A305349 Numbers k such that sopfr(k) = tau(k)^3.

183, 295, 583, 799, 943, 7042, 10978, 13581, 18658, 20652, 22402, 22898, 29698, 40162, 43522, 48442, 54778, 59362, 62338, 68098, 74938, 82618, 87418, 89722, 97282, 99298, 102202, 108418, 110842, 113122, 116602, 118498, 122362, 123322, 123778, 128482, 128698

Offset: 1

Author

Parker Grootenhuis, May 30 2018

Keywords

• nonn

Comments

Numbers k such that A001414(k) = A000005(k)^3.

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).

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

Links

• Robert Israel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000005, A001414, A078511, A305026.

Programs

Maple

filter:= proc(n) local F;
  F:= ifactors(n)[2];
  add(t[1]*t[2],t=F) = mul(t[2]+1,t=F)^3
end proc:
select(filter, [$1..200000]); # Robert Israel, Dec 10 2018

Mathematica

sopf[n_] := If[n==1,0,Plus@@Times@@@FactorInteger@ n];Select[Range[200000],sopf[#]==DivisorSigma[0,#]^3 &] (* Amiram Eldar, Nov 01 2018 *)

PARI

sopfr(n) = my(f=factor(n)); sum(k=1, matsize(f)[1], f[k, 1]*f[k, 2]);
isok(n) = sopfr(n) == numdiv(n)^3; \\ Michel Marcus, Nov 02 2018