A249609 a(n) is the smallest m, 1<=m<=n, such that binomial(n,m) is evil (A001969); a(n)=0 if there is no such m.

0, 0, 0, 1, 2, 1, 1, 0, 0, 1, 1, 3, 1, 2, 7, 1, 2, 1, 1, 3, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 4, 5, 1, 1, 3, 1, 3, 2, 1, 1, 3, 4, 1, 2, 1, 1, 6, 1, 2, 6, 1, 2, 1, 1, 3, 3, 1, 1, 2, 1, 2, 6, 1, 2, 1, 1, 3, 1, 3, 2, 1, 1, 2, 2, 1, 9, 1, 1, 2, 1, 4, 2, 1, 2, 1, 1

Offset

0,5

Comments

Conjecture: there are only five n: 0,1,2,7,8, for which all entries of the n-th Pascal row (A007318) are odious (A000069). Peter J. C. Moses verified the conjecture up to n = 10^6.

Positions of records are 0,3,4,11,14,76,...; see A249650.

Conjecture verified up to n = 5*10^10. - Michael S. Branicky, Jul 13 2024

Links

Peter J. C. Moses, Table of n, a(n) for n = 0..1000

Formula

a(n) = 1, iff n is evil.

Mathematica

evilQ:=EvenQ[First[DigitCount[#, 2]]]&;
Table[If[#>n, 0, #]&[NestWhile[#+1&, 1, !evilQ[Binomial[n, #]]&]], {n, 0, 100}] (* Peter J. C. Moses, Nov 03 2014 *)

Prog

(Python)
from math import comb
from itertools import count
def A249609(n):
    for m in range(1, n+1):
        if comb(n, m).bit_count()&1 == 0: return m
    return 0
print([A249609(n) for n in range(87)]) # Michael S. Branicky, Jul 13 2024

Crossrefs

Cf. A000069, A001969, A007318, A249650.

Sequence in context: A025908 A134404 A334134 * A232442 A107110 A061197

Adjacent sequences: A249606 A249607 A249608 * A249610 A249611 A249612

Keyword

nonn,base

Author

Vladimir Shevelev, Nov 02 2014

Extensions

More terms from Peter J. C. Moses, Nov 02 2014

Status

approved