%I #29 Aug 21 2024 01:56:03
%S 1,1,1,2,3,12,30,240,1050,16800,132300,4233600,61122600,3911846400,
%T 104886381600,13425456844800,674943865596000,172785629592576000,
%U 16407885372638760000,8400837310791045120000,1515727634953623371280000,1552105098192510332190720000
%N Consider Pascal's triangle A007318; a(n) = product of terms at +45 degrees slope with the horizontal.
%C The sum of the terms pertaining to the above product is the (n+1)-th Fibonacci number: 1 + 5 + 6 + 1 = 13.
%C n divides A073617(n+1) for n>=1; see the Mathematica section. [_Clark Kimberling_, Feb 29 2012]
%F a(n) = Product_{k=0..floor(n/2)} binomial(n-k,k).
%F a(2n+1)/a(2n-1) = binomial(2n,n); a(2n)/a(2n-2) = (1/2)*binomial(2n,n); (a(2n+1)*a(2n-2))/(a(2n)*a(2n-1)) = 2. - _John Molokach_, Sep 09 2013
%e For n=6 the diagonal is 1,5,6,1 and product of the terms = 30 hence a(6) = 30.
%p a:= n-> mul(binomial(n-i, i), i=0..floor(n/2)):
%p seq(a(n), n=0..21); # _Alois P. Heinz_, Nov 27 2023
%t p[n_] := Product[Binomial[n + 1 - k, k], {k, 1, Floor[(n + 1)/2]}]
%t Table[p[n], {n, 1, 20}] (* A073617(n+1) *)
%t Table[p[n]/n, {n, 1, 20}] (* A208649 *)
%t ( * _Clark Kimberling_, Feb 29 2012 *)
%Y Cf. A000045, A073618, A007685, A208649, A000984.
%K nonn
%O 0,4
%A _Amarnath Murthy_, Aug 07 2002
%E More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 22 2003