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Squarefree numbers n such that (sopf(n)*d(n))^2 = sigma(n) where sopf(n) = sum of prime factors of n and d(n) = number of divisors of n.
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%I #5 Jun 09 2013 11:32:10

%S 22446139,26116291,28097023,30236557,31090489,31124341,49941589,

%T 61137673,62224039,66960589,71334867,71585139,82266591,83045869,

%U 92346023,92837591,105183961,114762567,123563821,130399138,131494219,134156197,134867722,135095767,136026037

%N Squarefree numbers n such that (sopf(n)*d(n))^2 = sigma(n) where sopf(n) = sum of prime factors of n and d(n) = number of divisors of n.

%C Suggested by _N. J. A. Sloane_.

%H Donovan Johnson, <a href="/A226480/b226480.txt">Table of n, a(n) for n = 1..500</a>

%e n = 22446139 = 31*67*101*107. sopf(n) = 31+67+101+107 = 306. d(n) = 16. (sopf(n)*d(n))^2 = (306*16)^2 = 23970816 = sigma(n).

%Y Cf. A000005, A000203, A006532, A008472, A126028, A226479.

%K nonn

%O 1,1

%A _Donovan Johnson_, Jun 09 2013