A244973 -id:A244973 - cp's OEIS frontend

A245088 a(n) = (Sum_{k=0..n-1} (8k^2+12k+5)*A244973(k))/n^2.

5, -5, -9, 115, -397, 85, 6625, -36181, 63915, 377365, -3357175, 10579245, 12408269, -295386005, 1383403247, -1317964405, -23051599205, 159765455077, -392146366775, -1358686166755, 16622987639325, -64283150090725, -10522422920465, 1532928077704325, -8583492124492507

Offset: 1

Zhi-Wei Sun, Jul 11 2014

sign

Conjecture: All the terms are odd integers. Moreover, a(p) == 3 (mod p) for any prime p.

In 2015 Guo, Mao and Pan confirmed the conjecture partially by proving that n*a(n) is always an integer. - Zhi-Wei Sun, Nov 14 2015

			a(2) = -5 since (5*A244973(0)+(8+12+5)*A244973(1))/2^2 = (5*1+25*(-1))/4 = -5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..200
V. J. W. Guo, S.-M. Guo and H. Pan, Proof of a conjecture involving Sun polynomials, arXiv preprint arXiv:1511.04005 [math.NT], 2015.
Z.-W. Sun, Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k, preprint, arXiv:1407.0967 [math.NT], 2014-2015.
Z.-W. Sun, Congruences involving g_n(x)=sum_{k=0..n} binomial(n,k)^2*binomial(2k,k)*x^k, Ramanujan J., in press. Doi: 10.1007/s11139-015-9727-3.

Cf. A244973.

s[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k](-1)^k,{k,0,n}]
S[n_]:=Sum[(8k^2+12k+5)*s[k],{k,0,n-1}]/n^2
Table[S[n],{n,1,25}]

Conjecture: +n^2*(23656*n-2233)*(n-1)*a(n) +(n-1)*(23656*n^3+655645*n^2-2273556*n+1915600)*a(n-1) -2*(n-2) *(177048*n^3-3748011*n^2+13870329*n-15407140)*a(n-2) -2*(n-3) *(2020728*n^3-18138829*n^2+63885382*n-73802216)*a(n-3) -5*(n-4) *(782552*n^3-11993599*n^2+55520403*n-80051816)*a(n-4) +125*(66088*n-203507)*(n-5)*(n-4)^2*a(n-5)=0. - R. J. Mathar, Jul 12 2014
Recurrence (of order 3): (n-1)*n^2*(8*n^2 - 36*n + 41)*(8*n^2 - 20*n + 13)*a(n) =  -(n-1)*(8*n^2 - 36*n + 41)*(40*n^4 - 148*n^3 + 161*n^2 - 42*n + 4)*a(n-1) - (n-2)*(8*n^2 - 4*n + 1)*(152*n^4 - 1188*n^3 + 3427*n^2 - 4293*n + 1960)*a(n-2) + 25*(n-3)*(n-2)^2*(8*n^2 - 20*n + 13)*(8*n^2 - 4*n + 1)*a(n-3). - Vaclav Kotesovec, Jul 13 2014
Lim sup n->infinity |a(n)|^(1/n) = 5. - Vaclav Kotesovec, Jul 13 2014

A002893 a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(2*k,k).

Original entry on oeis.org

1, 3, 15, 93, 639, 4653, 35169, 272835, 2157759, 17319837, 140668065, 1153462995, 9533639025, 79326566595, 663835030335, 5582724468093, 47152425626559, 399769750195965, 3400775573443089, 29016970072920387, 248256043372999089

