A212861 Numbers n such that the sum of prime factors of n (counted with repetition) equals three times the largest prime divisor.

8, 24, 27, 125, 150, 160, 180, 343, 490, 588, 700, 840, 896, 945, 1008, 1134, 1331, 2197, 3388, 3718, 4840, 4913, 5445, 5808, 6292, 6534, 6859, 8085, 8624, 9464, 9625, 9702, 10647, 11550, 12167, 12274, 12320, 12675, 13520, 13750, 13860, 14784, 15015, 15028

Offset

1,1

Comments

The numbers prime(n)^3 are in the sequence.

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Formula

sopfr(n) = 3*gpf(n), where gpf = A006530. - Charles R Greathouse IV, May 29 2012

Example

150 is in the sequence because 150 = 2*3*5^2 => sum of prime divisors = 2+3 + 5*2 = 15 = 3*5 where 5 is the greatest prime divisor.

Maple

with(numtheory):A:= proc(n) local e, j; e := ifactors(n)[2]: add (e[j][1]*e[j][2], j=1..nops(e)) end: for m from 2 to 20000 do: x:=factorset(m):n1:=nops(x):if A(m)=3*x[n1] then printf(`%d, `, m):else fi:od:

Mathematica

spfQ[n_]:=Module[{f=FactorInteger[n]}, Total[Flatten[Table[#[[1]], #[[2]]]&/@ f]]==3*f[[-1, 1]]]; Select[Range[16000], spfQ] (* Harvey P. Dale, Jul 26 2016 *)

Prog

(PARI) is(n)=my(f=factor(n), k=#f[, 1]); k && sum(i=1, k, f[i, 1]*f[i, 2]) == 3*f[k, 1] \\ Charles R Greathouse IV, May 29 2012

Crossrefs

Cf. A001414, A163836.

Sequence in context: A048109 A068781 A365864 * A333961 A038524 A261394

Adjacent sequences: A212858 A212859 A212860 * A212862 A212863 A212864

Keyword

nonn

Author

Michel Lagneau, May 29 2012

Status

approved