A333564 a(n) = [x^n] ( c(x)/c(-x) )^n, where c(x) = (1 - sqrt( 1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

2, 8, 56, 384, 2752, 20096, 148864, 1114112, 8403968, 63787008, 486584320, 3727196160, 28649455616, 220869853184, 1707123245056, 13223868760064, 102636144295936, 797982357192704, 6213784327684096, 48452953790480384, 378291752487878656, 2956824500391378944

Offset: 1

Author

Peter Bala, Apr 07 2020

Keywords

• nonn
• easy

Comments

It can be shown that a(n) satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p. We conjecture that a(n) satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 3 and positive integers n and k. Some examples are given below.

More generally, for integer r and positive integer s, we conjecture that the sequence a(r,s;n) := 2^(r*n) * Sum_{k = 0..s*n-1} (-1)^(s*n-1+k)*( binomial(n+k,k) )^r satisfies the same congruences. This is the case r = s = 1.

Examples

			Examples of congruences:
a(11) - a(1) = 486584320 - 2 = 2*(11^3)*182789 == 0 ( mod 11^3 ).
a(2*11) - a(2) = 2956824500391378944 - 8 = (2^3)*(3^2)*(11^3)*
107*288357478039 == 0 ( mod 11^3 ).
a(5^2) - a(5) = 1420245922851693002752 - 2752 = (2^6)*(3^3)*(5^6)* 7*19*1123*352183001 == 0 ( mod 5^6 ).
		

Crossrefs

Cf. A000108, A014300, A119259, A333565, A349648.

Programs

Maple

seq( (2^n)*add( (-1)^(n-1+k)*binomial(n+k,k), k = 0..n-1), n = 1..25);
# alternative program
a := proc (n) option remember; `if`(n = 1, 2, `if`(n = 2, 8, `if`(n = 3, 56, ((3*n+4)*a(n-1)+(36*n-76)*a(n-2)+(32*n-80)*a(n-3))/n)))
end proc:
seq(a(n), n = 1..25);

Mathematica

a[n_] := -2^(n) Binomial[2n, n-1] Hypergeometric2F1[1, 2n +1, n + 2, 2];
Table[Simplify[a[n]], {n, 1, 22}] (* Peter Luschny, Apr 13 2020 *)
c[x_] := (1-Sqrt[1-4x])/(2x); ser[n_] := Series[(c[x]/c[-x])^n, {x, 0, 22}];
Table[SeriesCoefficient[ser[n], n], {n, 1, 22}] (* Peter Luschny, Apr 14 2020 *)

Python

from itertools import count, islice
def A333564_gen(): # generator of terms
    yield (a:=2)
    c = 1
    for n in count(1):
        yield a<>1
A333564_list = list(islice(A333564_gen(),20)) # Chai Wah Wu, Apr 26 2023

Formula

a(n) = 2^n * Sum_{k = 0..n-1} (-1)^(n-1+k)*binomial(n+k,k) = 2^n * A014300(n).

a(n) = (-1)^(n+1) + Sum_{k = 0..n-1} n^2/((n-k)*(2*n-k))*C(n-k,k)*C(3*n-2*k-1,n-k).

Congruences: a(p) == 2 ( mod p^3 ) for all prime p >= 3, follows from previous formula.

a(n) = Sum_{k = 1..n} (-1)^(n+k)*(3*k-1)*2^(k-1)*A000108(k-1).

a(n) = (1/2)*(A119259(n) - (-1)^n).

a(n) ~ 8^n / (3*sqrt(Pi*n)).

P-recursive with recurrence n*(3*n - 4)*a(n) = (21*n^2 - 40*n + 12)*a(n-1) + 4*(3*n - 1)*(2*n - 3)*a(n-2) with a(1) = 2 and a(2) = 8. Cf. A333565.

Alternative form: (a(n) + a(n-1))/(a(n) - a(n-2)) = P(n)/Q(n), where P(n) = 4*(3*n - 1)*(2*n - 3) and Q(n) = (21*n^2 - 40*n + 12).

Also, n*a(n) = (3*n + 4)*a(n-1) + 4*(9*n - 19)*a(n-2) + 16*(2*n - 5)*a(n-3) with a(1) = 2, a(2) = 8 and a(3) = 56.

O.g.f.: A(x) = 4*x/(1 - 8*x + (1 + 4*x)*sqrt(1 - 8*x)), which satisfies the differential equation (x + 1)*(4*x + 1)*(8*x - 1)*A'(x) + (16*x^2 - 4*x + 7)*A(x) + 2*(1 - 2*x) = 0.