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.

1, 2, 12, 92, 752, 6352, 54768, 478928, 4231424, 37680320, 337622912, 3040354176, 27492359936, 249463806464, 2270319909632, 20714443816192, 189418898063360, 1735482632719360, 15928224355854336, 146414296847992832, 1347721096376573952, 12421053168197722112

Offset: 0

Author

Peter Bala, Mar 23 2020

Keywords

• nonn
• easy

Comments

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.

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.

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.

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.

Examples

			Examples of congruences:
a(11) - a(1) = 3040354176 - 2 = 2*(11^2)*13*966419 == 0 ( mod 11^2 ).
a(3*7) - a(3) = 12421053168197722112 - 92 = (2^2)*(3^7)*5*(7^2)* 5795401942927 == 0 ( mod 7^2 ).
a(5^2) - a(5) = 90551762251592215396352 - 6352 = (2^4)*(5^4)*293* 30905038311123623 == 0 ( mod 5^4 ).
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1000

Crossrefs

Main diagonal of A378317.

Cf. A006318, A025227, A103885, A333090, A333472.

Programs

Maple

Sch := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := x → Sch(x/(1+x));
H := (x, n) -> series(G(x)^n, x, 51):
seq(coeff(H(x, n), x, n), n = 0..25)

Mathematica

Table[SeriesCoefficient[((1 - Sqrt[1- 4*x - 4*x^2])/(2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 28 2020 *)

PARI

a(n) = sum(k=0, n, binomial(n, k)*binomial(n+2*k-1, 2*k)); \\ Seiichi Manyama, Nov 24 2024

Formula

a(n) = [x^n] ( (1 - sqrt(1- 4*x - 4*x^2))/(2*x) )^n.

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

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+2*k-1,2*k). - Seiichi Manyama, Nov 24 2024