A333472 a(n) = [x^n] ( c (x/(1 + x)) )^n, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

1, 1, 3, 13, 59, 276, 1317, 6371, 31131, 153292, 759428, 3780888, 18900389, 94805959, 476945913, 2405454213, 12158471195, 61574325840, 312365992620, 1587052145492, 8074474510884, 41131551386120, 209760563456920, 1070822078321520, 5471643738383781, 27982867986637151

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 u(n) := [x^n] F(x)^n is known to satisfy the Gauss congruences: u(n*p^k) == u(n*p^(k-1)) ( mod p^k ) for any prime p and positive integers n and k. For certain power series F(x), stronger congruences may hold. 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) = c(x/(1 + x)) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 21*x^6 + ... (cf. A001006 and A086246) and conjecture that the associated sequence a(n) = [x^n] ( c(x/(1 + x)) )^n satisfies the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. Cf. A333473.

More generally, we conjecture that for any positive integer r and any integer s the sequence a(r,s;n) := [x^(r*n)] ( c(x/(1 + x)) )^(s*n) also satisfies the above congruences.

Note that the sequence b(n) := [x^n] c(x)^n = A025174(n) satisfies 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)*c(x/(1 + x)) )^n = A333093(n) appears to satisfy the same congruences.

Examples

			Examples of congruences:
a(11) - a(1) = 3780888 - 1 = (11^2)*31247 == 0 ( mod 11^2 ).
a(3*7) - a(3) = 41131551386120 - 13 = (7^2)*13*23671*2727841 == 0 ( mod 7^2 ).
a(5^2) - a(5) = 27982867986637151 - 276 = (5^4)*13*74687*46113049 == 0 ( mod 5^4 ).
		

Crossrefs

Cf. A000108, A001006, A086246, A106228, A219537, A333093, A333473.

Programs

Maple

Cat := x -> (1/2)*(1-sqrt(1-4*x))/x:
G := x -> Cat(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 + x - Sqrt[1 - 2*x - 3*x^2]) / (2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 29 2020 *)

Formula

a(n) = [x^n] ( (1 + x - sqrt(1 - 2*x - 3*x^2)) / (2*x) )^n.

a(n) ~ sqrt(((9386 + 1026*sqrt(57))^(1/3) + (9386 - 1026*sqrt(57))^(1/3) - 19)/228) * (((1261 + 57*sqrt(57))^(1/3) + (1261 - 57*sqrt(57))^(1/3) + 10)/6)^n / sqrt(Pi*n). - Vaclav Kotesovec, Mar 29 2020