A327965 "Tamed variant" of arithmetic derivative: a(0) = a(1) = 0; for n > 1, a(n) = A327938(A003415(n)).

0, 0, 1, 1, 1, 1, 5, 1, 3, 6, 7, 1, 1, 1, 9, 2, 2, 1, 21, 1, 6, 10, 13, 1, 11, 10, 15, 1, 2, 1, 31, 1, 5, 14, 19, 3, 15, 1, 21, 1, 17, 1, 41, 1, 3, 39, 25, 1, 7, 14, 45, 5, 14, 1, 3, 1, 23, 22, 31, 1, 23, 1, 33, 51, 3, 18, 61, 1, 18, 26, 59, 1, 39, 1, 39, 55, 5, 18, 71, 1, 11, 1, 43, 1, 31, 22, 45, 2, 35, 1, 123, 5, 6

Offset: 0

Author

Antti Karttunen, Oct 01 2019

Keywords

• nonn

Comments

Applying A327938 to the result of A003415(n) ensures that all terms stay in A048103, and that all iteration paths will (hopefully) terminate in zero. See A327966.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..20000
• Antti Karttunen, Data supplement: n, a(n) computed for n = 0..100000

Crossrefs

Cf. A003415, A048103, A327938, A327966, A327967.

Cf. also A327859.

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)));

Formula

a(0) = a(1) = 0; for n > 1, a(n) = A327938(A003415(n)).