_{4}$, $Q_{3}$, and $K_{2,2,2}$ are in good agreements with what are
expected from their MIS ground states, $\ket{{\rm MIS}(K_{4})}$, $\ket{{\rm
MIS}(Q_{3})}$, and $\ket{{\rm MIS}(K_{2,2,2})}$, respectively in Eqs. (2),
(5), and (6), as explained below.

Tetrahedron: In Fig. 4(a), we plot the experimental $K_{4}^{\prime}$
probabilities, $|\langle n\ket{{\rm Exp}(K_{4}^{\prime})}|^{2}$, where $n$ is
a decimal number enumerating the spin configuration of the $K_{4}^{\prime}$
atom-array of two wire atoms and four $K_{4}$ atoms in such a way that
$\ket{n=1}=\ket{00}_{W}\otimes\ket{0000}$,
$\ket{2}=\ket{00}_{W}\otimes\ket{0001}$, $\cdots$,
$\ket{64}=\ket{11}_{W}\otimes\ket{1111}$. The shaded region in Fig. 3(a)
corresponds to the spin configurations satisfying the AF condition of the
quantum wire. The max-population peaks in the AF region are identified to be
$\ket{21}=\ket{01}_{W}\ket{0100}$, $\ket{25}=\ket{01}_{W}\ket{1000}$,
$\ket{34}=\ket{10}_{W}\ket{0001}$, and $\ket{35}=\ket{10}_{W}\ket{0010}$,
agreeing well with the spin configurations of $\ket{{\rm MIS}(K_{4})}$ in Eq.
(2). It is noted that the significant populations in the first quadrant in
Fig. 3(b), which corresponds to $\ket{00}_{W}$, are the four spin-
configurations in the third term of Eq. (3), so they violate the AF condition.
In Fig. 4(d), after the AF quantum-wire condition is imposed, we plot the
probability distribution of $\ket{{\rm Exp}(K_{4})}$, which is the
normalization of $|\bra{{\rm AF}}_{W}\ket{{\rm Exp}(K_{4}^{\prime})}|^{2}$,
and compared it with a numerical simulation. The quantum evolution of the
$K_{4}^{\prime}$ atoms is numerically traced along the adiabatic control path
introduced in Sec. IV with a Lindbladian master equation [39] which takes into
account experimental noises from laser phase noises and bit-flip detection
errors of $P_{0\rightarrow 1^{\prime}}=0.12$, $P_{1\rightarrow
0^{\prime}}=0.09$ [40, 41]. The experimentally observed $K_{4}$ probability,
which is the sum of the populations of the four peaks,
$\ket{5}=\ket{01}_{W}\ket{0100}$, $\ket{9}=\ket{01}_{W}\ket{1000}$,
$\ket{18}=\ket{10}_{W}\ket{0001}$, and $\ket{19}=\ket{10}_{W}\ket{0010}$, is
$|\langle{\rm MIS}(K_{4})\ket{{\rm Exp(K_{4})}}|^{2}=0.33\pm 0.03$.

Cube: The experimental observed $Q_{3}^{\prime}$ probability distribution of
$|\langle n\ket{{\rm Exp}(Q_{3}^{\prime})}|^{2}$, for $n=1,\cdots,2^{16}$, are
plotted in Fig. 4(b). In the MIS phase of the $Q_{3}^{\prime}$ graph, we
expect

$\displaystyle\ket{{\rm
MIS}(Q_{3}^{\prime})}=(\ket{01}_{W_{1}}\ket{10}_{W_{2}}\ket{01}_{W_{3}}\ket{10}_{W_{4}}\ket{10101010}$
$\displaystyle\quad+\ket{10}_{W_{1}}\ket{01}_{W_{2}}\ket{10}_{W_{3}}\ket{01}_{W_{4}}\ket{01010101})/\sqrt{2},$
(7)

because the state of each wire-atom pair which couples a pair of $Q_{3}$ atoms
in $\ket{01}$ ($\ket{10}$) is determined to be $\ket{10}_{W}$
($\ket{01}_{W}$). We observed two max-populated spin-configurations,
$\ket{26283}$ and $\ket{39254}$, in Fig. 4(b), agreeing well with the two spin
configurations in $\ket{{\rm MIS}(Q_{3}^{\prime})}$. In Fig. 4(e), after
imposing the AF conditions of the quantum wires, we plot the experimental
$Q_{3}$ probability distribution of $\ket{{\rm Exp}(Q_{3})}$, of which the
result agrees well with $\ket{{\rm MIS}(Q_{3})}$ in Eq. (5). The
experimentally observed probability of $Q_{3}$ is $|\langle{\rm
MIS}(Q_{3})\ket{{\rm Exp}(Q_{3})}|^{2}=0.07\pm 0.01$, which is the sum of
populations of $\ket{1451}$ and $\ket{2646}$ in Fig. 4(e).

Octahedron: In Fig. 4(c), we plot $|\langle n\ket{{\rm
Exp}(K_{2,2,2}^{\prime})}|^{2}$, the experimental probability distribution of
$K_{2,2,2}^{\prime}$ which has twelve quantum-wire atoms and six $K_{2,2,2}$
atoms. The MIS ground state of the $K_{2,2,2}^{\prime}$ is a little
complicated. Among the three quantum wires, each of which has four wire atoms,
as in Fig. 1(c), two quantum wires, say $W_{1}$ and $W_{2}$ for the case when
atoms $3$,$6$ are Rydberg atoms, must be in the AF states,
$\ket{W_{1}}=\ket{1010}$ and $\ket{W_{2}}=\ket{0101}$, but then the other
quantum wire, $W_{3}$, can be in $\ket{1001}$ as well as in $\ket{0101}$ and
$\ket{1010}$, because they contribute the same energies. So, the MIS ground
state of $K_{2,2,2}^{\prime}$ is given by

