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%I #13 Oct 01 2019 19:51:47
%S 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,
%T 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,
%U 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
%N "Tamed variant" of arithmetic derivative: a(0) = a(1) = 0; for n > 1, a(n) = A327938(A003415(n)).
%C 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.
%H Antti Karttunen, <a href="/A327965/b327965.txt">Table of n, a(n) for n = 0..20000</a>
%H Antti Karttunen, <a href="/A327965/a327965.txt">Data supplement: n, a(n) computed for n = 0..100000</a>
%F a(0) = a(1) = 0; for n > 1, a(n) = A327938(A003415(n)).
%o (PARI)
%o 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
%o A327938(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]%f[k,1])); factorback(f); };
%o A327965(n) = if(n<=1,0,A327938(A003415(n)));
%Y Cf. A003415, A048103, A327938, A327966, A327967.
%Y Cf. also A327859.
%K nonn
%O 0,7
%A _Antti Karttunen_, Oct 01 2019
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