Offset: 0

N. J. A. Sloane

This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
a(n) is the 2n-th moment of the distance from the origin of a 3-step random walk in the plane. - Peter M. W. Gill (peter.gill(AT)nott.ac.uk), Feb 27 2004
a(n) is the number of Abelian squares of length 2n over a 3-letter alphabet. - Jeffrey Shallit, Aug 17 2010
Consider 2D simple random walk on honeycomb lattice. a(n) gives number of paths of length 2n ending at origin. - Sergey Perepechko, Feb 16 2011
Row sums of A318397 the square of A008459. - Peter Bala, Mar 05 2013
Conjecture: For each n=1,2,3,... the polynomial g_n(x) = Sum_{k=0..n} binomial(n,k)^2*binomial(2k,k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
a(n) is the sum of the squares of the coefficients of (x + y + z)^n. - Michael Somos, Aug 25 2018
a(n) is the constant term in the expansion of (1 + (1 + x) * (1 + y) + (1 + 1/x) * (1 + 1/y))^n. - Seiichi Manyama, Oct 28 2019

			G.f.: A(x) = 1 + 3*x + 15*x^2 + 93*x^3 + 639*x^4 + 4653*x^5 + 35169*x^6 + ...
G.f.: A(x) = 1/(1-3*x) + 6*x^2*(1-x)/(1-3*x)^4 + 90*x^4*(1-x)^2/(1-3*x)^7 + 1680*x^6*(1-x)^3/(1-3*x)^10 + 34650*x^8*(1-x)^4/(1-3*x)^13 + ... - _Paul D. Hanna_, Feb 26 2012

Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 0..1051 (terms 0..100 from T. D. Noe)
B. Adamczewski, J. P. Bell, and E. Delaygue, Algebraic independence of G-functions and congruences à la Lucas", arXiv preprint arXiv:1603.04187 [math.NT], 2016.
Rostislav Akhmechet, Mikhail Khovanov, and Melissa Zhang, Annular SL(2) and SL(3) web algebras, arXiv:2508.05605 [math.GT], 2025. See p. 48.
Gert Almkvist, The art of finding Calabi-Yau differential equations, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.
David H. Bailey, Jonathan M. Borwein, David Broadhurst and M. L. Glasser, Elliptic integral evaluations of Bessel moments, arXiv:0801.0891 [hep-th], 2008.
P. Barrucand, A combinatorial identity, Problem 75-4, SIAM Rev., 17 (1975), 168. Solution by D. R. Breach, D. McCarthy, D. Monk and P. E. O'Neil, SIAM Rev. 18 (1976), 303.
P. Barrucand, Problem 75-4, A Combinatorial Identity, SIAM Rev., 17 (1975), 168. [Annotated scanned copy of statement of problem]
Arnaud Beauville, Les familles stables de courbes sur P_1 admettant quatre fibres singulières, Comptes Rendus, Académie Sciences Paris, no. 294, May 24 1982.
Frits Beukers and Jan Stienstra, On the Picard-Fuchs Equation and the Formal Brauer Group of Certain Elliptic K3-Surfaces, Mathematische Annalen (1985), Vol. 271, pp. 269-304 (see Part III).
Artur Bille, Victor Buchstaber, Simon Coste, Satoshi Kuriki, and Evgeny Spodarev, Random eigenvalues of graphenes and the triangulation of plane, arXiv:2306.01462 [math.SP], 2023.
Jonathan M. Borwein, A short walk can be beautiful, 2015.
Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Random Walk Integrals, 2010.
Jonathan M. Borwein and Armin Straub, Mahler measures, short walks and log-sine integrals.
Jonathan M. Borwein, Armin Straub and Christophe Vignat, Densities of short uniform random walks, Part II: Higher dimensions, Preprint, 2015
Jonathan M. Borwein, Armin Straub and James Wan, Three-Step and Four-Step Random Walk Integrals, Exper. Math., 22 (2013), 1-14.
Charles Burnette and Chung Wong, Abelian Squares and Their Progenies, arXiv:1609.05580 [math.CO], 2016.
David Callan, A combinatorial interpretation for an identity of Barrucand, JIS 11 (2008) 08.3.4.
Shaun Cooper, Apéry-like sequences defined by four-term recurrence relations, arXiv:2302.00757 [math.NT], 2023.
M. Coster, Email, Nov 1990
Eric Delaygue, Arithmetic properties of Apery-like numbers, arXiv preprint arXiv:1310.4131 [math.NT], 2013.
C. Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9 (1960), 149-361.
Jeffrey S. Geronimo, Hugo J. Woerdeman, and Chung Y. Wong, The autoregressive filter problem for multivariable degree one symmetric polynomials, arXiv:2101.00525 [math.CA], 2021.
Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See C p. 2.
Victor J. W. Guo, Proof of two conjectures of Z.-W. Sun on congruences for Franel numbers, arXiv preprint arXiv:1201.0617 [math.NT], 2012.
Victor J. W. Guo, Guo-Shuai Mao and Hao Pan, Proof of a conjecture involving Sun polynomials, arXiv preprint arXiv:1511.04005 [math.NT], 2015.
E. Hallouin and M. Perret, A Graph Aided Strategy to Produce Good Recursive Towers over Finite Fields, arXiv preprint arXiv:1503.06591 [math.NT], 2015.
J. A. Hendrickson, Jr., On the enumeration of rectangular (0,1)-matrices, Journal of Statistical Computation and Simulation, 51 (1995), 291-313.
S. Herfurtner, Elliptic surfaces with four singular fibres, Mathematische Annalen, 1991. Preprint.
Pakawut Jiradilok and Elchanan Mossel, Gaussian Broadcast on Grids, arXiv:2402.11990 [cs.IT], 2024. See p. 27.
Davidson Noby Joseph and Igor Boettcher, Walking on Archimedean Lattices: Insights from Bloch Band Theory, arXiv:2507.12662 [cond-mat.stat-mech], 2025. See p. 18.
Tanya Khovanova and Konstantin Knop, Coins of three different weights, arXiv:1409.0250 [math.HO], 2014.
Murray S. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 148-149.
Bradley Klee, Checking Weierstrass data, 2023.
Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, 2016, 2:5
Mathematics Stack Exchange, sum involving the product of binomial coefficients, Nov 10 2016.
L. B. Richmond and Jeffrey Shallit, Counting abelian squares, Electronic J. Combinatorics 16 (1), #R72, June 2009. [From _Jeffrey Shallit_, Aug 17 2010]
Armin Straub, Arithmetic aspects of random walks and methods in definite integration, Ph. D. Dissertation, School Of Science And Engineering, Tulane University, 2012. - From _N. J. A. Sloane_, Dec 16 2012
Zhi-Hong Sun, Congruences for Apéry-like numbers, arXiv:1803.10051 [math.NT], 2018.
Zhi-Hong Sun, New congruences involving Apéry-like numbers, arXiv:2004.07172 [math.NT], 2020.
Zhi-Wei Sun, Connections between p = x^2+3y^2 and Franel numbers, J. Number Theory 133(2013), 2919-2928.
Zhi-Wei Sun, Congruences involving g_n(x) = Sum_{k=0..n} binomial(n,k)^2*binomial(2k,k)*x^k, Ramanujan J., in press. Doi: 10.1007/s11139-015-9727-3.
Brani Vidakovic, All roads lead to Rome--even in the honeycomb world, Amer. Statist., 48 (1994) no. 3, 234-236.
Yi Wang and Bao-Xuan Zhu, Proofs of some conjectures on monotonicity of number-theoretic and combinatorial sequences, arXiv preprint arXiv:1303.5595 [math.CO], 2013.
D. Zagier, Integral solutions of Apery-like recurrence equations. See line C in sporadic solutions table of page 5.

Cf. A000172, A002895, A000984, A006480, A087457, A274600, A318397, A244973.
Cf. A169714, A169715.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

[&+[Binomial(n, k)^2 * Binomial(2*k, k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 26 2018

series(1/GaussAGM(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3)), x=0, 42) # Gheorghe Coserea, Aug 17 2016
A002893 := n -> hypergeom([1/2, -n, -n], [1, 1], 4):
seq(simplify(A002893(n)), n=0..20); # Peter Luschny, May 23 2017