$\displaystyle\ket{{\rm MIS}(K_{2,2,2}^{\prime})}$ $\displaystyle=$
$\displaystyle\ket{1010}_{W_{1}}\ket{0101}_{W_{2}}\left(a(\ket{0101}+\ket{1010})+b\ket{1001}\right)_{W_{3}}\ket{001001}$
(8) $\displaystyle+$
$\displaystyle\left(a(\ket{0101}+\ket{1010})+b\ket{1001}\right)_{W_{1}}\ket{1010}_{W_{2}}\ket{0101}_{W_{3}}\ket{010010}$
$\displaystyle+$
$\displaystyle\ket{0101}_{W_{1}}\left(a(\ket{0101}+\ket{1010})+b\ket{1001}\right)_{W_{2}}\ket{1010}_{W_{3}}\ket{100100},$

where $a=\sqrt{20/243}$ and $b=\sqrt{41/243}$ are normalization factors. Due
to the limited statistics of the $K_{2,2,2}^{\prime}$ experiment, it is
unclear if there are 9 peaks in Fig. 4(c), as expected from Eq. 8; however,
after summing the events with the AF quantum-wire conditions, we observe three
clear peaks in Fig. 4(f), which correspond to the spin-configurations of the
MIS phase of $K_{2,2,2}$, agreeing well with $\ket{{\rm MIS}(K_{2,2,2})}$.
Experimentally observed $K_{2,2,2}$ probability is $|\langle{\rm
MIS}(K_{2,2,2})\ket{{\rm Exp}(K_{2,2,2})}|^{2}=0.39\pm 0.07$, which is the sum
of the populations of $\ket{10}=\ket{001001}$ and $\ket{19}=\ket{010010}$, and
$\ket{37}=\ket{100100}$.

##  VI Discussions and Outlook

Now we turn our attention to the scaling issue involved with the current
method of graph transformations from 3D surfaces to 2D planes. Our experiment
presented the transformation of the tetrahedron graph $K_{4}$ of $N=4$ atoms
to the quantum-wired graph $K_{4}^{\prime}$ of total $N^{\prime}=6$ atoms, the
cube graph $Q_{3}$ of $N=8$ to $Q_{3}^{\prime}$ of $N^{\prime}=16$, and the
octahedron graph $K_{2,2,2}$ of $N=6$ to $K_{2,2,2}^{\prime}$ of
$N^{\prime}=22$. In Fig. 5, we plot $N$ versus $N^{\prime}$ for these three
Platonic graphs and we also plot other examples of the remaining Platonic
graphs, the icosahedron and dodecahedron graphs, and two Fullerene graphs,
$C_{24}$ and $C_{60}$. We observe a linear relation between $N$ and
$N^{\prime}$, of which the scaling can be understood as follows: When we move
the vertices from the sphere of radius $R$ to the square plane of length $L$,
we expect $L\sim R$, because the distances of vertices along a chosen
circumference of the sphere can be chosen to be unchanged. The number $N$ of a
graph $G$ scales with $N\sim R^{2}/r_{B}^{2}$ and $N^{\prime}$ of $G^{\prime}$
scales with $N^{\prime}\sim L^{2}/r_{B}^{2}$. So, the quantum-wired graph
$G^{\prime}$ can be constructed with $N^{\prime}\sim N$ to transform $G$ from
3D to 2D.

Figure 5: Atom-number scaling of $N^{\prime}$, of a quantum-wire graph
$G^{\prime}$, versus $N$, of a target graph $G$

While the number of atoms required for the 3D-to-2D transformation is scaled
favorably, the actual $N^{\prime}$ in our quantum-wired graph experiments is
limited by, for example, the rearrangement probabiliy and bit-flip errors.
When $N^{\prime}$ atoms are to be rearranged to chosen positions, the
rearrangement probability of $N^{\prime}$ atoms is given by
$P_{r}=p^{N^{\prime}}$, where $p=\exp(-t/t_{0})=0.97$ is the single-atom
survival probability for an average trap lifetime of $t_{0}=16$ s and
rearrangement time of $t=0.6$ s, which gives $N^{\prime}<80$ for $P_{r}>0.01$.
The bit-flip errors, due to imperfections of the state preparation and
measurements as well as quasi-adiabatic controls, result in statistical
ambiguities of the ground-state probability,
$P_{g}^{\prime}\sim(1-P_{0\rightarrow 1})^{N^{\prime}/2}(1-P_{1\rightarrow
0})^{N^{\prime}/2}P_{g}$, and the probabilites of other states, $P_{\rm
others}\sim P_{0\rightarrow 1}(1-P_{0\rightarrow
1})^{N^{\prime}/2-1}(1-P_{1\rightarrow 0})^{N^{\prime}/2}P_{g}$. A criterion
of $|P_{g}^{\prime}-P_{\rm
others}|\sim\sqrt{P_{g}^{\prime}(1-P_{g}^{\prime})/M}$ demands about $M=100$
repeated experiments, which takes $\sim$ 2 minites for $N^{\prime}=25$, about
$M=2\times 10^{5}$ experiments or $\sim 3$ days for $N^{\prime}=80$. Both the
considered experimental limitations can be improved by the state of the art
Rydberg-atom technologies. One approach to increase the rearrangement
probability is using a cryogenic environment of Rydberg atoms [43], in which
the the trap lifetime is measured to be $t_{0}=6\times 10^{3}$ s. It is
expected that the single-atom survival probability $p=0.9999$ in the cryogenic
setup can achieve $N^{\prime}>1000$ for $P_{r}=0.9$. The bit-flip error are
significantly reduced with alkaline-earth atomic systems [42], where
$P_{0\rightarrow 1}$ and $P_{1\rightarrow}$ are less than 0.01.

