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Sum of A003415(d) over the proper divisors d of n, where A003415 is arithmetic derivative.
6

%I #6 Oct 02 2018 17:39:37

%S 0,0,0,1,0,2,0,5,1,2,0,11,0,2,2,17,0,13,0,13,2,2,0,39,1,2,7,15,0,23,0,

%T 49,2,2,2,54,0,2,2,49,0,27,0,19,16,2,0,115,1,19,2,21,0,61,2,59,2,2,0,

%U 98,0,2,18,129,2,35,0,25,2,31,0,170,0,2,20,27,2,39,0,149,34,2,0,120,2,2,2,79,0,120,2,31,2,2,2,307,0,25,22,92,0

%N Sum of A003415(d) over the proper divisors d of n, where A003415 is arithmetic derivative.

%H Antti Karttunen, <a href="/A319683/b319683.txt">Table of n, a(n) for n = 1..65537</a>

%F a(n) = Sum_{d|n, d<n} A003415(d).

%F a(n) = A319684(n) - 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 A319683(n) = sumdiv(n,d,(d<n)*A003415(d));

%Y Cf. A003415, A319684.

%Y Cf. also A300251, A300252, A305809, A317835.

%K nonn

%O 1,6

%A _Antti Karttunen_, Oct 02 2018