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Numbers n such that Sum_{i=1..j} 1/pn(i) + Sum_{i=1..k} 1/pd(i) is an integer, where pn are the prime factors of n and pd the prime factors of the arithmetic derivative of n, both counted with multiplicity.
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%I #10 Jan 10 2025 02:02:04

%S 4,27,256,1728,2401,3125,11664,72000,78732,200000,486000,531441,

%T 823543,1350000,3280500,9112500,22143375,52706752,56250000,61509375,

%U 156250000

%N Numbers n such that Sum_{i=1..j} 1/pn(i) + Sum_{i=1..k} 1/pd(i) is an integer, where pn are the prime factors of n and pd the prime factors of the arithmetic derivative of n, both counted with multiplicity.

%e Arithmetic derivative of 2401 is 1372. Prime factors of 2401 are 7^4; prime factors of 1372 are 2^2, 7^3 and 1/7 + 1/7 + 1/7 + 1/7 + 1/2 + 1/2 + 1/7 + 1/7 + 1/7 = 2.

%p with(numtheory); P:= proc(q) local a,b,c,n,p;

%p for n from 2 to q do a:=n*add(op(2,p)/op(1,p),p=ifactors(n)[2]);

%p b:=ifactors(a)[2]; c:=ifactors(n)[2]; if type(add(c[k][2]/c[k][1],k=1..nops(c))+add(b[k][2]/b[k][1],k=1..nops(b)),integer) then print(n); fi; od; end: P(10^9);

%Y Cf. A003415, A239491.

%K nonn,more

%O 1,1

%A _Paolo P. Lava_, Mar 20 2014

%E a(15)-a(21) from _Giovanni Resta_, Mar 20 2014