Table[Sum[Binomial[n,k]^2 Binomial[2k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Aug 19 2011 *)
a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {1/2, -n, -n}, {1, 1}, 4]]; (* Michael Somos, Oct 16 2013 *)
a[n_] := SeriesCoefficient[BesselI[0, 2*Sqrt[x]]^3, {x, 0, n}]*n!^2; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Dec 30 2013 *)
a[ n_] := If[ n < 0, 0, Block[ {x, y, z},  Expand[(x + y + z)^n] /. {t_Integer -> t^2, x -> 1, y -> 1, z -> 1}]]; (* Michael Somos, Aug 25 2018 *)

{a(n) = if( n<0, 0, n!^2 * polcoeff( besseli(0, 2*x + O(x^(2*n+1)))^3, 2*n))};

{a(n) = sum(k=0, n, binomial(n, k)^2 * binomial(2*k, k))}; /* Michael Somos, Jul 25 2007 */

{a(n)=polcoeff(sum(m=0,n, (3*m)!/m!^3 * x^(2*m)*(1-x)^m / (1-3*x+x*O(x^n))^(3*m+1)),n)} \\ Paul D. Hanna, Feb 26 2012

N = 42; x='x + O('x^N); v = Vec(1/agm(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3))); vector((#v+1)\2, k, v[2*k-1])  \\ Gheorghe Coserea, Aug 17 2016

def A002893(n): return simplify(hypergeometric([1/2,-n,-n], [1,1], 4))
[A002893(n) for n in range(31)] # G. C. Greubel, Jan 21 2023

a(n) = Sum_{m=0..n} binomial(n, m) * A000172(m). [Barrucand]
D-finite with recurrence: (n+1)^2 a(n+1) = (10*n^2+10*n+3) * a(n) - 9*n^2 * a(n-1). - Matthijs Coster, Apr 28 2004
Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0, 2*sqrt(x))^3. - Vladeta Jovovic, Mar 11 2003
a(n) = Sum_{p+q+r=n} (n!/(p!*q!*r!))^2 with p, q, r >= 0. - Michael Somos, Jul 25 2007
a(n) = 3*A087457(n) for n>0. - Philippe Deléham, Sep 14 2008
a(n) = hypergeom([1/2, -n, -n], [1, 1], 4). - Mark van Hoeij, Jun 02 2010
G.f.: 2*sqrt(2)/Pi/sqrt(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z))) *  EllipticK(8*z^(3/2)/(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z)))). - Sergey Perepechko, Feb 16 2011
G.f.: Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)*(1-x)^n / (1-3*x)^(3*n+1). - Paul D. Hanna, Feb 26 2012
Asymptotic: a(n) ~ 3^(2*n+3/2)/(4*Pi*n). - Vaclav Kotesovec, Sep 11 2012
G.f.: 1/(1-3*x)*(1-6*x^2*(1-x)/(Q(0)+6*x^2*(1-x))), where Q(k) = (54*x^3 - 54*x^2 + 9*x -1)*k^2 + (81*x^3 - 81*x^2 + 18*x -2)*k + 33*x^3 - 33*x^2 +9*x - 1 - 3*x^2*(1-x)*(1-3*x)^3*(k+1)^2*(3*k+4)*(3*k+5)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: G(0)/(2*(1-9*x)^(2/3)), where G(k) = 1 + 1/(1 - 3*(3*k+1)^2*x*(1-x)^2/(3*(3*k+1)^2*x*(1-x)^2 - (k+1)^2*(1-9*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 31 2013
a(n) = [x^(2n)] 1/agm(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3)). - Gheorghe Coserea, Aug 17 2016
0 = +a(n)*(+a(n+1)*(+729*a(n+2) -1539*a(n+3) +243*a(n+4)) +a(n+2)*(-567*a(n+2) +1665*a(n+3) -297*a(n+4)) +a(n+3)*(-117*a(n+3) +27*a(n+4))) +a(n+1)*(+a(n+1)*(-324*a(n+2) +720*a(n+3) -117*a(n+4)) +a(n+2)*(+315*a(n+2) -1000*a(n+3) +185*a(n+4)) +a(n+3)*(+80*a(n+3) -19*a(n+4))) +a(n+2)*(+a(n+2)*(-9*a(n+2) +35*a(n+3) -7*a(n+4)) +a(n+3)*(-4*a(n+3) +a(n+4))) for all n in Z. - Michael Somos, Oct 30 2017
G.f. y=A(x) satisfies: 0 = x*(x - 1)*(9*x - 1)*y'' + (27*x^2 - 20*x + 1)*y' + 3*(3*x - 1)*y. - Gheorghe Coserea, Jul 01 2018
Sum_{k>=0} binomial(2*k,k) * a(k) / 6^(2*k) = A086231 = (sqrt(3)-1) * (Gamma(1/24) * Gamma(11/24))^2 / (32*Pi^3). - Vaclav Kotesovec, Apr 23 2023
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=3*(-1 + 3*x)*T(x) + (1 - 20*x + 27*x^2)*T'(x) + x*(-1 + x)*(-1 + 9*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = (3/64)*(1 + 3*x)*(1 - 15*x + 75*x^2 + 3*x^3);
g3 = -(1/512)*(-1 + 6*x + 3*x^2)*(1 - 12*x + 30*x^2 - 540*x^3 + 9*x^4);
which determine an elliptic surface with four singular fibers. (End)
a(n) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(3*k, 2*n) (Almkvist, p. 16). - Peter Bala, May 22 2025

