A016837 Primes reached after k iterations of sum of n and its prime divisors = t (where t replaces n in each iteration).

23, 11, 23, 17, 11, 23, 23, 23, 17, 47, 19, 41, 23, 23, 47, 53, 41, 59, 29, 31, 47, 71, 47, 47, 41, 71, 71, 89, 71, 167, 83, 47, 53, 47, 71, 113, 59, 71, 71, 269, 83, 131, 59, 167, 71, 167, 59, 149, 167, 71, 167, 191, 83, 71, 167, 79, 89, 179, 251, 227, 167, 149, 149, 83, 269, 239, 89, 167, 251, 263, 251, 251, 113, 239, 149, 167

Offset

2,1

Comments

Patrick asked what composite would produce 666 or 313 iterations. Carlos has also been working on the problem and asks if there is a run of 3 primes produced by consecutive composites. So original idea belongs to Patrick. This sequence was calculated by Enoch.

Links

Robert Israel, Table of n, a(n) for n = 2..10000

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 940

Formula

Factor n, add n and its prime divisors. Sum = t, t replaces n, repeat until a prime is produced.

Example

Starting from 4, 4=2*2, so 4+2+2=8. 8=2*2*2 so 8+2+2+2=14. 14=2*7 so 14+2+7=23, prime is 23 in 3 iterations.

Maple

f:= proc(n) option remember; local t;
  t:= n + add(f[1]*f[2], f=ifactors(n)[2]);
  if isprime(t) then return t
  else f(t)
  fi;
end proc:
map(f, [$2 .. 100]); # Robert Israel, Jul 24 2015

Mathematica

a[n_] := a[n] = Module[{t, f = FactorInteger[n]}, t = n + f[[All, 1]].f[[All, 2]]; If[PrimeQ[t], Return[t], a[t]]];
Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Sep 16 2022 *)

Prog

(PARI) sfpn(n) = {my(f = factor(n)); n + sum(k=1, #f~, f[k, 1]*f[k, 2]); }
a(n) = {while (! isprime(t=sfpn(n)), n=t); t; } \\ Michel Marcus, Jul 24 2015

Crossrefs

Cf. A096461, A018845.

Sequence in context: A281924 A054574 A363371 * A226218 A294087 A323137

Adjacent sequences: A016834 A016835 A016836 * A016838 A016839 A016840

Keyword

easy,nonn

Author

Enoch Haga, Carlos Rivera, Patrick De Geest

Extensions

Corrected by Michel Marcus and Robert Israel, Jul 24 2015

Status

approved