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

Author

N. J. A. Sloane

Keywords

• nonn

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

Crossrefs

Partial sums of A014137.

Cf. A000108, A005043, A106270.

Programs

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 *)

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 */

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

Extensions

More terms from Alexander Adamchuk, Jul 04 2006