A079727 a(n) = 1 + C(2,1)^3 + C(4,2)^3 + ... + C(2n,n)^3.

1, 9, 225, 8225, 351225, 16354233, 805243257, 41229480825, 2172976383825, 117106008311825, 6423711336265041, 357470875526646609, 20131502573232075025, 1145190201805448075025, 65706503254247744075025

Offset: 0

Author

Benoit Cloitre, Feb 17 2003

Keywords

• nonn
• easy

Comments

a(n) seems to have an interesting congruence property: For p prime, a(p) == 8 (mod p) if and only if p == 3, 5, 7, or 13 (mod 14); i.e., iff p = 7 or p is in A003625.

From Peter Bala, Jul 12 2024: (Start)

Zhi-Wei Sun (2010) conjectured that if p is an odd prime such that the Legendre symbol (p/7) = -1 (i.e., if p == 3, 5, 6 (mod 7)) then a(p-1) == 0 (mod p^2). Otherwise, if (p/7) = 1 then a(p-1) == 4*x^2 - 2*p (mod p^2) where p = x^2 + 7*y^2 with x, y in Z.

The author’s twin brother Zhi_Hong Sun confirmed the conjecture in the case (p/7) = -1.

Conjectures: if prime p is in A003625 then

1) a(p^2) == 8 + p^2 (mod p^3)

2) a(p*(p-1)) == p^2 (mod p^3)

3) a((p^2-1)/2) == p^2 (mod p^4) (all checked up to p = 101).

4) if n is a product of distinct primes from A003625 then a((n-1)/2) is divisible by n^2. (End)

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..556
• Zhi-Hong Sun, Congruences concerning Legendre polynomials II", Theorem 3.2, arXiv:1012.3898v2 [math.NT], 2010-2012.
• Zhi-Wei Sun, Open conjectures on congruences, Part A, conjecture A1, arXiv:0911.5665v59 [math.NT], 2009-2011.

Crossrefs

Cf. A002476.

Cf. Sum_{k = 0..n} binomial(2*k, k)^m: A006134 (m=1), A115257 (m=2), this sequence (m=3).

Programs

Magma

[&+[Binomial(2*k, k)^3: k in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Nov 16 2016

Mathematica

Table[Sum[Binomial[2 k, k]^3, {k, 0, n}], {n, 0, 14}] (* Michael De Vlieger, Nov 15 2016 *)

Maxima

makelist(sum(binomial(2*k,k)^3,k,0,n),n,0,12); /* Emanuele Munarini, Nov 15 2016 */

PARI

a(n)=sum(k=0,n,binomial(2*k,k)^3)

Formula

a(n) = Sum_{k=0..n} binomial(2*k,k)^3.

G.f.: hypergeom([1/2, 1/2, 1/2], [1, 1], 64*x)/(1-x). - Vladeta Jovovic, Feb 18 2003

G.f.: hypergeom([1/4,1/4],[1],64*x)^2/(1-x). - Mark van Hoeij, Nov 17 2011

Recurrence: (n+2)^3*a(n+2)-(5*n+8)*(13*n^2+38*n+28)*a(n+1)+8*(2n+3)^3*a(n)=0. - Emanuele Munarini, Nov 15 2016

a(n) ~ 2^(6*n+6) / (63*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Nov 16 2016