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A333473 a(n) = [x^n] ( S(x/(1 + 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 A333473 #23 Nov 24 2024 08:26:37
%S A333473 1,2,12,92,752,6352,54768,478928,4231424,37680320,337622912,
%T A333473 3040354176,27492359936,249463806464,2270319909632,20714443816192,
%U A333473 189418898063360,1735482632719360,15928224355854336,146414296847992832,1347721096376573952,12421053168197722112
%N A333473 a(n) = [x^n] ( S(x/(1 + 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 A333473 Let F(x) = 1 + f(1)*x + f(2)*x^2 + ... be a power series with integer coefficients. The associated sequence s(n) := [x^n] F(x)^n is known to satisfy the Gauss congruences: s(n*p^k) == s(n*p^(k-1)) ( mod p^(k) ) for any prime p and positive integers n and k. For certain power series F(x) we may get stronger congruences. Examples include F(x) = (1 + x)^2, F(x) = 1/(1 - x) and F(x) = c(x), where c(x) is the o.g.f. of the Catalan numbers A000108. The associated sequences (with some differences of offset) are A000984, A001700 and A025174, respectively.
%C A333473 Here we take F(x) = S(x/(1 + x)) = 1 + 2*x + 4*x^2 + 12*x^3 + 40*x^4 + 154*x^5 + 544*x^6 + ...(see A025227), where S(x) is the o.g.f. of the large Schröder numbers A006318. We conjecture that the associated sequence a(n) = [x^n] ( S(x/(1 + x)) )^n satisfies the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. Cf. A333472.
%C A333473 More generally, we conjecture that for a positive integer r and integer s, the sequence a(r,s;n) := [x^(r*n)] ( S(x/(1 + x)) )^(s*n) also satisfies the above congruences.
%C A333473 Note the sequence b(n) := [x^n] ( S(x) )^n = A103885(n) appears to satisfy the stronger congruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. The sequence d(n) := [x^n] ( (1 + x)*S(x/(1 + x)) )^n = A333090(n) also appears to satisfy the same congruences.
%H A333473 Seiichi Manyama, <a href="/A333473/b333473.txt">Table of n, a(n) for n = 0..1000</a>
%F A333473 a(n) = [x^n] ( (1 - sqrt(1- 4*x - 4*x^2))/(2*x) )^n.
%F A333473 a(n) ~ sqrt(((sqrt(2) + 1)^(2/3) + (sqrt(2) - 1)^(2/3) - 1)/3) * ((3*(71 + 8*sqrt(2))^(1/3) + 3*(71 - 8*sqrt(2))^(1/3) + 13))^n / (sqrt(Pi*n) * 2^(2*n+1)). - _Vaclav Kotesovec_, Mar 28 2020
%F A333473 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+2*k-1,2*k). - _Seiichi Manyama_, Nov 24 2024
%e A333473 Examples of congruences:
%e A333473 a(11) - a(1) = 3040354176 - 2 = 2*(11^2)*13*966419 == 0 ( mod 11^2 ).
%e A333473 a(3*7) - a(3) = 12421053168197722112 - 92 = (2^2)*(3^7)*5*(7^2)* 5795401942927 == 0 ( mod 7^2 ).
%e A333473 a(5^2) - a(5) = 90551762251592215396352 - 6352 = (2^4)*(5^4)*293* 30905038311123623 == 0 ( mod 5^4 ).
%p A333473 Sch := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
%p A333473 G := x → Sch(x/(1+x));
%p A333473 H := (x, n) -> series(G(x)^n, x, 51):
%p A333473 seq(coeff(H(x, n), x, n), n = 0..25)
%t A333473 Table[SeriesCoefficient[((1 - Sqrt[1- 4*x - 4*x^2])/(2*x))^n, {x, 0, n}], {n, 0, 25}] (* _Vaclav Kotesovec_, Mar 28 2020 *)
%o A333473 (PARI) a(n) = sum(k=0, n, binomial(n, k)*binomial(n+2*k-1, 2*k)); \\ _Seiichi Manyama_, Nov 24 2024
%Y A333473 Main diagonal of A378317.
%Y A333473 Cf. A006318, A025227, A103885, A333090, A333472.
%K A333473 nonn,easy
%O A333473 0,2
%A A333473 _Peter Bala_, Mar 23 2020