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Search: a018845 -id:a018845
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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).
+10
3
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.
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
KEYWORD
easy,nonn
EXTENSIONS
Corrected by Michel Marcus and Robert Israel, Jul 24 2015
STATUS
approved

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