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

1,1

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.

Links

Table of n, a(n) for n=1..75.

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).

Prog

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

Crossrefs

Cf. A090368.

Sequence in context: A070604 A268592 A127146 * A063609 A040139 A317312

Adjacent sequences: A306533 A306534 A306535 * A306537 A306538 A306539

Keyword

nonn

Author

Jinyuan Wang, Feb 22 2019

Status

approved