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

0,3

Comments

Conjecture: from all n, zero is eventually reached.

Links

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

Formula

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

a(p) = 2 for all primes p.

Prog

(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)));

Crossrefs

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

Sequence in context: A029233 A147981 A051888 * A366580 A088019 A301508

Adjacent sequences: A327963 A327964 A327965 * A327967 A327968 A327969

Keyword

nonn

Author

Antti Karttunen, Oct 01 2019

Status

approved