OFFSET
1,1
COMMENTS
This is a subsequence of A181595. [Proof: sigma(m) = (2^t-1)*(2^t-2) leads to an abundance of m which is 2.]
Numbers m such that the sum of the even divisors of m equals the square of the odd divisors of m.
Proof: let s0 the sum of the even divisors and s1 the sum of the odd divisors.
s1 = 2^t-2 because 2^t-3 is prime.
s0 = 2 + 4 + 8 + ... + 2^(t-1) + (2^t - 3)(2 + 4 + 8 + ... + 2^(t-1)) = (2^t - 2)^2 => s0 = s1^2. - Michel Lagneau, Apr 17 2013
LINKS
Eric Chen, Table of n, a(n) for n = 1..28
MAPLE
with(numtheory):for n from 1 to 600000 do:x:=divisors(n):n0:=nops(x):s0:=0:s1:=0:for k from 1 to n0 do:if irem(x[k], 2)=0 then s0:=s0+ x[k]:else s1:=s1+ x[k]:fi:od:if s0=s1^2 then print(n):else fi:od: # Michel Lagneau, Apr 17 2013
PROG
(PARI) for(k=1, 200, if(ispseudoprime(2^k-3), print1(2^(k-1)*(2^k-3), ", "))) \\ Eric Chen, Jun 13 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Nov 06 2010
EXTENSIONS
Edited and extended by D. S. McNeil, Nov 18 2010
Definition simplified by R. J. Mathar, Nov 18 2010
STATUS
approved