A231813 Number of iterations of A046665(n) = (greatest prime divisor of n) - (least prime divisor of n) [with A046665(1) = 0] required to reach zero.

0, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 3, 2, 2, 1, 2, 3, 2, 1, 2, 1, 2, 2, 3, 1, 2, 1, 2, 3, 2, 1, 2, 3, 2, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 3, 3, 2, 1, 2, 1, 3, 2, 2, 2, 2, 1, 2, 1, 4, 1, 2, 3, 2, 3, 2, 1, 2, 3, 3, 3, 3, 3, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2

Offset: 0

Author

Clark Kimberling, Dec 11 2013

Keywords

• nonn

Examples

			A046665(6) = 3 - 2, and A046665(1) = 0, so a(6) = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..16384 (terms 1..1000 from Clark Kimberling)
• Antti Karttunen, Data supplement: n, a(n) computed for n = 0..100000

Crossrefs

Cf. A006530, A020639, A046665, A233510.

Programs

Mathematica

z = 400; h[n_] := h[n] = FactorInteger[n][[-1, 1]] - FactorInteger[n][[1, 1]]; t[n_] := Drop[FixedPointList[h, n], -2]; Table[t[n], {n, 1, z}]; a = Table[Length[t[n]], {n, 1, z}]

PARI

A046665(n) = if(1==n,0, my(f = factor(n), lpf = f[1, 1], gpf = f[#f~, 1]); (gpf-lpf));
A231813(n) = if(0==n,0, 1+A231813(A046665(n))); \\ Antti Karttunen, Jan 03 2019

Formula

a(0) = 0; for n > 0, a(n) = 1 + a(A046665(n)). - Antti Karttunen, Jan 03 2019

Extensions

Name edited, term a(0)=0 prepended and more terms added by Antti Karttunen, Jan 03 2019