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A333482 a(n) = [x^(2*n)] S(x)^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

%I A333482 #20 May 01 2023 09:28:47
%S A333482 1,6,304,17718,1093760,69690006,4530426640,298634382374,
%T A333482 19886739416064,1334658881073894,90125657301992304,
%U A333482 6116315760393531094,416791616968522726784,28500344434239455360758,1954614576511349850157392,134392738169746273774331718,9260873342398000417556078592
%N A333482 a(n) = [x^(2*n)] S(x)^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.
%C A333482 Compare with the sequence A103885(n) = [x^n] S(x)^n. See also A333481.
%C A333482 The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
%C A333482 We conjecture that the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) hold for prime p >= 5 and positive integers n and k.
%C A333482 More generally, we conjecture that for any positive integer a and any integer b, the sequence u(a,b;n) := [x^(a*n)] S(x)^(b*n) also satisfies the above congruences.
%F A333482 a(n) = (1/3) * Sum_{k = 0..2*n} C(3*n,k)*C(5*n-k-1,3*n-1) for n >= 1.
%F A333482 P-recursive: P(n)*s(n+1) = 2*(1023893*n^8 - 1278327*n^6 + 474507*n^4 - 57533*n^2 + 1620)*s(n) + P(-n)*s(n-1), where P(n) = 3*(n + 1)*(2*n + 1)*(3*n + 1)*(3*n + 2)*(533*n^4 - 1066*n^3 + 736*n^2 - 203*n + 18).
%F A333482 a(n) ~ (1921 + 533*sqrt(13))^n / (13^(1/4) * sqrt(Pi*n) * 2^(n+1) * 3^(3*n + 1/2)). - _Vaclav Kotesovec_, Mar 28 2020
%e A333482 Examples of congruences:
%e A333482 a(17) - a(1) = 639400846289617183203551941830 - 6 = (2^6)*3*(17^3)* 677836910460361523003969 == 0 ( mod 17^3 ).
%e A333482 a(2*7) - a(2) = 1954614576511349850157392 - 304 = (2^5)*(7^4)*12408377* 2050236754217 == 0 ( mod 7^3 ).
%e A333482 a(5^2) - a(5) = 346904370487885277935868635823142219775940006 - 69690006 = (2^4)*(5^8)*911*39097145981*1558354721574649484551883 == 0 ( mod 5^6 )
%p A333482 [1, seq((1/3)*add(binomial(3*n,k)*binomial(5*n-k-1,3*n-1), k = 0..3*n), n = 1..25)];
%p A333482 # alternative program
%p A333482 S := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
%p A333482 G := (x, n) -> series(S(x)^n, x, 82):
%p A333482 seq(coeff(G(x, n), x, 2*n), n = 0..25);
%t A333482 Join[{1}, Table[Binomial[5*n-1, 3*n-1] * Hypergeometric2F1[-3*n, -2*n, 1 - 5*n, -1]/3, {n,1,20}]] (* _Vaclav Kotesovec_, Mar 28 2020 *)
%Y A333482 Cf. A006318, A103885, A333481.
%K A333482 nonn,easy
%O A333482 0,2
%A A333482 _Peter Bala_, Mar 24 2020