A275027 a(n) = Sum_{k=0..n} C(n,k)^2C(n-k,k), where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

Original entry on oeis.org

1, 1, 5, 19, 85, 401, 1931, 9605, 48469, 248365, 1286605, 6726875, 35441275, 187935775, 1002122525, 5369287019, 28889315669, 156015203845, 845330354321, 4593724615175, 25029614166685, 136704935601785, 748273234994675, 4103928115592365, 22549175326327675, 124105065258631651, 684100888645922051, 3776354280849020005

Offset: 0

Zhi-Wei Sun, Nov 12 2016

nonn

Conjecture: For any prime p > 5 and positive integer n, the number (a(p*n)-a(n))/(p*n)^3 is always a p-adic integer.
The author has proved that for any prime p > 5 and positive integer n the number (a(p*n)-a(n))/(p^3*n^2) is always a p-adic integer.
As a(n) = Sum_{k=0..n} C(n,k)*C(n,2k)*C(2k,k) and C(2k,k) = 2*C(2k-1,k-1) for k = 1,2,3,..., we see that a(n) is always odd. We guess that a(n) is congruent to one of 0, 1, -1 modulo 5.
Diagonal of the rational function 1 / ((1 - x)*(1 - y)*(1 - z) - x^2*y*z). - Ilya Gutkovskiy, Apr 23 2025

			a(2) = 5 since a(2) = Sum_{k=0,1,2}C(2,k)^2*C(2-k,k) = C(2,0)^2*C(2,0) + C(2,1)^2*C(1,1) = 1 + 4 = 5.

Zhi-Wei Sun, Table of n, a(n) for n = 0..200
Joel A. Henningsen and Armin Straub, Generalized Lucas congruences and linear p-schemes, arXiv:2111.08641 [math.NT], 2021.
Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Cf. A000984, A005258, A208425, A244973, A277640.

a[n_]:=a[n]=Sum[Binomial[n,k]^2*Binomial[n-k,k],{k,0,n/2}]
Table[a[n],{n,0,27}]
a[n_] := HypergeometricPFQ[{-n, 1/2 - n/2, -n/2}, {1, 1}, -4];
Table[a[n], {n, 0, 27}] (* Peter Luschny, Mar 21 2018 *)

a(n) = sum(k=0, n, binomial(n,k)^2*binomial(n-k,k)); \\ Michel Marcus, Nov 13 2016

