A361715 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2*binomial(n+k-1,k).

0, 1, 9, 82, 745, 6876, 64764, 621860, 6070761, 60085720, 601493134, 6078225792, 61907445340, 634751002718, 6545478537810, 67830084149832, 705950951578089, 7375212511115184, 77310175072063914, 812839577957617640, 8569327793354169870, 90562666708303706642, 959212007563384494522, 10180245921386807485152

Offset: 0

Author

Peter Bala, Mar 23 2023

Keywords

• nonn
• easy

Comments

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).

Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 5.

Compare with the Apéry numbers A005258(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k), which satisfy the weaker supercongruences A005258(p^r) == A005258(p^(r-1)) (mod p^(3*r)) for all primes p >= 5.

Crossrefs

Cf. A005258, A103882, A361712, A361713, A361714, A361717.

Programs

Maple

seq( add( binomial(n,k)^2*binomial(n+k-1,k), k = 0..n-1), n = 0..25);
#faster alternative program
P(n) := 145*n^4 - 1217*n^3 + 3763*n^2 - 5079*n + 2532:
Q(n) := (n - 1)*(n - 2)*(2175*n^6 - 20140*n^5 + 73132*n^4 - 131786*n^3 + 122789*n^2 - 55626*n + 9936):
R(n) := (n - 2)*(6235*n^7 - 67846*n^6 + 304860*n^5 - 731294*n^4 + 1008701*n^3 - 798060*n^2 + 335340*n - 58320):
a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 elif n = 2 then 9 else (Q(n)*a(n-1) - R(n)*a(n-2) - 2*(n - 1)*(n - 3)^2*(2*n - 5)*P(n+1)*a(n-3))/((n - 1)*(n - 2)*n^2*P(n)) end if; end:
seq(a(n), n = 0..25);
# Alternative:
A361715 := n -> hypergeom([-n, -n, n], [1, 1], 1) - binomial(2*n-1, n):
seq(simplify(A361715(n)), n = 0..23); # Peter Luschny, Mar 27 2023

Mathematica

Table[Sum[Binomial[n,k]^2 Binomial[n+k-1,k],{k,0,n-1}],{n,0,30}] (* Harvey P. Dale, Nov 01 2023 *)

Formula

a(n) = A103882(n) - binomial(2*n-1,n) = (3*A005258(n) + A005258(n-1))/5 - binomial(2*n-1,n) for n >= 1.

a(n) ~ sqrt(sqrt(5)/10 + 1/4)*(5*sqrt(5)/2 + 11/2)^n/(Pi*n)

P-recursive:

(n - 1)*(n - 2)*n^2*P(n)*a(n) = Q(n)*a(n - 1) - R(n)*a(n-2) - 2*(n - 1)*(n - 3)^2*(2*n - 5)*P(n+1)*a(n-3) with a(0) = 0, a(1) = 1 and a(2) = 9 and where

P(n) = 145*n^4 - 1217*n^3 + 3763*n^2 - 5079*n + 2532,

Q(n) = (n - 1)*(n - 2)*(2175*n^6 - 20140*n^5 + 73132*n^4 - 131786*n^3 + 122789*n^2 - 55626*n + 9936) and

R(n) = (n - 2)*(6235*n^7 - 67846*n^6 + 304860*n^5 - 731294*n^4 + 1008701*n^3 - 798060*n^2 + 335340*n - 58320).

a(n) = hypergeom([-n, -n, n], [1, 1], 1) - binomial(2*n-1, n). This is another way to write the first formula. - Peter Luschny, Mar 27 2023