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A014140
Apply partial sum operator twice to Catalan numbers.
6
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
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.
Sequence in context: A190528 A203611 A176604 * A271788 A103439 A147321
KEYWORD
nonn
EXTENSIONS
More terms from Alexander Adamchuk, Jul 04 2006
STATUS
approved