|
%I #10 Nov 11 2019 09:30:04
%S 2,3,3,3,3,5,5,3,3,7,7,5,5,9,9,3,3,3,11,3,3,5,5,5,5,15,7,9,9,17,17,3,
%T 3,3,5,5,5,5,9,3,3,23,7,13,13,9,9,5,5,19,11,3,3,5,21,9,9,15,23,17,17,
%U 33,33,3,3,3,3,3,7,5,5,3,3,7,7,21,21,17,9,3,3
%N For any n > 0, let m = 2*n - 1 (m is the n-th odd number); a(n) is the least k > 1 such that m AND (m^k) = m (where AND denotes the bitwise AND operator).
%C The sequence is well defined: for any n > 0:
%C - let x be such that 2*n-1 < 2^x,
%C - hence gcd(2*n-1, 2^x) = 1,
%C - and a(n) <= 1 + ord_{2^x}(2*n-1) (where ord_u(v) is the multiplicative order of v modulo u).
%H Rémy Sigrist, <a href="/A329245/b329245.txt">Table of n, a(n) for n = 1..8192</a>
%e For n = 7:
%e - m = 2*7 - 1 = 13,
%e - 13 AND (13^2) = 9,
%e - 13 AND (13^3) = 5,
%e - 13 AND (13^4) = 1,
%e - 13 AND (13^5) = 13,
%e - hence a(7) = 5.
%o (PARI) a(n) = my (m=2*n-1, mk=m); for (k=2, oo, if (bitand(m, mk*=m)==m, return (k)))
%Y Cf. A253719.
%K nonn,base
%O 1,1
%A _Rémy Sigrist_, Nov 09 2019
|