In summary, we have performed quantum simulation of Ising-like spins on three
Platonic graphs, the tetrahedron, cube, and octahedron graphs, to
experimentally obtain their many-body ground states in antiferromagnetic
phases. While these graphs are 3D structures due to their coupling structures,
we used Rydberg atom wires to transform them to 2D structures, in which the
quantum wires adjusted the coupling strengths of stretched edges to be the
same as those of the unstretched edges. With the 2D quantum-wired graphs, we
control the atom-array Hamiltonians adiabatically from their paramagnetic
phase to the AF phases and obtained their many-body AF ground-state spin
configurations. The results are in good agreements with the symmetry analysis
and numerical simulations. In our estimations, our approach of using quantum
wires to transform 3D planar-graph structures could achieve, in conjunction
with alkaline-earth atom technologies in cryogenic environments, more than
1000-atom quantum simulations.

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Combinatorial identities with multiple harmonic-like numbers

Kunle Adegoke
Department of Physics and Engineering Physics
Obafemi Awolowo University, 220005 Ile-Ife, Nigeria
<EMAIL_ADDRESS>

Robert Frontczak
Independent Researcher
Reutlingen, Germany
<EMAIL_ADDRESS>

###### Abstract

Multiple harmonic-like numbers are studied using the generating function
approach. A closed form is stated for binomial sums involving these numbers
and two additional parameters. Several corollaries and examples are presented
which are immediate consequences of the main result. Finally, combinatorial
identities involving harmonic-like numbers and other prominent sequences like
hyperharmonic numbers and odd harmonic numbers are offered.

2010 Mathematics Subject Classification: 05A19, 11B73, 11B75.

_Keywords:_ Multiple harmonic-like number; Harmonic number; binomial
transform.

##  1 Preliminaries

Cheon and El-Mikkawy [8, 9] defined multiple harmonic-like numbers by

$H_{n}(m)=\sum_{1\leq k_{1}+k_{2}+\cdots+k_{m}\leq n}\frac{1}{k_{1}k_{2}\cdots
k_{m}}$ (1)

with $H_{n}(0)=1$ for $n\geq 0$ and $H_{0}(m)=0$ for $m\geq 1$. They showed
that the generating function of $H_{n}(m)$ equals

$H(z)=\sum_{n=0}^{\infty}H_{n}(m)z^{n}=\frac{(-\ln(1-z))^{m}}{1-z}.$ (2)

For $m=1$ these numbers reduce to ordinary harmonic numbers $H_{n}$ as

$H_{n}(1)=\sum_{1\leq k_{1}\leq n}\frac{1}{k_{1}}=H_{n},\qquad H_{0}=0.$ (3)

For $m=2$ we see that

$\displaystyle H_{n}(2)$ $\displaystyle=\sum_{1\leq k_{1}+k_{2}\leq
n}\frac{1}{k_{1}k_{2}}=\sum_{j=1}^{n}\sum_{k_{1}+k_{2}=j}\frac{1}{k_{1}k_{2}}$
$\displaystyle=\sum_{j=1}^{n}\sum_{k_{1}=1}^{j-1}\frac{1}{k_{1}(j-k_{1})}=\sum_{j=1}^{n}\frac{2}{j}H_{j-1}$
$\displaystyle=\sum_{j=1}^{n}\frac{H_{n-j}}{j}=H_{n}^{2}-H_{n}^{(2)},$

where in the last line a result of Kargin and Can [13] was used and where
$H_{n}^{(2)}$ are the second-order harmonic numbers, i.e.,

$H_{n}^{(2)}=\sum_{j=1}^{n}\frac{1}{j^{2}}.$

The brute force computation of $H_{n}(3)$ is tedious. The result is

$H_{n}(3)=\sum_{j=1}^{n}\frac{1}{j}\sum_{l=1}^{n-j}\frac{H_{n-j-l}}{l}.$

The next lemma is helpful.

######  Lemma 1.

For all $n\geq 1$ and $m\geq 0$ we have the identity

$H_{n}(m+1)=\sum_{j=1}^{n}\frac{H_{n-j}(m)}{j}.$ (4)

###### Proof.

Using (2) we have

$\displaystyle\sum_{n=0}^{\infty}H_{n}(m+1)z^{n}$
$\displaystyle=(1-z)\frac{-\ln(1-z)}{1-z}\frac{(-\ln(1-z))^{m}}{1-z}$
$\displaystyle=(1-z)\left(\sum_{n=0}^{\infty}H_{n}z^{n}\right)\left(\sum_{n=0}^{\infty}H_{n}(m)z^{n}\right)$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{j=0}^{n}H_{j}H_{n-j}(m)z^{n}-\sum_{n=1}^{\infty}\sum_{j=0}^{n-1}H_{j}H_{n-1-j}(m)z^{n}.$

Extracting and comparing the coefficients of $z^{n}$ we obtain for all $n\geq
1$

$\displaystyle H_{n}(m+1)$
$\displaystyle=\sum_{j=0}^{n-1}H_{j}\left(H_{n-j}(m)-H_{n-1-j}(m)\right)$
$\displaystyle=\sum_{j=1}^{n}H_{n-j}(m)(H_{j}-H_{j-1})$
$\displaystyle=\sum_{j=1}^{n}\frac{H_{n-j}(m)}{j}$

as claimed. ∎

Multiple harmonic-like numbers were studied recently by Chen and Guo in the
papers [6] and [7]. For instance, in [6] several summation formulae involving
harmonic-like numbers and other combinatorial numbers were derived. In the
paper [7], a certain sequence $A_{\alpha}(n,k)$ was studied and, as a part of
this study, additional interesting combinatorial identities involving
harmonic-like numbers were presented.

In this paper, we continue the work on harmonic-like numbers applying the
generating function approach. Our first main result is a closed form for
binomial sums involving these numbers and two additional parameters
$a,b\in\mathbb{C}$. Several corollaries and examples are presented which are
immediate consequences of the main result. Finally, combinatorial identities
involving harmonic-like numbers and other prominent sequences like
hyperharmonic numbers and odd harmonic numbers are offered.

