A213020 Smallest number k such that the sum of prime factors of k (counted with multiplicity) is n times a prime.
2, 4, 8, 15, 21, 35, 33, 39, 65, 51, 57, 95, 69, 115, 86, 87, 93, 155, 212, 111, 122, 123, 129, 215, 141, 235, 158, 159, 265, 371, 177, 183, 194, 427, 201, 335, 213, 219, 365, 511, 237, 395, 249, 415, 446, 267, 278, 623, 964, 291, 302, 303, 309, 515, 321, 327
Offset: 1
Keywords
Examples
a(19) = 212 because 212 = 2^2 * 53 => sum of prime factors = 2*2+53 = 57 = 19*3 where 3 is prime.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..5000
Programs
-
Maple
sopfr:= proc(n) option remember; add(i[1]*i[2], i=ifactors(n)[2]) end: a:= proc(n) local k, p; for k from 2 while irem (sopfr(k), n, 'p')>0 or not isprime(p) do od; k end: seq (a(n), n=1..100); # Alois P. Heinz, Jun 03 2012 -
Mathematica
sopfr[n_] := Sum[Times @@ f, {f, FactorInteger[n]}]; a[n_] := For[k = 2, True, k++, If[PrimeQ[sopfr[k]/n], Return[k]]]; Array[a, 100] (* Jean-François Alcover, Nov 13 2020 *) -
PARI
sopfr(n) = my(f=factor(n)); sum(k=1,#f~,f[k,1]*f[k,2]); \\ A001414 isok(k, n) = my(dr = divrem(sopfr(k), n)); (dr[2]==0) && isprime(dr[1]); a(n) = {my(k=2); while (!isok(k, n), k++); k;} \\ Michel Marcus, Nov 13 2020
Comments