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%I #20 Oct 10 2019 04:08:56
%S 183,295,583,799,943,7042,10978,13581,18658,20652,22402,22898,29698,
%T 40162,43522,48442,54778,59362,62338,68098,74938,82618,87418,89722,
%U 97282,99298,102202,108418,110842,113122,116602,118498,122362,123322,123778,128482,128698
%N Numbers k such that sopfr(k) = tau(k)^3.
%C Numbers k such that A001414(k) = A000005(k)^3.
%C 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)^3 which is odd as tau(k).
%C A squarefree number k = p_1*...*p_j is in the sequence if p_1 + ... + p_j = 8^j. It is likely that 8^j is the sum of j distinct primes for all j >= 2. - _Robert Israel_, Dec 10 2018
%H Robert Israel, <a href="/A305349/b305349.txt">Table of n, a(n) for n = 1..1000</a>
%p filter:= proc(n) local F;
%p F:= ifactors(n)[2];
%p add(t[1]*t[2],t=F) = mul(t[2]+1,t=F)^3
%p end proc:
%p select(filter, [$1..200000]); # _Robert Israel_, Dec 10 2018
%t sopf[n_] := If[n==1,0,Plus@@Times@@@FactorInteger@ n];Select[Range[200000],sopf[#]==DivisorSigma[0,#]^3 &] (* _Amiram Eldar_, Nov 01 2018 *)
%o (PARI) sopfr(n) = my(f=factor(n)); sum(k=1, matsize(f)[1], f[k, 1]*f[k, 2]);
%o isok(n) = sopfr(n) == numdiv(n)^3; \\ _Michel Marcus_, Nov 02 2018
%Y Cf. A000005, A001414, A078511, A305026.
%K nonn
%O 1,1
%A _Parker Grootenhuis_, May 30 2018
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