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Primes reached after k iterations of sum of n and its prime divisors = t (where t replaces n in each iteration).
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%I #30 Sep 16 2022 04:03:43

%S 23,11,23,17,11,23,23,23,17,47,19,41,23,23,47,53,41,59,29,31,47,71,47,

%T 47,41,71,71,89,71,167,83,47,53,47,71,113,59,71,71,269,83,131,59,167,

%U 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

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

%C 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.

%H Robert Israel, <a href="/A016837/b016837.txt">Table of n, a(n) for n = 2..10000</a>

%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=940">Encyclopedia of Combinatorial Structures 940</a>

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

%e 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.

%p f:= proc(n) option remember; local t;

%p t:= n + add(f[1]*f[2],f=ifactors(n)[2]);

%p if isprime(t) then return t

%p else f(t)

%p fi;

%p end proc:

%p map(f, [$2 .. 100]); # _Robert Israel_, Jul 24 2015

%t a[n_] := a[n] = Module[{t, f = FactorInteger[n]}, t = n + f[[All, 1]].f[[All, 2]]; If[PrimeQ[t], Return[t], a[t]]];

%t Table[a[n], {n, 2, 100}] (* _Jean-François Alcover_, Sep 16 2022 *)

%o (PARI) sfpn(n) = {my(f = factor(n)); n + sum(k=1, #f~, f[k,1]*f[k,2]);}

%o a(n) = {while (! isprime(t=sfpn(n)), n=t); t;} \\ _Michel Marcus_, Jul 24 2015

%Y Cf. A096461, A018845.

%K easy,nonn

%O 2,1

%A _Enoch Haga_, _Carlos Rivera_, _Patrick De Geest_

%E Corrected by _Michel Marcus_ and _Robert Israel_, Jul 24 2015