A238529 a(0) = a(1) = 0, and for n > 1, a(n) = number of iterations of A238525 (n modulo sopfr(n)) needed to reach either 0 or 1. Here sopfr(n) is the sum of the prime factors of n, with multiplicity, A001414.

0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 3, 2, 2, 1, 2, 2, 2, 1, 2, 1, 3, 1, 2, 1, 2, 2, 2, 2, 1, 1, 3, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 1, 1, 1, 3, 3, 2, 2, 2, 1, 2, 3, 3, 1, 1, 2, 2, 2, 2, 1, 3, 2, 2, 3, 2, 2, 2, 1, 2, 3, 2, 1, 3, 1, 3, 1

Offset: 0

Author

J. Stauduhar, Feb 28 2014

Keywords

• nonn

Comments

Previous name was: Recursive depth of n modulo sopfr(n), where sopfr(n) is the sum of the prime factors of n, with multiplicity.

Indices of records are 0, 2, 8, 22, 166, ... (A238530) - David A. Corneth & Antti Karttunen, Oct 20 2017

Examples

			a(2) = 1, because 2 mod sopfr(2) = 2 mod 2 = 0, and further recursion (0 mod sopfr(0)) is undefined.
a(8) = 2, because 8 mod sopfr(8) = 8 mod 6 = 2, and 2 mod sopfr(2) is defined as above, giving 8 a recursive depth of 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..16384

Crossrefs

Cf. A001414, A238525, A238530.

Programs

Mathematica

Array[-1 + Length@ NestWhileList[Mod[#, Total@ Flatten@ Map[ConstantArray[#1, #2] & @@ # &, FactorInteger@ #]] &, #, # > 1 &] &, 105, 0] (* Michael De Vlieger, Oct 20 2017 *)

PARI

A001414(n) = { my(f=factor(n)); sum(k=1, matsize(f)[1], f[k, 1]*f[k, 2]); };
A238525(n) = (n%A001414(n));
A238529(n) = if(n<=1,0,1+A238529(A238525(n))); \\ Antti Karttunen, Oct 20 2017

Sage

def a(n):
    d = 0
    while n>1:
        n = n % sum([f[0]*f[1] for f in factor(n)])
        d = d+1
   return d
# Ralf Stephan, Mar 09 2014

Formula

a(0) = a(1) = 0; for n > 1, a(n) = 1 + a(A238525(n)). - Antti Karttunen, Oct 20 2017

Extensions

More terms from Ralf Stephan, Mar 09 2014

Terms a(0) = a(1) = 0 prepended and name changed by Antti Karttunen, Oct 20 2017