In what follows we will need the definition of the Stirling numbers of the
first kind, $s(n,k)$. These numbers is defined by the generating function

$\sum_{n=k}^{\infty}s(n,k)\frac{z^{n}}{n!}=\frac{\ln^{k}(1+z)}{k!}.$ (5)

Some particular values are

$\displaystyle s(n,k)$ $\displaystyle=0\qquad\mbox{for}\,\,n<k,$
$\displaystyle s(n,0)$ $\displaystyle=\begin{cases}1,&n=0;\\\ 0,&n\geq 1;\\\
\end{cases}$ $\displaystyle s(n,1)$ $\displaystyle=(-1)^{n-1}(n-1)!$
$\displaystyle s(n,2)$ $\displaystyle=(-1)^{n}(n-1)!H_{n-1}.$

##  2 Binomial sums involving $H_{n}(m)$

For $a,b\in\mathbb{C}$ let $S_{n}(a,b,m)$ be defined by

$S_{n}(a,b,m)=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}H_{k}(m).$ (6)

Then we have the following result.

######  Theorem 1.

For all $n\geq 0$ we have

$S_{n}(a,b,m)=\sum_{j=0}^{m}\binom{m}{j}\sum_{k=0}^{n}H_{k}(j)(a+b)^{k}\frac{(m-j)!}{(n-k)!}(-1)^{n-k}b^{n-k}s(n-k,m-j),$
(7)

where $s(n,k)$ are the Stirling numbers of the first kind.

###### Proof.

Let $S(z)$ denote the generating function of $S_{n}(a,b,m)$. Then (see [3,
16])

$\displaystyle S(z)$ $\displaystyle=\sum_{n=0}^{\infty}S_{n}(a,b,m)z^{m}$
$\displaystyle=\frac{1}{1-bz}H\left(\frac{az}{1-bz}\right)$
$\displaystyle=\frac{1}{1-(a+b)z}\left(-\ln\left(\frac{1-(a+b)z}{1-bz}\right)\right)^{m}$
$\displaystyle=(-1)^{m}\frac{(\ln(1-(a+b)z)-\ln(1-bz))^{m}}{1-(a+b)z}$
$\displaystyle=\sum_{j=0}^{m}\binom{m}{j}\frac{(-\ln(1-(a+b)z))^{j}}{1-(a+b)z}\ln^{m-j}(1-bz).$

The result follows from (2) and (5) in combination with the Cauchy product. ∎

######  Remark.

When $m=0$ then using $H_{n}(0)=1,n\geq 0$,

$S_{n}(a,b,0)=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=(a+b)^{n}.$

When $m=1$ then we get

$\displaystyle S_{n}(a,b,1)$
$\displaystyle=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}H_{k}$
$\displaystyle=\sum_{k=0}^{n}\frac{(-1)^{n-k}}{(n-k)!}(a+b)^{k}b^{n-k}s(n-k,1)+\sum_{k=0}^{n}\frac{(-1)^{n-k}}{(n-k)!}H_{k}(a+b)^{k}b^{n-k}s(n-k,0)$
$\displaystyle=H_{n}(a+b)^{n}-\sum_{k=0}^{n-1}(a+b)^{k}b^{n-k}\frac{1}{n-k},$
(8)

which reproduces Boyadzhiev’s main result (Proposition 6) from [2].

######  Corollary 2.

For $n,m\geq 0$ we have

$\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}H_{k}(m)=(-1)^{n}\frac{m!}{n!}s(n,m)$ (9)

and

$\sum_{k=m}^{n}\binom{n}{k}\frac{s(k,m)}{k!}=\frac{1}{m!}H_{n}(m).$ (10)

###### Proof.

The first result follows immediately by setting $a=-1$ and $b=1$ in (7) and
simplifying. The second identity is the inverse binomial transform of the
first (for information on the binomial transform consult [3]). ∎

######  Remark.

The identity (9) was discovered first by Chen and Guo (see [7, Corollary 8]).
The second identity also appears in that paper (with a typo, see the proof of
Proposition 13).

######  Corollary 3.

For $n,m\geq 0$ we have

$\sum_{k=0}^{n}\binom{n}{k}H_{k}(m)=\sum_{j=0}^{m}\binom{m}{j}\sum_{k=0}^{n}H_{k}(j)(-1)^{n-k}2^{k}\frac{(m-j)!}{(n-k)!}s(n-k,m-j).$
(11)

In particular, we recover the classical identity [2]

$\sum_{k=0}^{n}\binom{n}{k}H_{k}=2^{n}\left(H_{n}-\sum_{k=1}^{n}\frac{1}{2^{k}k}\right).$
(12)

###### Proof.

Set $a=b$ in (7) and simplify. ∎

######  Corollary 4.

For $n\geq 0$ we have

$S_{n}(a,b,2)=H_{n}(2)(a+b)^{n}+2\sum_{k=1}^{n}(a+b)^{n-k}b^{k}\frac{H_{k-1}-H_{n-k}}{k}.$
(13)

###### Proof.

Set $m=2$ in (7) and simplify. ∎

Some consequences of Corollary 4 will be stated as examples.

######  Example 1.

We have

$\sum_{k=0}^{n}\binom{n}{k}H_{k}(2)=2^{n}\left(H_{n}(2)+2\sum_{k=1}^{n}\frac{H_{k-1}-H_{n-k}}{2^{k}k}\right).$
(14)

######  Example 2.

From

$\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}H_{k}(2)=\frac{2}{n}H_{n-1}$ (15)

we deduce that

$\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}H_{k}^{(2)}=-\frac{H_{n}}{n},$ (16)

where we used that (see for instance [1])

$\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}H_{k}^{2}=\frac{H_{n}}{n}-\frac{2}{n^{2}}.$

The inverse binomial relation also yields

$\sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1}\frac{H_{k}}{k}=H_{n}^{(2)}.$ (17)

######  Example 3.

We have

$\sum_{k=0}^{n}\binom{n}{k}2^{k}(-1)^{n-k}H_{k}(2)=H_{n}(2)+2\sum_{k=1}^{n}(-1)^{k}\frac{H_{k-1}-H_{n-k}}{k}.$
(18)

