

A242992


Least k>n/2, k<n, such that 2^(nk)1 divides 2^k2, or 0 if no such k exists.


0



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,4


COMMENTS

Related to the search of primitive weird numbers A006037 of the form 2^(k1)*Q*R with Q=2^n1 and R=(2^k*QQ1)/(Q+12^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 auch a k=a(n)>0, since k=n1 trivially satisfies the condition (2^(nk)1 = 1).
For odd indices n>2, k=a(n)=(n+1)/2 since this is the least k>n/2 and 2^(k1)1 divides 2^k2.


LINKS

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


FORMULA

For n = 2m1 > 2, a(n) = m.
For all n, a(n) <= n1, and equality holds when n1 is a prime.


CROSSREFS

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



