A115081 - OEIS

%I #5 Mar 30 2012 18:36:55

%S 1,1,3,11,50,257,1467,9081,60272,424514,3151226,24510411,198870388,

%T 1676878231,14648843341,132228263355,1230505582380,11782173683640,

%U 115878367974480,1168833058344870,12075008262774120,127608480923659770

%N Column 0 of triangle A115080.

%C Also equals row sums of triangle A125080.

%F a(n) = Sum_{k=0..[n/2]} A000108(n-k)*A001147(k)*C(n,2*k), where A000108 is the Catalan numbers and A001147 is the double factorials.

%F a(n) = Sum_{k=0..[n/2]} A000108(n-k)*A000108(k)*(k+1)!*C(n,2k)/2^k where A000108(n) = C(2n,n)/(n+1) are the Catalan numbers. a(n) = Sum_{k=0..n} (-1)^(n-k)*n!/k!*A115082(k) . - _Paul D. Hanna_, Feb 19 2007

%e At n=5, a(5) = Sum_{k=0..2} A000108(5-k)*A001147(k)*C(5,2*k) so that a(5) = 42*1*C(5,0) + 14*1*C(5,2) + 5*3*C(5,4) = 42*1*1 + 14*1*10 + 5*3*5 = 42 + 140 + 75 = 257.

%o (PARI) {a(n)=sum(k=0,n\2,binomial(2*n-2*k,n-k)/(n-k+1)*binomial(2*k,k)*k!/2^k*binomial(n,2*k))}

%o (PARI) {a(n)=sum(k=0,n\2,(2*n-2*k)!*n!/k!/(n-k)!/(n-k+1)!/(n-2*k)!/2^k )}

%Y Cf. A115080, A115082 (column 1), A115083 (column 2), A115084 (row sums); A115086.

%Y Cf. A125080 (related triangle); A000108, A001147.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jan 13 2006, Nov 19 2006