######  Example 4.

We have

$\sum_{k=0}^{n}\binom{n}{k}2^{k}H_{k}(2)=3^{n}\left(H_{n}(2)+2\sum_{k=1}^{n}\frac{H_{k-1}-H_{n-k}}{3^{k}k}\right).$
(19)

######  Example 5.

Let $F_{n}$ and $L_{n}$ be the Fibonacci and Lucas numbers, respectively (see
Koshy [15] or Vajda [17]). Then we have

$\sum_{k=0}^{n}\binom{n}{k}F_{k}H_{k}(2)=H_{n}(2)F_{2n}+2\sum_{k=1}^{n}F_{2(n-k)}\frac{H_{k-1}-H_{n-k}}{k},$
(20)
$\sum_{k=0}^{n}\binom{n}{k}L_{k}H_{k}(2)=H_{n}(2)L_{2n}+2\sum_{k=1}^{n}L_{2(n-k)}\frac{H_{k-1}-H_{n-k}}{k},$
(21)

and

$\sum_{k=0}^{n}\binom{n}{k}(-1)^{k+1}F_{k}H_{k}(2)=H_{n}(2)F_{n}+2\sum_{k=1}^{n}F_{n-k}\frac{H_{k-1}-H_{n-k}}{k},$
(22)
$\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}L_{k}H_{k}(2)=H_{n}(2)L_{n}+2\sum_{k=1}^{n}L_{n-k}\frac{H_{k-1}-H_{n-k}}{k}.$
(23)

######  Corollary 5.

For $n\geq 0$ we have

$\displaystyle S_{n}(a,b,3)$ $\displaystyle=H_{n}(3)(a+b)^{n}$
$\displaystyle\quad-3\sum_{k=1}^{n}(a+b)^{n-k}b^{k}\frac{H_{k-1}^{2}-H_{k-1}^{(2)}-2H_{k-1}H_{n-k}+H_{n-k}^{2}-H_{n-k}^{(2)}}{k}.$
(24)

###### Proof.

Set $m=3$ in (7) and simplify using

$s(n,3)=\frac{1}{2}(-1)^{n-1}(n-1)!\left(H_{n-1}^{2}-H_{n-1}^{(2)}\right).$

∎

##  3 Combinatorial identities from clever telescoping

######  Lemma 2.

If $(a_{n})_{n\geq 1}$ is a sequence, then

$\sum_{k=1}^{n}H_{k}\left({a_{k+1}-a_{k}}\right)=H_{n}a_{n+1}-\sum_{k=1}^{n}\frac{a_{k}}{k}.$

###### Proof.

This is an immediate consequence of the recurrence relation of the harmonic
numbers. ∎

Lemma 2 is broadly applicable and allows to instantly state a myriad of
interesting results. We give three examples as a warmup. Setting
$a_{n}=H_{n-1}$ gives the known result

$\sum_{k=1}^{n}\frac{{H_{k-1}}}{k}=\frac{1}{2}\left({H_{n}^{2}-H_{n}^{(2)}}\right).$

The choice $a_{n}=n$ yields the well-known harmonic number summation formula

$\sum_{k=1}^{n}H_{k}=(n+1)H_{n}-n.$

Let $a_{n}=F_{n+1}$ be the sequence of (shifted) Fibonacci numbers. Then we
get the following Fibonacci-harmonic number identity:

$\sum_{k=1}^{n}H_{k}F_{k}=H_{n}F_{n+2}-\sum_{k=1}^{n}\frac{{F_{k+1}}}{k}.$

######  Theorem 6.

If $m$ and $n$ are non-negative integers, then

$\sum_{k=1}^{n}H_{k}\sum_{j=m}^{k}{\binom{{k-1}}{{j-1}}\frac{{s(j,m)}}{{j!}}}=\frac{1}{{m!}}H_{n}(m)H_{n}-\frac{1}{{m!}}\sum_{k=1}^{n}\frac{{H_{k-1}(m)}}{k}.$
(25)

###### Proof.

Use Lemma 2 with $a_{k}=H_{k-1}\left(m\right)$ while noting from (10) that

$H_{k}(m)-H_{k-1}(m)=m!\sum_{j=m}^{k}\binom{{k-1}}{{j-1}}\frac{s(j,m)}{j!}.$
(26)

∎

Setting $m=1$ in (25) gives the known result

$\sum_{k=1}^{n}\frac{{H_{k}}}{k}=\frac{1}{2}\left(H_{n}^{2}+H_{n}^{(2)}\right),$

while $m=2$ gives

$2\sum_{k=1}^{n}H_{k}\sum_{j=1}^{k}{(-1)^{j}\binom{{k-1}}{{j-1}}\frac{{H_{j-1}}}{j}}=H_{n}^{3}-H_{n}^{(2)}H_{n}-\sum_{k=1}^{n}\frac{{H_{k-1}^{2}-H_{k-1}^{(2)}}}{k}.$

######  Theorem 7.

If $m$ and $n$ are positive integers, then

$\sum_{k=1}^{n}H_{k}\sum_{j=m}^{n-k+1}\binom{{n-k}}{{j-1}}\frac{s(j,m)}{j!}=\frac{1}{m!}H_{n+1}(m+1).$
(27)

In particular,

$\sum_{k=1}^{n}\frac{{H_{k}}}{{n-k+1}}=H_{n+1}^{2}-H_{n+1}^{(2)}$ (28)

and

$\sum_{k=1}^{n}H_{k}\sum_{j=2}^{n-k+1}\binom{{n-k}}{{j-1}}\frac{{\left({-1}\right)^{j}}}{j}H_{j-1}=\frac{1}{2}\sum_{k=1}^{n+1}\frac{1}{k}\sum_{j=1}^{n+1-k}\frac{{H_{n-k-j+1}}}{j}.$
(29)

###### Proof.

Set $a_{k}=H_{n-k+1}\left(m\right)$ in Lemma 2 and use (4) and (26). ∎

######  Lemma 3.

If $(a_{n})_{n\geq 0}$ is a sequence, then

$\sum_{k=1}^{n}\frac{a_{k}-a_{k-1}}{k}=\sum_{k=1}^{n}\frac{{a_{k}}}{{k(k+1)}}-a_{0}+\frac{a_{n}}{n+1}.$
(30)

