%I
%S 3,1,1,1,0,2,6,11,4,0,3,4,5,3,6,24,4,0,4,5,3,5,17,10,2,3,6,21,
%T 18,7,5,105,0,3,0,6,8,10,9,7,7,1,14,1,1,1,5,23,47,5,4,20,11,19,10,
%U 10,20,0,5,0,49,3,20,347,29,1,5,0,3,4,3,13,1,18,9,1,23,1,12,36,54,0,75,16,2,5,40
%N A236269(n)  A236313(n).
%C Known formulas for first differences of Stanley sequences are sums with one term always being A236313(n), so it makes sense in order to find a formula for the first differences of the Stanley sequence S[0,4] to subtract A236313(n) from that and look if something shows.
%C a(n) is negative for n = 2,4,8,12,16,24,32,40,48,56,64,66,72... and it appears that n is always even if a(n) is negative. It also seems that a(n) is always negative if n is a power of two, with a(2^m) = 1,1,11,24,105,347,1073,3260,9839,29467,85479,265530...
%H Ralf Stephan, <a href="/A236774/b236774.txt">Table of n, a(n) for n = 1..4599</a>
%o (PARI)
%o NAP(sv,N)=local(v,vv,m,k,l,sl,vvl);sl=length(sv);vvl=min(N*N,10^6);v=vector(N);vv=vector(vvl);for(k=1,sl,v[k]=sv[k];for(l=1,k1,vv[2*v[k]v[l]]=1));m=v[sl]+1;for(k=sl+1,N,while(m<=vvl&&vv[m],m=m+1);if(m>vvl,return(v));for(l=1,k1,sl=2*mv[l];if(sl<=vvl,vv[sl]=1));vv[m]=1;v[k]=m);v
%o v=NAP([0,4],5000)
%o a(n)=v[n+1]v[n](3^valuation(n,2)+1)/2
%K sign
%O 1,1
%A _Ralf Stephan_, Jan 31 2014
