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The least m > 1 such that the arithmetic derivative of m*n is a multiple of the arithmetic derivative of n.
5

%I #10 Oct 10 2019 20:32:28

%S 2,2,2,2,6,2,8,4,10,2,4,2,14,12,4,2,10,2,8,21,22,2,24,4,26,2,12,2,30,

%T 2,6,33,34,8,6,2,18,12,40,2,42,2,8,22,46,2,10,4,14,32,33,2,8,12,56,24,

%U 30,2,56,2,62,40,8,65,66,2,65,69,70,2,22,2,45,24,32,65,78,2,24,4,82,2,30,24,50,16,88,2,42,32,20,40

%N The least m > 1 such that the arithmetic derivative of m*n is a multiple of the arithmetic derivative of n.

%H Antti Karttunen, <a href="/A328236/b328236.txt">Table of n, a(n) for n = 2..16385</a>

%H Antti Karttunen, <a href="/A328236/a328236.txt">Data supplement: n, a(n) computed for n = 2..65537</a>

%e Arithmetic derivative of 6 is 6' = A003415(6) = 5. Taking arithmetic derivatives of successive multiples of six 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) = 6.

%o (PARI)

%o A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));

%o A328236(n) = { my(d=A003415(n)); for(m=2,oo,if(!(A003415(n*m)%d), return(m))); };

%Y Cf. A003415, A328235, A328238 (gives the corresponding quotients).

%K nonn

%O 2,1

%A _Antti Karttunen_, Oct 08 2019