###### Proof.

This is a consequence of

$\frac{1}{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}.$

∎

Again, many results can be derived from Lemma 3. For instance, setting
$a_{k}=H_{k}+p$ ($p$ a non-negative integer) gives

$\sum_{k=1}^{n}\frac{{H_{k+p}}}{k(k+1)}=\begin{cases}H_{n}^{(2)}-\frac{H_{n}}{n+1},&p=0;\\\
\\\ \frac{H_{n}+H_{p}-H_{n+p}}{p}+H_{p}-\frac{H_{n+p}}{n+1},&p\geq
1;\end{cases}$ (31)

where we used the result

$\sum_{k=1}^{n}\frac{1}{k(k+p)}=\begin{cases}H_{n}^{(2)},&p=0;\\\ \\\
\frac{H_{n}+H_{p}-H_{n+p}}{p},&p\geq 1.\end{cases}$

The result (31) is most likely not new.

######  Theorem 8.

If $m$ is a non-negative integer and $n$ is a positive integer, then

$\sum_{k=1}^{n}{\frac{{H_{n-k}\left(m\right)}}{{k\left({k+1}\right)}}}=H_{n}\left(m\right)+H_{n}\left({m+1}\right)-H_{n+1}\left({m+1}\right).$
(32)

In particular,

$\sum_{k=1}^{n}{\frac{{H_{n-k}}}{{k\left({k+1}\right)}}}=H_{n}+H_{n}^{2}-H_{n}^{(2)}-H_{n+1}^{2}+H_{n+1}^{(2)}$
(33)

and

$\sum_{k=1}^{n}{\frac{{H_{n-k}^{2}-H_{n-k}^{(2)}}}{{k(k+1)}}}=H_{n}^{2}-H_{n}^{(2)}+\sum_{k=1}^{n}{\frac{1}{k}\sum_{j=1}^{n-k}{\frac{{H_{n-k-j}}}{j}}}-\sum_{k=1}^{n+1}{\frac{1}{k}\sum_{j=1}^{n+1-k}{\frac{{H_{n+1-k-j}}}{j}}}.$
(34)

###### Proof.

Use $a_{k}=H_{n-k}(m)$ in Lemma 3 and invoke Lemma 1. ∎

######  Lemma 4.

Let $(a_{n})_{n\geq 0}$ be a sequence. If $r$ is a complex number and $n$ is a
non-negative number, then

$\sum_{k=0}^{n}(-1)^{k}\binom{{r-1}}{k}(a_{k+1}-a_{k})=(-1)^{n}\binom{{r-1}}{n}a_{n+1}-\sum_{k=0}^{n}(-1)^{k}\binom{r}{k}a_{k}.$
(35)

###### Proof.

This is a variation on an identity of Kollár [14, Lemma 1]. ∎

######  Theorem 9.

If $r$ is a complex number and $m$ and $n$ are positive integers, then

$\begin{split}\sum_{k=0}^{n}{(-1)^{k}\binom{{r-1}}{k}\sum_{j=m}^{k+1}{\binom{{k}}{{j-1}}\frac{{s(j,m)}}{{j!}}}}&=(-1)^{n}\binom{{r-1}}{n}\frac{1}{{m!}}H_{n+1}\left(m\right)\\\
&\qquad-\frac{1}{{m!}}\sum_{k=0}^{n}{(-1)^{k}\binom{{r}}{k}H_{k}\left(m\right)}.\end{split}$
(36)

In particular,

$\sum_{k=0}^{n}{\frac{{\left({-1}\right)^{k}}}{{k+1}}\binom{{r-1}}{k}}=\left({-1}\right)^{n}\binom{{r-1}}{n}H_{n+1}-\sum_{k=0}^{n}{\left({-1}\right)^{k}\binom{{r}}{k}H_{k}}$
(37)

and

$\begin{split}\sum_{k=0}^{n}{\left({-1}\right)^{k}\binom{{r-1}}{k}\sum_{j=2}^{k+1}{\left({-1}\right)^{j}\binom{{k}}{{j-1}}\frac{{H_{j-1}}}{j}}}&=\left({-1}\right)^{n}\binom{{r-1}}{n}\frac{1}{2}\left({H_{n+1}^{2}-H_{n}^{(2)}}\right)\\\
&\qquad-\frac{1}{2}\sum_{k=0}^{n}{\left({-1}\right)^{k}\binom{{r}}{k}\left({H_{k}^{2}-H_{k}^{(2)}}\right)}.\end{split}$
(38)

###### Proof.

Use $a_{k}=H_{k}\left(m\right)$ in Lemma 4. ∎

##  4 More identities involving $H_{n}(m)$ and other sequences

Hyperharmonic numbers $H_{n,p}$ (or $h_{n}^{(p)}$), $p\geq 1,$ are another
generalization of harmonic numbers ([4, 10, 11]). They are defined by

$H_{n,p}=\sum_{i=1}^{n}H_{i,p-1}\qquad\mbox{with}\qquad
H_{n,0}=\frac{1}{n},\,H_{0,p}=0,\,H_{n,1}=H_{n}.$ (39)

They can be written in the compact form

$H_{n,p+1}=\binom{n+p}{n}(H_{n+p}-H_{p}),\qquad p=0,1,2,\ldots,$ (40)

######  Theorem 10.

For all $n,p,m\geq 0$ we have

$\sum_{k=0}^{n}\binom{k+p}{k}H_{n-k}(m)(H_{k+p}-H_{p})=\sum_{k=0}^{n}\binom{k+p}{k}H_{n-k}(m+1).$
(41)

