A260667 a(n) = (1/n^2) * Sum_{k=0..n-1} (2k+1)*S(k,n)^2, where S(k,x) denotes the polynomial Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).

1, 37, 1737, 102501, 6979833, 523680739, 42129659113, 3572184623653, 315561396741609, 28807571694394593, 2701627814373536601, 259121323945378645947, 25330657454041707496017, 2516984276442279642274311, 253667099464270541534450025, 25884030861250181046253181349, 2670255662315910532447096232073

Offset: 1

Author

Zhi-Wei Sun, Nov 14 2015

Keywords

• nonn

Comments

Conjecture: For k = 0,1,2,... define S(k,x):= Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).

(i) For any integer n > 0, the polynomial (1/n^2) * Sum_{k=0..n-1}(2k+1)*S(k,x)^2 is integer-valued (and hence a(n) is always integral).

(ii) Let r be 0 or 1, and let x be any integer. Then, for any positive integers m and n, we have the congruence

Sum_{k=0..n-1} (-1)^(k*r)*(2k+1)*S(k,x)^(2m) == 0 (mod n).

(iii) For any odd prime p, we have Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p)(1-7*p^3*B_{p-3}) (mod p^4), where (a/p) is the Legendre symbol, and B_0,B_1,B_2,... are Bernoulli numbers. Also, for any prime p > 3 we have Sum_{k=0..p-1} S(k,-1/3)^2 == p - (14/3)*(p/3)*p^3*B_{p-2}(1/3) (mod p^4), where B_n(x) denotes the Bernoulli polynomial of degree n; Sum_{k=0..p-1} S(k,-1/4)^2 == (2/p)*p - 26*(-2/p)*p^3*E_{p-3} (mod p^4), where E_0,E_1,E_2,... are Euler numbers; Sum_{k=0..p-1} S(k,-1/6)^2 == (3/p)*p - (155/12)*(-1/p)*p^3*B_{p-2}(1/3) (mod p^4).

Our conjecture is motivated by a conjecture of Kimoto and Wakayama which states that Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p) (mod p^3) for any odd prime p. The Kimoto-Wakayama conjecture was confirmed by Long, Osburn and Swisher in 2014.

For more related conjectures, see Sun's paper arXiv.1512.00712. - Zhi-Wei Sun, Dec 03 2015

Examples

			a(2) = 37 since (1/2^2) * Sum_{k=0..1} (2k+1)*S(k,2)^2 = (S(0,2)^2 + 3*S(1,2)^2)/4 = (1^2 + 3*7^2)/4 = 148/4 = 37.
G.f. = x + 37*x^2 + 1737*x^3 + 102501*x^4 + 6979833*x^5 + 523680739*x^6 + ...
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..100
• K. Kimoto and M. Wakayama, Apéry-like numbers arising from special values of spectral zeta function for non-commutative harmonic oscillators, Kyushu J. Math. 60(2006), no.2, 383-404. (Cf. the formula (6.19).)
• L. Long, R. Osburn and H. Swisher, On a conjecture of Kimoto and Wakayama, arXiv:1404.4723 [math.NT], 2014.
• Z.-W. Sun, Supercongruences involving products of two binomial coefficients, arXiv:1011.6676 [math.NT], 2010-2013; Finite Fields Appl. 22(2013), 24-44.
• 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.
• Z.-W. Sun, Supercongruences involving dual sequences, arXiv:1502.00712 [math.NT], 2015.

Crossrefs

Cf. A000290, A027641, A027642, A122045.

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.)

Programs

Maple

# Implementing Mark van Hoeij's formula.
c := n -> binomial(2*n, n)/(n + 1):
h := n -> simplify(hypergeom([-n,-n,-n], [1,-2*n], 1)):
b := n -> c(n)^2*((n+11)*(2+4*n)^2*h(n+1)^2-2*(n+1)*(11*n+16)*(1+2*n)*h(n)*h(n+1)-h(n)^2*(n+1)^3)/(25*(n+2)):
a := n -> b(n-1): seq(a(n), n = 1..17);  # Peter Luschny, Nov 11 2022

Mathematica

S[k_,x_]:=S[k,x]=Sum[Binomial[k,j]Binomial[x,j]Binomial[x+j,j],{j,0,k}]
a[n_]:=a[n]=Sum[(2k+1)*S[k,n]^2,{k,0,n-1}]/n^2
Do[Print[n," ",a[n]],{n,1,17}]

Formula

a(n) ~ phi^(10*n + 3) / (10 * Pi^2 * n^3), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Nov 06 2021

Conjecture: a(p-1) == 1 (mod p^3) for all primes p >= 5. - Peter Bala, Aug 15 2022

a(n) = ((n+10)*A005258(n)^2 - (11*n+5)*A005258(n)*A005258(n-1) - n*A005258(n-1)^2)/(25*(n+1)). - Mark van Hoeij, Nov 11 2022