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

1,1

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.

Links

Table of n, a(n) for n=1..22.

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.

Example

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 ).

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 *)

Prog

(Python)
from itertools import count, islice
def A333564_gen(): # generator of terms
    yield (a:=2)
    c = 1
    for n in count(1):
        yield a<<n+1
        a=(3*n+5)*(c:=c*((n<<2)+2)//(n+2))-a>>1
A333564_list = list(islice(A333564_gen(), 20)) # Chai Wah Wu, Apr 26 2023

Crossrefs

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

Sequence in context: A306087 A113248 A353611 * A291314 A354248 A366267

Adjacent sequences: A333561 A333562 A333563 * A333565 A333566 A333567

Keyword

nonn,easy

Author

Peter Bala, Apr 07 2020

Status

approved