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

1,1

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

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 *)

Prog

(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

Crossrefs

Cf. A000005, A001414, A078511, A305026.

Sequence in context: A217608 A264779 A133609 * A030465 A147270 A147124

Adjacent sequences: A305346 A305347 A305348 * A305350 A305351 A305352

Keyword

nonn

Author

Parker Grootenhuis, May 30 2018

Status

approved