A290649 The largest number z less than or equal to 3n+1 such that binomial(z,n) is odd.

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

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^{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

Sequence in context: A137315 A139795 A064829 * A118259 A060845 A059478

Adjacent sequences: A290646 A290647 A290648 * A290650 A290651 A290652

Keyword

nonn,easy

Author

Donald M Davis, Aug 08 2017

Status

approved