A327966 Number of iterations of "tamed variant of arithmetic derivative", A327965 needed to reach 0 from n, or -1 if zero is never reached.

0, 1, 2, 2, 2, 2, 3, 2, 3, 4, 3, 2, 2, 2, 5, 3, 3, 2, 5, 2, 4, 4, 3, 2, 3, 4, 4, 2, 3, 2, 3, 2, 3, 6, 3, 3, 4, 2, 5, 2, 3, 2, 3, 2, 3, 3, 5, 2, 3, 6, 4, 3, 6, 2, 3, 2, 3, 4, 3, 2, 3, 2, 7, 4, 3, 6, 3, 2, 6, 5, 3, 2, 3, 2, 3, 3, 3, 6, 3, 2, 3, 2, 3, 2, 3, 4, 4, 3, 4, 2, 4, 3, 4, 4, 7, 4, 3, 2, 7, 4, 4, 2, 4, 2, 3, 3, 3, 2, 3, 2, 4, 4, 4, 2, 3, 3, 4, 4, 3, 4, 3

Offset: 0

Author

Antti Karttunen, Oct 01 2019

Keywords

• nonn

Comments

Conjecture: from all n, zero is eventually reached.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..90609

Crossrefs

Cf. A003415, A256750, A327938, A327965, A327967 (indices of the records).

Programs

PARI

A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
A327938(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]%f[k,1])); factorback(f); };
A327965(n) = if(n<=1,0,A327938(A003415(n)));
A327966(n) = { my(k=0); while(n>0, k++; n = A327965(n)); (k); };
\\ Or alternatively, as a recurrence:
A327966(n) = if(!n,0,1+A327966(A327965(n)));

Formula

a(0) = 0; for n > 0, a(n) = 1 + a(A327965(n)).

a(p) = 2 for all primes p.