A032357 Convolution of Catalan numbers and powers of -1.

1, 0, 2, 3, 11, 31, 101, 328, 1102, 3760, 13036, 45750, 162262, 580638, 2093802, 7601043, 27756627, 101888163, 375750537, 1391512653, 5172607767, 19293659253, 72188904387, 270870709263, 1019033438061, 3842912963391, 14524440108761

Offset: 0

Author

Wolfdieter Lang

Keywords

• easy
• nonn

Comments

Absolute value of the alternating sum of Catalan Numbers. - Alexander Adamchuk, Jul 03 2006

Sums of two consecutive terms are a(n-1) + a(n) = 1, 2, 5, 14, 42, ... = A000108(n) (Catalan Numbers). The prime p divides a((p-3)/2) for p = 11, 19, 29, 31, 41, 59, 61, 71, ... = A045468 (Primes congruent to {1, 4} mod 5). Prime p divides a(2*p+1) for p = 5, 11, 19, 29, 31, 41, 59, 61, 71, ... = A038872 (Primes congruent to {0, 1, 4} mod 5). Also odd primes where 5 is a square mod p. - Alexander Adamchuk, Jul 03 2006

Hankel transform is F(2*n+1), where F = A000045. - Paul Barry, Jul 22 2008

Equals INVERTi transform of A000958. - Gary W. Adamson, Apr 10 2009

Inverse binomial transform of A002212. - Philippe Deléham, Sep 17 2009

Number of singleton and plus-decomposable (2143, 2413, 3142)-avoiding permutations with no +bonds (ascents by 1), with offset 1. Equivalently, number of (2143, 2413, 3142)-avoiding permutations that start with 1 or end with n (top entry). E.g., 132 and 213 for n = 3; 1324, 1432, 3214 for n = 4. - Alexander Burstein, May 22 2015

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Shalosh B. Ekhad and Mingjia Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, 2017.

Crossrefs

Cf. A000045, A000108, A000958, A001622, A002212, A014137, A014138, A033297, A038872, A045468, A064739.

Programs

Maple

rec:= (n+1)*a(n) +3*(-n+1)*a(n-1) +2*(-2*n+1)*a(n-2)=0:
A:= gfun:-rectoproc({rec,a(0)=1,a(1)=0},a(n),remember):
seq(A(n),n=0..50); # Robert Israel, May 22 2015

Mathematica

Table[Sum[(-1)^(k+n)*CatalanNumber[k],{k,0,n}],{n,0,60}] (* Alexander Adamchuk, Jul 03 2006 *)
Round@Table[(-1)^n/GoldenRatio + CatalanNumber[n + 1] Hypergeometric2F1[1, n + 3/2, n + 3, -4], {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Oct 02 2016 *)
Table[(CatalanNumber[n] (2 + (n + 1) Hypergeometric2F1[1, -n, 1/2, 5/4]) - (-1)^n)/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)

Sage

def A032357():
    f, c, n = 1, 1, 1
    while True:
        yield f
        n += 1
        c = c * (4*n - 6) // n
        f = c - f
a = A032357()
print([next(a) for  in range(27)]) # _Peter Luschny, Nov 30 2016

Formula

G.f.: c(x)/(1 + x), where c(x) is the g.f. for the Catalan numbers A000108.

a(n) = Sum_{k=0..n} (-1)^(n-k)*C(k), where C(k) = A000108(k).

a(n) = ((-1)^(n+1) - binomial(2*(n+1), n+1)*Sum_{k=0..n+1} (-5)^k*binomial(n+1, k)/binomial(2*k, k))/2.

a(n) = C(2*n, n)/(n+1) - a(n-1) = A000108(n) - a(n-1) with a(0) = 1. - Labos Elemer, Apr 26 2003

Conjecture: (n+1)*a(n) + 3*(-n+1)*a(n-1) + 2*(-2*n+1)*a(n-2) = 0. - R. J. Mathar, Nov 30 2012

Conjecture is true since the g.f. satisfies (x - 3*x^2 - 4*x^3)*g'(x) + (1 - 6*x^2)*g(x) = 1. - Robert Israel, May 22 2015

a(n) = (-1)^n/A001622 + A000108(n+1)*hypergeom([1, n + 3/2], [n + 3], -4). - Vladimir Reshetnikov, Oct 02 2016

a(n) ~ 2^(2*n + 2) / (5*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 03 2016

a(n) = (A000108(n) * (2 + (n + 1)*hypergeom([1,-n], [1/2], 5/4)) - (-1)^n)/2. - Vladimir Reshetnikov, Oct 03 2016

Extensions

More terms from Christian G. Bower, Apr 15 1998

More terms from Alexander Adamchuk, Jul 03 2006