OFFSET
1,1
COMMENTS
Smallest k such that sopfr(k) = n*p, p prime.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..5000
EXAMPLE
a(19) = 212 because 212 = 2^2 * 53 => sum of prime factors = 2*2+53 = 57 = 19*3 where 3 is prime.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jun 02 2012
STATUS
approved