A014140 Apply partial sum operator twice to Catalan numbers.

1, 3, 7, 16, 39, 104, 301, 927, 2983, 9901, 33615, 116115, 406627, 1440039, 5147891, 18550588, 67310955, 245716112, 901759969, 3325067016, 12312494483, 45766188970, 170702447097, 638698318874, 2396598337975, 9016444758528, 34003644251233, 128524394659942, 486793096819011

Offset

0,2

Comments

From Alexander Adamchuk, Jul 04 2006: (Start)

p divides a(p-1) and a((p-3)/2) for primes in A002476.

p divides a((p-5)/2) for primes in A068228.

p^2 divides a(p^2-1) for all primes p > 3. (End)

Equals triangle A106270(unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Apr 02 2009

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Formula

1*C(n) + 2*C(n-1) + 3*C(n-2) + ... + (n+1-k)*C(k) + ... + n*C(1) + (n+1)*C(0), where C(k) = (2k)!/(k!*(k+1)!) is Catalan Number A000108(k). - Alexander Adamchuk, Jul 04 2006

From Alexander Adamchuk, Jul 04 2006: (Start)

a(n) = Sum_{m=0..n} Sum_{k=0..m} (2k)!/(k!*(k+1)!).

a(n) = Sum_{k=0..n} (n+1-k)*(2k)!/(k!*(k+1)!). (End)

G.f.: 1/(1-x)^2*(1-sqrt(1-4*x))/(2*x). - Vladimir Kruchinin, Oct 14 2016

a(n) = Sum_{k=0..n} binomial(n+2,k+2)*r(k), where r(k) are the Riordan numbers A005043. - Vladimir Kruchinin, Oct 14 2016

a(n) ~ 2^(2*n+4) / (9*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2016

Maple

b:= proc(n) option remember; `if`(n<0, [0$2], (q->(f->
      [f[2]+q, q]+f)(b(n-1)))(binomial(2*n, n)/(n+1)))
    end:
a:= n-> b(n)[1]:
seq(a(n), n=0..28);  # Alois P. Heinz, Feb 13 2022

Mathematica

Table[Sum[Sum[(2k)!/k!/(k+1)!, {k, 0, m}], {m, 0, n}], {n, 0, 50}] Table[Sum[(n+1-k)*(2k)!/k!/(k+1)!, {k, 0, n}], {n, 0, 50}] (* Alexander Adamchuk, Jul 04 2006 *)

Prog

(PARI)
sm(v)={my(s=vector(#v)); s[1]=v[1]; for(n=2, #v, s[n]=v[n]+s[n-1]); s; }
C(n)=binomial(2*n, n)/(n+1);
sm(sm(vector(66, n, C(n-1))))
/* Joerg Arndt, May 04 2013 */

Crossrefs

Partial sums of A014137.

Cf. A000108, A005043, A106270.

Sequence in context: A190528 A203611 A176604 * A271788 A103439 A147321

Adjacent sequences: A014137 A014138 A014139 * A014141 A014142 A014143

Keyword

nonn

Author

N. J. A. Sloane

Extensions

More terms from Alexander Adamchuk, Jul 04 2006

Status

approved