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