A333561 a(n) = Sum_{j = 0..2*n} binomial(n+j-1,j)*2^j.
1, 7, 129, 2815, 65537, 1579007, 38862849, 970522623, 24494735361, 623210135551, 15956734640129, 410649406472191, 10612705274626049, 275241225206890495, 7159857331658817537, 186731505521384226815, 4880983719142471237633, 127836403093194475044863
Offset: 0
Examples
Examples of supercongruences: a(11) - a(1) = 410649406472191 - 7 = (2^3)*3*(11^3)*12855290711 == 0 ( mod 11^3 ). a(3*7) - a(3) = 61103847305642669128888090623 - 2815 = (2^8)*(7^5)* 87326419*162627033103121 == 0 ( mod 7^3 ). a(5^2) - a(5) = 29754989698128108780761000609579007 - 1579007 = (2^11)*(5^6)*179*751*10267*673710468794491483 == 0 ( mod 5^6 ).
Programs
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Maple
seq(add( binomial(n+j-1,j)*2^j, j = 0..2*n), n = 0..25);
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Mathematica
Table[(-1)^n - 2^(2*n+1) * Binomial[3*n, 2*n+1] * Hypergeometric2F1[1, 3*n+1, 2*n+2, 2], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *) -
PARI
a(n) = sum(j = 0, 2*n, binomial(n+j-1,j)*2^j); \\ Michel Marcus, Mar 28 2020
Formula
Conjectural o.g.f.: 1/(1 + x) + 8*x*f'(4*x)/(2*f(4*x) - 1), where f(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... is the o.g.f. of A001764.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 89*x^2 + 1447*x^3 + ... appears to be the o.g.f. of A062747.
Conjectural recurrence: n*(n - 1)*(2*n - 1)*(3098*n - 6455)*a(n) = (n - 1)*(172988*n^3 - 585840*n^2 + 550321*n - 169824)*a(n-1) - 12*(11825*n^4 - 168518*n^3 + 627675*n^2 - 853766*n + 350744)*a(n-2) - 36*(n - 3)*(3*n - 7)*(3*n - 8)*(991*n - 724)*a(n-3) with a(1) = 7, a(2) = 129, a(3) = 2815.
From Vaclav Kotesovec, Mar 28 2020: (Start)
a(n) ~ 3^(3*n + 1/2) / (4*sqrt(Pi*n)).
Recurrence: n*(2*n - 1)*(7*n^2 - 20*n + 14)*a(n) = (364*n^4 - 1411*n^3 + 1818*n^2 - 868*n + 120)*a(n-1) + 6*(3*n - 5)*(3*n - 4)*(7*n^2 - 6*n + 1)*a(n-2). (End)
From Peter Bala, Mar 05 2022: (Start)
a(n) = Sum_{k = 0..2*n} binomial(3*n, 2*n-k)*binomial(n+k-1,k).
a(n) = [x^(2*n)] ( (1 + x^3)/(1 - x) )^n.
The o.g.f. satisfies the algebraic equation (108*x^3 + 212*x^2 + 100*x - 4)*A(x)^3 - (216*x^2 + 208*x - 8)*A(x)^2 + (48*x^2 + 155*x - 5)*A(x) + 8*x^2 - 40*x + 1 = 0. (End)
a(n) = binomial(3*n, 2*n)*hypergeom([-2*n, n], [n + 1], -1). - Peter Luschny, Mar 07 2022
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