OFFSET
1,1
COMMENTS
For numbers k that satisfy the condition, tau(k) will always be even because tau(k) is odd only if k is a square, but if k is a square then sopfr(k) is even (because every prime appears with an even exponent) and thus it cannot be equal to tau(k)^2 which is odd as tau(k). - Giovanni Resta, May 24 2018
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..1000
MATHEMATICA
Rest@ Select[Range[10^5], Total[Times @@@ FactorInteger@ #] == DivisorSigma[0, #]^2 &] (* Michael De Vlieger, May 27 2018 *)
PROG
(PARI) isok(n) = my(f=factor(n)); sum(k=1, #f~, f[k, 1]*f[k, 2]) == numdiv(n)^2; \\ Michel Marcus, May 24 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Parker Grootenhuis, May 23 2018
STATUS
approved