



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, 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, 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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

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.
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...


LINKS

Ralf Stephan, Table of n, a(n) for n = 1..4599


PROG

(PARI)
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
v=NAP([0, 4], 5000)
a(n)=v[n+1]v[n](3^valuation(n, 2)+1)/2


CROSSREFS

Sequence in context: A321444 A104608 A241499 * A110245 A230003 A136093
Adjacent sequences: A236771 A236772 A236773 * A236775 A236776 A236777


KEYWORD

sign


AUTHOR

Ralf Stephan, Jan 31 2014


STATUS

approved