a(n) = Sum_{k=0..n}C(n,k)*C(n,2k)*C(2k,k).
By the Zeilberger algorithm, we have the recurrence (n+3)^2*(23n+25)*a(n+3) = 25*(n+1)^2*(23n+48)*a(n) + (391n^3+1989n^2+3288n+1750)*a(n+1) + (46n^3+280n^2+ 519n+265)*a(n+2) for all n >= 0.
a(n) = hypergeom([-n, 1/2 - n/2, -n/2], [1, 1], -4). - Peter Luschny, Mar 21 2018
a(n) ~ c * d^n / (Pi*n), where d = 5.729031537980930837932235459792820714... is the real root of the equation -25 - 17*d - 2*d^2 + d^3 = 0 and c = 1.107089291883984657933126801836156175486638498732... is the positive real root of the equation -125 + 1048*c^2 - 2576*c^4 + 1472*c^6 = 0. - Vaclav Kotesovec, Jun 09 2019
G.f.: hypergeom([1/12, 5/12],[1],-1728*(25*x^3+17*x^2+2*x-1)*x^7/(1-4*x-10*x^2+4*x^3+25*x^4)^3)/(1-4*x-10*x^2+4*x^3+25*x^4)^(1/4). - Mark van Hoeij, Nov 28 2024

A208425 Expansion of Sum_{n>=0} (3n)!/n!^3 x^(2n)/(1-x)^(3n+1).

Original entry on oeis.org

1, 1, 7, 25, 151, 751, 4411, 24697, 146455, 862351, 5195257, 31392967, 191815339, 1177508515, 7276161907, 45154764025, 281492498455, 1761076827895, 11055132835705, 69600761349175, 439370198255401, 2780265190892641, 17631718101804517, 112038660509078695

Offset: 0

Paul D. Hanna, Feb 26 2012

nonn

Compare g.f. to: Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)/(1-2*x)^(3*n+1), which is a g.f. of the Franel numbers (A000172).
From Zhi-Wei Sun, Nov 12 2016: (Start)
Conjecture: (i) For any prime p > 3 and positive integer n, the number (a(p*n)-a(n))/(p*n)^3 is always a p-adic integer.
(ii) For any prime p == 1 (mod 3), we have Sum_{k=0..p-1}a(k) == C(2(p-1)/3,(p-1)/3) (mod p^2). For any prime p == 2 (mod 3), we have Sum_{k=0..p-1}a(k) == 2p/C(2(p+1)/3,(p+1)/3) (mod p^2).
We have proved part (i) of this conjecture for n = 1. (End)
Diagonal of rational functions 1/(1 - x*y - y*z - x*z - x*y*z), 1/(1 - x*y + y*z + x*z - x*y*z). - Gheorghe Coserea, Jul 03 2018
Number of paths from (0,0,0) to (n,n,n) using steps (1,1,0), (1,0,1), (0,1,1), and (1,1,1). - William J. Wang, Dec 07 2020
Diagonal of the rational function 1/(1 - (x^2 + y^2 + z^2 + x*y*z)). - Seiichi Manyama, Jul 04 2025

			G.f.: A(x) = 1 + x + 7*x^2 + 25*x^3 + 151*x^4 + 751*x^5 + 4411*x^6 +...
where
A(x) = 1/(1-x) + 6*x^2/(1-x)^4 + 90*x^4/(1-x)^7 + 1680*x^6/(1-x)^10 + 34650*x^8/(1-x)^13 + 756756*x^10/(1-x)^16 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
A. Bostan, S. Boukraa, J.-M. Maillard and J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, arXiv preprint arXiv:1507.03227 [math-ph], 2015.
Hao Pan and Zhi-Wei Sun, Supercongruences for central trinomial coefficients, arXiv:2012.05121 [math.NT], 2020.
Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Cf. A000172, A001850, A208426, A244973, A274783.

