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A055515 a(n) = (2^n - 1)/product(2^p - 1) where the product is over all distinct primes p that divide n.

%I A055515 #20 Jan 18 2025 14:31:51
%S A055515 1,1,1,5,1,3,1,85,73,11,1,195,1,43,151,21845,1,12483,1,11275,2359,683,
%T A055515 1,798915,1082401,2731,19173961,704555,1,1649373,1,1431655765,599479,
%U A055515 43691,8727391,3272356035,1,174763,9588151,11822705675,1,1649061309,1
%N A055515 a(n) = (2^n - 1)/product(2^p - 1) where the product is over all distinct primes p that divide n.
%H A055515 John Tyler Rascoe, <a href="/A055515/b055515.txt">Table of n, a(n) for n = 1..1000</a>
%F A055515 For p prime, a(p) = 1. - _Michel Marcus_, May 18 2014
%F A055515 For p prime, a(p^2) = A051156(n). - _Michel Marcus_, May 18 2014
%e A055515 a(12) = (2^12 -1)/((2^2 -1) (2^3 -1)) = 195.
%t A055515 Table[(2^n-1)/Times@@(2^#-1&/@FactorInteger[n][[;;,1]]),{n,50}] (* _Harvey P. Dale_, Jan 18 2025 *)
%o A055515 (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 A055515 Cf. A055977.
%K A055515 easy,nonn
%O A055515 1,4
%A A055515 _Leroy Quet_, Jul 03 2000