OFFSET
0,2
COMMENTS
This sequence is related to and derived from a problem in algebraic topology which asks you to find the largest a+b such that (x+z)^a * (y+z)^b is nonzero mod 2 when x^{n+1}=y^{n+1}=z^{n+1}=0. See the paper at the url below for more information.
LINKS
D. M. Davis, Bounds for higher topological complexity of real projective spaces implied by BP, preprint. See Section 3.
Donald M. Davis, A lower bound for higher topological complexity of real projective space, arXiv:1709.04443 [math.AT], 2017.
FORMULA
If n=2^e + d with 0<= d < 2^e, then a(n)=min(3* 2^e + a(d), 2^{e+2}-1). I can prove this, and also the following explicit formula. If n=A+B, where A==0 mod 2^{e+2}, and 2^e + 2^{e-1} <= B < 2^{e+1}, then a(n)=3A + 2^{e+2}-1. (This is saying that the highest adjacent 1's in the binary expansion of n are in positions e and e-1. If there are none, then a(n)=3n+1 if n is even, and 3n if n is odd.)
PROG
(PARI) a(n) = my(z=3*n+1); while(z > 0, if(Mod(binomial(z, n), 2)==1, return(z)); z--); 0 \\ Felix Fröhlich, Aug 08 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Donald M Davis, Aug 08 2017
STATUS
approved