A054108 a(n) = (-1)^(n+1)*Sum_{k=0..n+1}(-1)^k*binomial(2*k,k).

1, 5, 15, 55, 197, 727, 2705, 10165, 38455, 146301, 559131, 2145025, 8255575, 31861025, 123256495, 477823895, 1855782325, 7219352975, 28125910825, 109720617995, 428537256445, 1675561707275, 6557869020325, 25689734662775

Offset: 0

Author

Clark Kimberling

Keywords

• nonn

Comments

p divides a((p-3)/2) for p in A045468 (primes congruent to {1, 4} mod 5). - Alexander Adamchuk, Jul 05 2006

The sequence 1,1,5,15,55,... has general term sum{k=0..n, (-1)^(n-k)*C(2k,k)}. Its Hankel transform is A082761. - Paul Barry, Apr 10 2007

From Paul Barry, Mar 29 2010: (Start)

The sequence 1,1,5,15,... has g.f. 1/((1+x)*sqrt(1-4x)).

The doubled sequence 1,1,1,1,5,5,... has e.g.f. dif(int((sin(x-t)+cos(x-t))*Bessel_I(0,2t),t,0,x),x). (End)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Crossrefs

T(2n, n), array T as in A054106.

Cf. A066796, A000984, A054109, A006134, A045468.

Programs

Mathematica

Table[Sum[(-1)^(k+n)*((2k)!/(k!)^2),{k,0,n}], {n,1,50}] (* Alexander Adamchuk, Jul 05 2006 *)
CoefficientList[Series[(1/Sqrt[1-4*x]/(1+x)-1)/x, {x, 0, 20}], x]
(* or *)
Table[(-1)^(n+1)*Sum[(-1)^k*Binomial[2*k, k], {k, 0, n+1}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 06 2012 *)
Round@Table[Binomial[2 (n + 2), n + 2] Hypergeometric2F1[1, n + 5/2, n + 3, -4] - (-1)^n/Sqrt[5], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)

PARI

a(n)=(-1)^(n+1)*sum(k=0,n+1,(-1)^k*binomial(2*k,k))

Python

from math import comb
def A054108(n): return (1 if n % 2 else -1)*sum((-1 if k % 2 else 1)*comb(2*k,k) for k in range(n+2)) # Chai Wah Wu, Jan 19 2022

Formula

a(n) = C(2n, n) - a(n-1) with a(0)=1. - Labos Elemer, Apr 26 2003

C(2n,n) - C(2n-2,n-1) + ... +(-1)^(k+n)*C(2k,k)+ ... + (-1)^(1+n)*C(2,1) + (-1)^n*C(0,0), where C(2k,k)=(2k)!/(k!)^2 - central binomial coefficients A000984[k]. - Alexander Adamchuk, Jul 05 2006

a(n) = Sum_{k=0..n} (-1)^(k+n)*((2k)!/(k!)^2). - Alexander Adamchuk, Jul 05 2006

G.f.: (1/sqrt(1-4*x)/(1+x)-1)/x = (-1 + 2/(U(0)-2*x))/(1+x) where U(k)= 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 27 2012

a(n) ~ 2^(2*n+4)/(5*sqrt(Pi*n)). - Vaclav Kotesovec, Nov 06 2012

Recurrence: (n+1)*a(n) = (3*n+1)*a(n-1) + 2*(2*n+1)*a(n-2). - Vaclav Kotesovec, Nov 06 2012

Extensions

Formula from Benoit Cloitre, Sep 29 2002

Definition corrected by Vaclav Kotesovec, Nov 06 2012