A333563 a(n) = [x^n] G(x)^n, where G(x) is the o.g.f. of A079489.

1, 3, 53, 1056, 22181, 480003, 10588508, 236720424, 5344683429, 121590541641, 2782821611053, 64001191118956, 1477895865330092, 34243264651422596, 795729752353810824, 18537154747116799056, 432781371485493257637, 10123439350286679005973

Offset: 0

Author

Peter Bala, Apr 12 2020

Keywords

• nonn
• easy

Comments

It can be shown that a(n) satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for prime p and positive integers n and k.

The o.g.f. G(x) of A079489 is given by G(x) = c(sqrt(x))*c(-sqrt(x)), where c(x) = ( (1 - sqrt(1 - 4*x))/(2*x) ) is the o.g.f. of the Catalan numbers A000108. It is known that the sequence b(n) := [x^n] c(x)^n = 1/3*binomial(3*n,n) satisfies the supercongruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k - see Meštrović, equation 39. We conjecture that the present sequence satisfies the same congruences. Some examples are given below.

More generally, if r > 0 and s are integers then the sequence a(r,s;n) := [x^(r*n)] G(x)^(s*n) may also satisfy the above congruences.

Examples

			Examples of congruences:
a(17) - a(1) = 10123439350286679005973 - 3 = 2(3^3)*5*(17^3)* 7631634401766047  == 0 ( mod 17^3 ).
a(3*5) - a(3) = 18537154747116799056 - 1056 = (2^4)*3*(5^3)*13* 237655830091241 == 0 ( mod 5^3 ).
a(5^2) - a(5) = 952866706104433648666617525245628 - 480003 = 3*(5^7)*17* 3642302759*65659247842693913 == 0 ( mod 5^6 ).
		

Links

• R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Crossrefs

Cf. A000108, A079489, A333564, A333715.

Programs

Maple

c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G:= x -> c(sqrt(x))*c(-sqrt(x)):
H:= series(G(x)^n, x, 26):
seq(coeff(H, x, n), n = 0..25);

Mathematica

Join[{1}, Table[n^2 * Sum[(-1)^k/((n + 2*k)*(5*n - 2*k))*Binomial[n + 2*k, k] * Binomial[5*n - 2*k, 2*n - k], {k, 0, 2*n}], {n, 1, 20}]] (* Vaclav Kotesovec, Apr 20 2020 *)
Join[{1}, Table[Binomial[5*n, 2*n] * HypergeometricPFQ[{1/2 + n/2, -3*n, -2*n, n/2}, {1/2 - 5*n/2, 1 - 5*n/2, 1 + n}, -1]/5, {n, 1, 20}]] (* Vaclav Kotesovec, May 16 2020 *)

PARI

a(n) = if (n==0, 1, n^2 * sum(k = 0, 2*n, (-1)^k/((n+2*k)*(5*n-2*k))*binomial(n+2*k,k)*binomial(5*n-2*k, 2*n-k))); \\ Michel Marcus, May 16 2020

Formula

a(n) = n^2 * Sum_{k = 0..2*n} (-1)^k/((n+2*k)*(5*n-2*k))*C(n+2*k,k)* C(5*n-2*k, 2*n-k) for n >= 1.

a(p) == 3 ( mod p^3) for prime p >= 3, follows from the above formula.

P-recursive: 6*n*(n - 1)*(2*n - 1)*(3*n - 1)*(3*n - 2)*(2117*n^4 - 12615*n^3 + 27976*n^2 - 27348*n + 9936)*a(n) = -(n - 1)*(27269077*n^8 - 217031969*n^7 + 722440183*n^6 - 1304402267*n^5 + 1384804360*n^4 - 874884704*n^3 + 315932544*n^2 - 57998736*n + 3913920)*a(n-1) + 48*(6*n - 7)*(6*n - 8)*(6*n - 9)*(6*n - 10)*(6*n - 11)*(2117*n^4 - 4147*n^3 + 2833*n^2 - 773*n + 66)*a(n-2) with a(1) = 3, a(2) = 53.

[We note that the sequence u(n) := n^2 * Sum_{k = 0..2*n} 1/((n+2*k)*(5*n-2*k))*C(n+2*k,k)*C(5*n-2*k, 2*n-k) = (1/3)*C(6*n,2*n) is known to satisfy the congruences u(n*p^k) == u(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k - see Meštrović, equation 39. If, in the binomial sum formulas for a(n) and u(n) given above, we restrict the summation range to k = 0..n then we conjecture that the resulting pair of sequences satisfy the same congruences.]

a(n) ~ sqrt(1/24 + 1/(8*sqrt(73))) * ((2117*sqrt(73) - 12881)/216)^n / sqrt(Pi*n). - Vaclav Kotesovec, Apr 20 2020

a(n) = (1/5)*binomial(5*n, 2*n)*hypergeom([1/2 + n/2, -3*n, -2*n, n/2], [1/2 - 5*n/2, 1 - 5*n/2, 1 + n], -1) for n >= 1. - Vaclav Kotesovec, May 16 2020