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
Keywords
Examples
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.
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Programs
-
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 *) -
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
Formula
sopfr(n) = 3*gpf(n), where gpf = A006530. - Charles R Greathouse IV, May 29 2012
Comments