

A290649


The largest number z less than or equal to 3n+1 such that binomial(z,n) is odd.


0



1, 3, 7, 7, 13, 15, 15, 15, 25, 27, 31, 31, 31, 31, 31, 31, 49, 51, 55, 55, 61, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 63, 97, 99, 103, 103, 109, 111, 111, 111, 121, 123, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127
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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

Table of n, a(n) for n=0..59.
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^{e1} <= 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 e1. 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 FrÃ¶hlich, Aug 08 2017


CROSSREFS

Sequence in context: A073881 A137315 A139795 * A118259 A060845 A059478
Adjacent sequences: A290646 A290647 A290648 * A290650 A290651 A290652


KEYWORD

nonn,easy


AUTHOR

Donald M Davis, Aug 08 2017


STATUS

approved



