A306536 The smallest integer k such that floor((2*n)^k/k) is an odd number.

12, 3, 5, 5, 3, 7, 10, 3, 5, 11, 3, 7, 5, 3, 17, 12, 3, 5, 5, 3, 11, 10, 3, 5, 7, 3, 7, 5, 3, 11, 11, 3, 5, 5, 3, 13, 10, 3, 5, 7, 3, 10, 5, 3, 17, 7, 3, 5, 5, 3, 11, 10, 3, 5, 7, 3, 10, 5, 3, 7, 7, 3, 5, 5, 3, 15, 7, 3, 5, 12, 3, 10, 5, 3, 7

Offset: 1

Author

Jinyuan Wang, Feb 22 2019

Keywords

• nonn

Comments

For n > 0, there are infinitely many numbers k such that floor(n^k/k) is an odd number. For odd number n, it's obvious because k can be n^j for any j >= 0. For n = 2, k can be 3*2^j for any j >= 1.

For even number n > 2, k can be A090368(n/2)^j for any j >= 1. Proof: if odd number k makes n^k == 1 (mod k), then n^k = 1 + k*(odd number t), so floor(n^k/k) = t is an odd number. A090368(n/2)^j (j>0) is such k.

Crossrefs

Cf. A090368.

Programs

PARI

a(n) = {k=1; while((2*n)^k\k%2==0, k++); k; }

Formula

a(2^j) <= 12. (This is because floor((2*2^j)^12/12) = floor(2^(12j+10)/3) = (2^(12j+10) - 1)/3 is an odd integer for all j >= 0. - Jianing Song, Feb 24 2019)

For n > 1, a(n) <= A090368(n).