Cf. A081798, A344560.

series(hypergeom([1/3, 2/3], [1], 27*x^2/(1 - x)^3)/(1 - x), x=0, 25): seq(coeff(%, x, n), n=0..23);  # Mark van Hoeij, May 20 2013
a := n -> hypergeom([1/2 - n/2, -n/2, n + 1], [1, 1], 4); seq(simplify(a(n)), n=0..23);  # Peter Luschny, Jan 11 2025

nmax = 20; CoefficientList[Series[Sum[(3*n)!/n!^3 * x^(2*n)/(1-x)^(3*n+1), {n, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 05 2016 *)

{a(n)=polcoeff(sum(m=0,n, (3*m)!/m!^3*x^(2*m)/(1-x+x*O(x^n))^(3*m+1)),n)}
for(n=0,25,print1(a(n),", "))

Conjecture: n^2*(3*n-5)*a(n) +(-9*n^3+24*n^2-17*n+4) *a(n-1) -(3*n-4) *(24*n^2-56*n+27)*a(n-2) -(3*n-2)*(n-2)^2*a(n-3)=0. - R. J. Mathar, Mar 10 2016
a(n) ~ sqrt(1/2 + sqrt(13)*cos(arctan(53*sqrt(3)/19)/3)/6) * (1 + 6*cos(Pi/9))^n / (Pi*n). - Vaclav Kotesovec, Jul 05 2016
It is easy to show that a(n) = Sum_{k=0..n}C(n,k)*C(n-k,k)*C(n+k,k) = Sum_{k=0..n}C(n+k,k)*C(n,2k)*C(2k,k). By this formula and the Zeilberger algorithm, we confirm the recurrence conjectured by R. J. Mathar. - Zhi-Wei Sun, Nov 12 2016
G.f. y=A(x) satisfies: 0 = x*(x + 2)*(x^3 + 24*x^2 + 3*x - 1)*y'' + (3*x^4 + 56*x^3 + 147*x^2 + 12*x - 2)*y' + (x^3 + 9*x^2 + 42*x + 2)*y. - Gheorghe Coserea, Jul 03 2018
a(n) = hypergeom([1/2 - n/2, -n/2, n + 1], [1, 1], 4). - Peter Luschny, Jan 11 2025

A278405 a(n) = Sum_{k=0..n} binomial(n,2k)^2*binomial(n-k,k).

Original entry on oeis.org

1, 1, 2, 19, 110, 476, 2477, 15093, 86830, 485290, 2826902, 16857116, 100034453, 594833357, 3574477090, 21611465819, 130955824174, 796195223398, 4860425688176, 29760574848750, 182655048136510, 1123720751229858, 6929124085148938, 42811398244528788

Offset: 0

Zhi-Wei Sun, Nov 20 2016

nonn

Conjecture: For any prime p > 5 and positive integer n, the number (a(p*n)-a(n))/(p*n)^3 is always a p-adic integer.
We have proved that for any prime p > 5 and positive integer n the number (a(p*n)-a(n))/(p^3*n^2) is always a p-adic integer.
Diagonal of the rational function 1 / ((1 + x)*(1 - x)*(1 - y)*(1 - z) - x*y*z). - Ilya Gutkovskiy, Apr 23 2025

			a(3) = 19 since a(3) = C(3,2*0)^2*C(3-0,0) + C(3,2*1)^2*C(3-1,1) = 1 + 3^2*2 = 19.
G.f. = 1 + x + 2*x^2 + 19*x^3 + 110*x^4 + 476*x^5 + 2477*x^6 + 15093*x^7 + ...

Zhi-Wei Sun, Table of n, a(n) for n = 0..200
Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Cf. A208425, A244973, A275027, A277640, A278415.

a[n_]:=a[n]=Sum[Binomial[n,2k]^2*Binomial[n-k,k],{k,0,n/2}]
Table[a[n],{n,0,27}]

A278415 a(n) = Sum_{k=0..n} binomial(n, 2k)binomial(n-k, k)(-1)^k.

Original entry on oeis.org

1, 1, 0, -5, -16, -24, 15, 197, 576, 724, -1200, -8832, -22801, -21293, 76440, 408795, 922368, 499104, -4446588, -19025060, -37012416, -1673992, 245604832, 880263936, 1441226991, -908700649, -13088509200, -40222012703, -52991533744, 88167061704, 678172355415, 1805175708261, 1747974632448, -6237554623536, -34300087628480

Offset: 0

Zhi-Wei Sun, Nov 21 2016

sign

Conjecture: For any prime p > 3 and positive integer n, the number (a(p*n)-a(n))/(p*n)^2 is always a p-adic integer.
We are able to show that for any prime p > 3 and positive integer n the number (a(p*n)-a(n))/(p^2*n) is always a p-adic integer.
See also A275027 and A278405 for similar conjectures.

			a(3) = -5 since a(3) = C(3, 2*0)*C(3-0, 0)(-1)^0 + C(3,2*1)*C(3-1,1)(-1)^1 = 1 - 6 = -5.