In particular,

$\sum_{k=0}^{n}\binom{k+p}{k}(H_{k+p}-H_{p})=\sum_{k=0}^{n}\binom{k+p}{k}H_{n-k}$
(42)

and

$\sum_{k=0}^{n}\binom{k+p}{k}H_{n-k}(H_{k+p}-H_{p})=\sum_{k=0}^{n}\binom{k+p}{k}\left(H_{n-k}^{2}-H_{n-k}^{(2)}\right).$
(43)

###### Proof.

Let $B(z)$ be the generating function of $H_{n,p+1}$. We can calculate

$\displaystyle H(z)\cdot B(z)$
$\displaystyle=\left(\sum_{n=0}^{\infty}H_{n}(m)z^{n}\right)\left(\sum_{n=0}^{\infty}H_{n,p+1}z^{n}\right)$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=0}^{n}H_{k}(m)H_{n-k,p+1}z^{n}$
$\displaystyle=\frac{(-\ln(1-z))^{m}}{1-z}\frac{-\ln(1-z)}{(1-z)^{p+1}}$
$\displaystyle=\frac{1}{(1-z)^{p+1}}\frac{(-\ln(1-z))^{m+1}}{1-z}$
$\displaystyle=\left(\sum_{n=0}^{\infty}\binom{n+p}{n}z^{n}\right)\left(\sum_{n=0}^{\infty}H_{n}(m+1)z^{n}\right)$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{k+p}{k}H_{n-k}(m+1)z^{n}$

and the statement follows. ∎

The next theorem contains an identity involving $H_{n}(m)$ and odd harmonic
numbers $O_{n}$, the latter being defined by

$O_{n}=\sum_{k=1}^{n}\frac{1}{2k-1},\quad O_{0}=0.$

Obvious relations between harmonic numbers $H_{n}$ and odd harmonic numbers
$O_{n}$ are given by

$H_{2n}=\frac{1}{2}H_{n}+O_{n}\qquad\text{and}\qquad
H_{2n-1}=\frac{1}{2}H_{n-1}+O_{n}.$

Additional relations are contained in the next lemma.

######  Lemma 5.

If $n$ is an integer, then

$\displaystyle H_{n-1/2}-H_{-1/2}=2O_{n},$ (44) $\displaystyle
H_{n-1/2}-H_{1/2}=2\left({O_{n}-1}\right),$ (45) $\displaystyle
H_{n+1/2}-H_{-1/2}=2O_{n+1},$ (46) $\displaystyle
H_{n+1/2}-H_{1/2}=2\left({O_{n+1}-1}\right),$ (47) $\displaystyle
H_{n+1/2}-H_{n-1/2}=\frac{2}{{2n+1}},$ (48) $\displaystyle
H_{n-1/2}-H_{-3/2}=2\left({O_{n}-1}\right),$ (49) $\displaystyle
H_{n+1/2}-H_{-3/2}=2\left({O_{n+1}-1}\right).$ (50)

######  Lemma 6.

If $p$ and $r$ are non-negative integers, then

$H_{r,p+1/2}=\frac{1}{{2^{2r-1}}}\binom{{2p}}{p}^{-1}\binom{{2(r+p)}}{{r+p}}\binom{{r+p}}{r}\left({O_{r+p}-O_{p}}\right).$
(51)

###### Proof.

Identity (45) gives

$H_{r+p-1/2}-H_{r-1/2}=2\left({O_{r+p}-O_{p}}\right).$ (52)

Using this and

$\binom{{r+p-1/2}}{r}=\frac{1}{{2^{2r}}}\binom{{2p}}{p}^{-1}\binom{{2(r+p)}}{{r+p}}\binom{{r+p}}{r}$
(53)

in (40) gives (51). ∎

######  Theorem 11.

If $n$ and $p$ are non-negative integers, then

$\begin{split}&\sum_{k=1}^{n}{\frac{1}{{2^{2k}}}\binom{{2\left({k+p}\right)}}{{k+p}}\binom{{k+p}}{k}\left({O_{k+p}-O_{p}}\right)}\\\
&=\frac{1}{{2^{2n+1}}}\,\frac{{p+1}}{{2p+1}}\binom{{2\left({n+p+1}\right)}}{{n+p+1}}\binom{{n+p+1}}{n}\left({O_{n+p+1}-O_{p+1}}\right).\end{split}$
(54)

In particular,

$\sum_{k=1}^{n}{\frac{{O_{k}}}{{2^{2k}}}\binom{{2k}}{k}}=\frac{{n+1}}{{2^{2n+1}}}\binom{{2(n+1)}}{{n+1}}\left({O_{n+1}-1}\right).$
(55)

###### Proof.

Write $p+1+1/2$ for $p$ in (39) to obtain

$\sum_{k=1}^{n}{H_{k,p+1/2}}=H_{n,(p+1)+1/2},$

and use (51). ∎

######  Theorem 12.

For all $n,m\geq 0$ we have

$\sum_{k=0}^{n}\binom{2k}{k}\frac{O_{k}H_{n-k}(m)}{4^{k}}=\frac{1}{2}\sum_{k=0}^{n}\binom{2k}{k}\frac{H_{n-k}(m+1)}{4^{k}}.$
(56)

In particular,

$\sum_{k=0}^{n}\binom{2k}{k}\frac{O_{k}}{4^{k}}=\frac{1}{2}\sum_{k=0}^{n}\binom{2k}{k}\frac{H_{n-k}}{4^{k}}$
(57)

and

$\sum_{k=0}^{n}\binom{2k}{k}\frac{O_{k}H_{n-k}}{4^{k}}=\frac{1}{2}\sum_{k=0}^{n}\binom{2k}{k}\frac{1}{4^{k}}\left(H_{n-k}^{2}-H_{n-k}^{(2)}\right).$
(58)

###### Proof.

We use the fact that [5, 12]

$\sum_{n=0}^{\infty}\binom{2n}{n}z^{n}=\frac{1}{\sqrt{1-4z}}.$

It is also known that [5]

$O(z)=\sum_{n=0}^{\infty}\binom{2n}{n}O_{n}z^{n}=\frac{1}{2}\sqrt{1-4z}\left(\frac{-\ln(1-4z)}{1-4z}\right).$

