%I #6 Aug 28 2013 04:49:15
%S 1,0,1,0,1,1,1,1,1,2,2,2,3,3,5,4,7,6,10,10,14,15,20,24,30,35,45,53,69,
%T 79,104,120,157,184,236,281,356,431,540,656,821,1001,1252,1525,1908,
%U 2328,2909,3557,4434,5436,6762
%N The backwards antidiagonal sums of triangle A228570.
%C The a(n) equal the backwards antidiagonal sums of triangle A228570.
%F a(n) = sum(A228570(n-k, n-2*k), k=0..floor(n/2)).
%F G.f.: (1/2)*(1/(1-x^2-x^5) + (1+x^2+x^5)/(1-x^4-x^10)).
%p f := x -> (1/((1-x^2-x^5)) + (1+x^2+x^5)/(1-x^4-x^10))/2 : seq(coeff(series(f(x), x, n+1), x, n), n=0..50); # End first program
%p a := proc(n): (A001687(n+1) + x(n) + x(n-2) + x(n-5))/2 end: A001687 := proc(n) option remember: if n=0 then 0 elif n=1 then 1 elif n=2 then 0 elif n=3 then 1 elif n=4 then 0 else procname(n-2) + procname(n-5) fi: end: x := proc(n) local x: if n <0 then return(0) fi: if type(n, even) then A001687((n+2)/2) else 0 fi: end: seq(a(n), n=0..50); # End second program
%Y Cf. A228570, A102543, A192928, A001687.
%K nonn,easy
%O 0,10
%A _Johannes W. Meijer_, Aug 26 2013