A048124 Becomes prime or 4 after exactly 2 iterations of f(x) = sum of prime factors of x.

8, 9, 21, 25, 30, 32, 35, 36, 42, 50, 57, 60, 64, 72, 81, 85, 86, 93, 102, 111, 115, 121, 122, 138, 145, 146, 159, 164, 174, 182, 187, 194, 215, 219, 235, 236, 237, 253, 258, 260, 265, 266, 282, 284, 287, 289, 302, 303, 308, 312, 318, 319, 326, 329, 338, 346

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Comments

f(x) = sum of prime factors with multiplicity, so that f(1500) = 2+2+3+5+5+5 = 22.

Numbers k such that A002217(k) = 3. - Andrew Howroyd, Sep 15 2019

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002217, A048125, A048126, A048127, A048128, A048129, A048130, A048131, A048132.

Programs

Mathematica

okQ[n_]:=Module[{lst=NestList[Total[Times@@@FactorInteger[#]]&,n,2]}, !PrimeQ[First[lst]] &&!PrimeQ[lst[[2]]]&&First[lst]!=4&&lst[[2]]!=4&&(PrimeQ[Last[lst]]||Last[lst]==4)]; Select[Range[400], okQ] (* Harvey P. Dale, Mar 23 2011 *)

PARI

sopfr(n)={my(f=factor(n)); sum(i=1, #f~, f[i,1]*f[i,2])}
ok(n)={forstep(k=2, 1, -1, n=sopfr(n); if(n==4||isprime(n), return(k==1))); 0}
select(ok, [1..500]) \\ Andrew Howroyd, Sep 14 2019