A242992 Least k>n/2, k<n, such that 2^(n-k)-1 divides 2^k-2, or 0 if no such k exists.

0, 0, 0, 2, 3, 3, 5, 4, 7, 5, 7, 6, 11, 7, 13, 8, 11, 9, 17, 10, 19, 11, 15, 12, 23, 13, 21, 14, 19, 15, 29, 16, 31, 17, 23, 18, 29, 19, 37, 20, 27, 21, 41, 22, 43, 23, 31, 24, 47, 25, 43, 26, 35, 27, 53, 28, 45, 29, 39, 30, 59, 31, 61, 32, 43, 33, 53, 34, 67, 35, 47, 36, 71, 37, 73, 38, 51, 39, 67, 40, 79, 41, 55, 42, 83, 43, 69, 44, 59, 45, 89, 46, 79

Offset

0,4

Comments

Related to the search of primitive weird numbers A006037 of the form 2^(k-1)*Q*R with Q=2^n-1 and R=(2^k*Q-Q-1)/(Q+1-2^k). (Of course only primes n can lead to a (Mersenne) prime Q (cf. A000043), and R must also be prime to get a weird number.)

For n>2, there always exists such a k=a(n)>0, since k=n-1 trivially satisfies the condition (2^(n-k)-1 = 1).

For odd indices n>2, k=a(n)=(n+1)/2 since this is the least k>n/2 and 2^(k-1)-1 divides 2^k-2.

Links

Table of n, a(n) for n=0..92.

Formula

For n = 2m-1 > 2, a(n) = m.

For all n, a(n) <= n-1, and equality holds when n-1 is a prime.

Conjecture: a(n) = n - A032742(n-1), for n > 2. - Ridouane Oudra, Mar 17 2024

Crossrefs

Cf. A032742.

Sequence in context: A064916 A062854 A057859 * A029579 A106647 A130157

Adjacent sequences: A242989 A242990 A242991 * A242993 A242994 A242995

Keyword

nonn

Author

M. F. Hasler, Aug 17 2014

Status

approved