A075660 - OEIS

A075660

Let f(n) = abs(lpd(n)-gpf(n)), where lpd(n) is the largest proper divisor of n and gpf(n) is the greatest prime factor of n. Sequence gives number of iterations for f(n) to reach zero.

1, 2, 3, 1, 2, 1, 2, 3, 1, 1, 2, 4, 5, 1, 1, 2, 3, 2, 3, 3, 1, 1, 2, 2, 1, 1, 2, 3, 4, 2, 3, 2, 1, 1, 1, 2, 3, 1, 1, 2, 3, 2, 3, 3, 2, 1, 2, 2, 1, 4, 1, 6, 7, 3, 1, 2, 1, 1, 2, 2, 3, 1, 2, 3, 1, 2, 3, 4, 1, 4, 5, 2, 3, 1, 4, 4, 1, 2, 3, 2, 3, 1, 2, 2, 1, 1, 1, 2, 3, 3, 1, 3, 1, 1, 1, 3, 4, 3, 2, 3

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

LINKS

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Jason Earls, Smarandache iterations of the first kind on functions involving divisors and prime factors, in Smarandache Notions Journal (2004), Vol. 14.1, page 259.

EXAMPLE

a(12)=4 because 12 -> 3 -> 2 -> 1 -> 0.

MATHEMATICA

Array[-1 + Length@ NestWhileList[Function[n, Abs[If[n == 1, 0, #[[-2]]] - SelectFirst[Reverse@ #, PrimeQ]] &@ Divisors[n]], #, # > 0 &] &, 100] (* Michael De Vlieger, Mar 28 2018 *)

PROG

(PARI) lpd(n)=n/factor(n)[1, 1];

gpf(n)=my(f=factor(n)[, 1]); f[#f];

f(n)=abs(lpd(n)-gpf(n));

a(n)=my(k=1); while(n=f(n), k++); k \\ Charles R Greathouse IV, May 30 2014

CROSSREFS

Sequence in context: A325494 A295561 A076423 * A270788 A190496 A193926

Adjacent sequences: A075657 A075658 A075659 * A075661 A075662 A075663

KEYWORD

nonn

AUTHOR

Jason Earls, Sep 23 2002

STATUS

approved