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A328238 Quotient A003415(n*m)/A003415(n) for the least m > 1 for which such a quotient is an integer. Here A003415(x) is the arithmetic derivative of x.

%I A328238 #7 Oct 10 2019 20:32:53
%S A328238 4,5,3,7,12,9,16,10,20,13,7,15,28,42,6,19,16,21,18,42,44,25,48,14,52,
%T A328238 3,26,31,60,33,8,66,68,43,9,39,56,51,80,43,84,45,19,37,92,49,13,18,24,
%U A328238 236,46,55,16,67,112,138,88,61,116,63,124,124,12,130,132,69,82,138,140,73,28,75,119,84,108,142,156,81,44,7,164
%N A328238 Quotient A003415(n*m)/A003415(n) for the least m > 1 for which such a quotient is an integer. Here A003415(x) is the arithmetic derivative of x.
%H A328238 Antti Karttunen, <a href="/A328238/b328238.txt">Table of n, a(n) for n = 2..16385</a>
%H A328238 Antti Karttunen, <a href="/A328238/a328238.txt">Data supplement: n, a(n) computed for n = 2..65537</a>
%F A328238 a(n) = A003415(n*A328236(n)) / A003415(n).
%e A328238 Arithmetic derivative of 6 is 6' = A003415(6) = 5. Taking arithmetic derivatives of its successive multiples, we obtain 12' = 16, 18' = 21, 24' = 44, 30' = 31, and not until with A003415(6*6) = 36' = 60 we obtain a multiple of 5. Thus a(6) = 60/5 = 12.
%o A328238 (PARI)
%o A328238 A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
%o A328238 A328238(n) = { my(d=A003415(n),t); for(m=2,oo,if(!((t=A003415(n*m))%d), return(t/d))); };
%Y A328238 Cf. A003415, A328237.
%Y A328238 Cf. A328236 (gives the corresponding m).
%K A328238 nonn
%O A328238 2,1
%A A328238 _Antti Karttunen_, Oct 08 2019