OFFSET
0,2
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.
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
MATHEMATICA
Table[Sum[Binomial[2 k, k]^3, {k, 0, n}], {n, 0, 14}] (* Michael De Vlieger, Nov 15 2016 *)
PROG
(PARI) a(n)=sum(k=0, n, binomial(2*k, k)^3)
(Maxima) makelist(sum(binomial(2*k, k)^3, k, 0, n), n, 0, 12); /* Emanuele Munarini, Nov 15 2016 */
(Magma) [&+[Binomial(2*k, k)^3: k in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Nov 16 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Benoit Cloitre, Feb 17 2003
STATUS
approved