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Numbers of the form m*(2^k-1), where m = 2^(k-1)*(2^k-1) is a perfect number (A000396).
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%I #41 Jun 20 2017 23:06:21

%S 18,196,15376,1032256,274810802176,1125882727038976,72057319160283136,

%T 4951760152529835082242850816,

%U 6129982163463555428116476125461573244012649752219877376

%N Numbers of the form m*(2^k-1), where m = 2^(k-1)*(2^k-1) is a perfect number (A000396).

%C The associated exponents k are in A000043: 2, 3, 5, 7, 13, 17, 19 ,31, 61, ...

%C One can prove that, if m = 2^(k-1)*(2^k-1) is a perfect number, then m*2^k and m*(2^k-1) are both in A181595. Thus every even term in A000396 is a difference of two terms in A181595.

%F If odd perfect numbers do not exist, then a(n) = A181710(n) - A000396(n).

%F a(n) = A019279(n)*(A000668(n))^2 if there are no odd superperfect numbers. - _César Aguilera_, Jun 13 2017

%e With k=3, m = 2^(k-1)*(2^k - 1) = 2^2*(8 - 1) = 28 is a perfect number (A000396), so m*(2^k - 1) = 28*7 = 196 is in the sequence. - _Michael B. Porter_, Jul 19 2016

%Y Cf. A000043, A000396, A181595, A181596, A181701, A181710.

%K nonn

%O 1,1

%A _Vladimir Shevelev_, Nov 07 2010

%E Definition condensed by _R. J. Mathar_, Dec 05 2010