Zhi-Wei Sun, Table of n, a(n) for n = 0..400
Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Cf. A208425, A244973, A277640, A275027, A278405.

a[n_]:=Sum[Binomial[n,2k]Binomial[n-k,k](-1)^k,{k,0,n}]
Table[a[n],{n,0,34}]

A336634 Sum_{n>=0} a(n) * x^n / (n!)^2 = exp(-x) * BesselI(0,2*sqrt(x))^2.

Original entry on oeis.org

1, 1, 0, -4, 14, -18, -168, 1920, -11898, 27398, 582896, -13028904, 183020620, -2061910004, 17930433744, -65293856160, -1965585556410, 69343044999750, -1519055329884960, 26755366818127560, -374375460816570780, 2924763867241325220

Offset: 0

Ilya Gutkovskiy, Jul 28 2020

sign

Robert Israel, Table of n, a(n) for n = 0..450

Cf. A000984, A009940, A244973, A336293.

rec:= n*a(n) = -(3*n^2 - 7*n + 3)*a(n - 1) + (7 - 3*n)*(n - 1)^2*a(n - 2) - (n - 1)^2*(n - 2)^2*a(n - 3):
f:= gfun:-rectoproc({rec,a(0)=1,a(1)=1,a(2)=0},a(n),remember):
map(f, [$0..30]); # Robert Israel, Jul 30 2020

nmax = 21; CoefficientList[Series[Exp[-x] BesselI[0, 2 Sqrt[x]]^2, {x, 0, nmax}], x] Range[0, nmax]!^2
Table[(-1)^n n! HypergeometricPFQ[{1/2, -n}, {1, 1}, 4], {n, 0, 21}]
Table[Sum[(-1)^(n - k) Binomial[n, k]^2 Binomial[2 k, k] (n - k)!, {k, 0, n}], {n, 0, 21}]

a(n) = sum(k=0, n, (-1)^(n-k) * binomial(n,k)^2 * binomial(2*k,k) * (n-k)!); \\ Michel Marcus, Jul 30 2020

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k)^2 * binomial(2*k,k) * (n-k)!.

D-finite with recurrence: n*a(n) = -(3*n^2 - 7*n + 3)*a(n - 1) + (7 - 3*n)*(n - 1)^2*a(n - 2) - (n - 1)^2*(n - 2)^2*a(n - 3). - Robert Israel, Jul 30 2020

cp's OEIS Frontend

A245088 a(n) = (Sum_{k=0..n-1} (8*k^2+12*k+5)*A244973(k))/n^2.

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A002893 a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(2*k,k).

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A275027 a(n) = Sum_{k=0..n} C(n,k)^2*C(n-k,k), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

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A208425 Expansion of Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)/(1-x)^(3*n+1).

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A278405 a(n) = Sum_{k=0..n} binomial(n,2k)^2*binomial(n-k,k).

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A278415 a(n) = Sum_{k=0..n} binomial(n, 2k)*binomial(n-k, k)*(-1)^k.

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A336634 Sum_{n>=0} a(n) * x^n / (n!)^2 = exp(-x) * BesselI(0,2*sqrt(x))^2.

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A245088 a(n) = (Sum_{k=0..n-1} (8k^2+12k+5)*A244973(k))/n^2.

A275027 a(n) = Sum_{k=0..n} C(n,k)^2C(n-k,k), where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

A208425 Expansion of Sum_{n>=0} (3n)!/n!^3 x^(2n)/(1-x)^(3n+1).

A278415 a(n) = Sum_{k=0..n} binomial(n, 2k)binomial(n-k, k)(-1)^k.