%I #18 Nov 12 2024 09:06:32
%S 1,1,1,5,1,3,1,85,73,11,1,195,1,43,151,21845,1,12483,1,11275,2359,683,
%T 1,798915,1082401,2731,19173961,704555,1,1649373,1,1431655765,599479,
%U 43691,8727391,3272356035,1,174763,9588151,11822705675,1,1649061309,1
%N a(n) = (2^n - 1)/product(2^p - 1) where the product is over all distinct primes p that divide n.
%H John Tyler Rascoe, <a href="/A055515/b055515.txt">Table of n, a(n) for n = 1..1000</a>
%F For p prime, a(p) = 1. - _Michel Marcus_, May 18 2014
%F For p prime, a(p^2) = A051156(n). - _Michel Marcus_, May 18 2014
%e a(12) = (2^12 -1)/((2^2 -1) (2^3 -1)) = 195.
%o (PARI) a(n) = my(f = factor(n)); (2^n-1)/prod(i=1, #f~, 2^f[i, 1] -1); \\ _Michel Marcus_, May 18 2014
%Y Cf. A055977.
%K easy,nonn,changed
%O 1,4
%A _Leroy Quet_, Jul 03 2000