This yields

$\displaystyle O(z)\cdot H(4z)$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{2k}{k}O_{k}4^{n-k}H_{n-k}(m)z^{n}$
$\displaystyle=\frac{1}{2\sqrt{1-4z}}\,\frac{(-\ln(1-4z))^{m+1}}{1-4z}$
$\displaystyle=\frac{1}{2}\left(\sum_{n=0}^{\infty}\binom{2n}{n}z^{n}\right)\left(\sum_{n=0}^{\infty}H_{n}(m+1)4^{n}z^{n}\right)$
$\displaystyle=\frac{1}{2}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{2k}{k}4^{n-k}H_{n-k}(m+1)z^{n}$

and the proof is completed. ∎

Note that from (55) and (57), we also have

$\sum_{k=0}^{n}\binom{2k}{k}\frac{H_{n-k}}{2^{2k}}=\frac{{n+1}}{{2^{2n}}}\binom{{2(n+1)}}{{n+1}}\left({O_{n+1}-1}\right).$
(59)

The next theorem, based on Theorem 10, generalizes Theorem 12.

######  Theorem 13.

If $n$ and $p$ are non-negative integers, then

$\begin{split}&\sum_{k=0}^{n}{\frac{1}{{2^{2k}}}\binom{{2(k+p)}}{{k+p}}\binom{{k+p}}{k}H_{n-k}\left(m\right)\left({O_{k+p}-O_{p}}\right)}\\\
&\qquad=\frac{1}{2}\sum_{k=0}^{n}{\frac{1}{{2^{2k}}}\binom{{2(k+p)}}{{k+p}}\binom{{k+p}}{k}H_{n-k}\left({m+1}\right)}.\end{split}$
(60)

In particular, setting $m=0$ and using (54) gives the following generalization
of (59):

$\begin{split}&\sum_{k=0}^{n}{\frac{1}{{2^{2k}}}\binom{{2(k+p)}}{{k+p}}\binom{{k+p}}{k}H_{n-k}}\\\
&\qquad=\frac{1}{{2^{2n}}}\,\frac{{p+1}}{{2p+1}}\binom{{2\left({n+p+1}\right)}}{{n+p+1}}\binom{{n+p+1}}{n}\left({O_{n+p+1}-O_{p+1}}\right).\end{split}$
(61)

###### Proof.

Write $p-1/2$ for $p$ in (41) and use (52) and (53). ∎

######  Lemma 7.

If $(a_{n})_{n\geq 0}$ is a sequence, then

$\sum_{k=1}^{n}k(a_{k}-a_{k-1})=na_{n}-\sum_{k=1}^{n}a_{k-1}.$

######  Theorem 14.

If $n\in\mathbb{N}_{0}$ and $p\in\mathbb{C}\setminus\mathbb{Z}^{-}$ then

$\sum_{k=1}^{n}{kH_{k,p}}=nH_{n,p+1}-H_{n-1,p+2}.$ (62)

###### Proof.

Identity (40) gives the following recurrence relation:

$H_{k,p+1}-H_{k-1,p+1}=H_{k,p}.$ (63)

Use $a_{k}=H_{k,p+1}$ in Lemma 7, keeping (63) in mind. ∎

######  Theorem 15.

If $n$ and $p$ are non-negative integers, then

$\begin{split}&\sum_{k=1}^{n}{\frac{k}{{2^{2k}}}\binom{{2(k+p)}}{{(k+p)}}\binom{{k+p}}{k}\left({O_{k+p}-O_{p}}\right)}\\\
&\qquad=\frac{n}{{2^{2n}}}\binom{{2(p+1)}}{{p+1}}^{-1}\binom{{2p}}{p}\binom{{2(n+p+1)}}{{n+p+1}}\binom{{n+p+1}}{n}\left({O_{n+p+1}-O_{p+1}}\right)\\\
&\qquad\qquad-\frac{1}{{2^{2n-2}}}\binom{{2(p+2)}}{{p+2}}^{-1}\binom{{2p}}{p}\binom{{2(n+p+1)}}{{n+p+1}}\binom{{n+p+1}}{{n-1}}\left({O_{n+p+1}-O_{p+2}}\right).\end{split}$
(64)

In particular,

$\begin{split}\sum_{k=1}^{n}{\frac{k}{{2^{2k}}}\binom{{2k}}{k}O_{k}}&=\frac{{n(n+1)}}{{2^{2n+1}}}\binom{{2(n+1)}}{{n+1}}\left({O_{n+1}-1}\right)\\\
&\qquad-\frac{{n(n+1)}}{{2^{2n}3}}\binom{{2(n+1)}}{{n+1}}\left({O_{n+1}-\frac{4}{3}}\right).\end{split}$
(65)

###### Proof.

Write $p+1/2$ for $p$ in (62) and use (51). ∎

######  Theorem 16.

If $n$ and $p$ are non-negative integers, then

$\begin{split}&\sum_{k=1}^{n}{\binom{{2k}}{k}^{-1}\binom{{2(k+p)}}{{k+p}}\binom{{k+p}}{k}\left({O_{k+p}-O_{k}}\right)}\\\
&\qquad=\frac{1}{4}\binom{{2n}}{n}^{-1}\binom{{2\left({n+p+1}\right)}}{{n+p+1}}\binom{n+p+1}{n}\left({O_{n+p+1}-O_{n}}\right)-\frac{1}{4}\binom{{2\left({p+1}\right)}}{{p+1}}O_{p+1}.\end{split}$
(66)

###### Proof.

Rearrange (63), interchange $k$ and $p$ and write $k+1/2$ for $k$ to obtain

$H_{p,k+1/2}=H_{p+1,k+1/2}-H_{p+1,k-1/2}$

which telescopes to give

$\sum_{k=1}^{n}{H_{p,k+1/2}}=H_{p+1,n+1/2}-H_{p+1,1/2},$

from which (66) follows, in view of (51). ∎

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