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%I #20 Nov 03 2017 03:40:48
%S 0,1,256,512,1152,1728,1920,3072,3456,7776,11664,12800,12960,20736,
%T 23328,52488,72000,78732,81920,86400,87480,100352,110208,124800,
%U 139968,153216,157464,200000,219520,263424,294912,321024,336000,354294,400000,486000,486720,531441
%N Numbers k = a * b, such that k' = a' * b' where k', a' and b' are the arithmetic derivatives of k, a and b.
%C A046311 is a subset of this sequence.
%C Some numbers admit more than one couple of divisors a, b: 3456 = 8 * 432 = 54 * 64 and 3456' = 15552 = 8' * 432' = 12 * 1296 = 54' * 64' = 81 * 192.
%C Apart from the first term, squares of A165558 are part of the sequence. In A165558 n' = 2 * n and therefore (n^2)' = 2 * n * n' = 2 * n * 2 * n = (2 * n)^2. Thus n^2 = n * n and (n^2)' = n' * n'.
%H Paolo P. Lava, <a href="/A294153/b294153.txt">Table of n, a(n) for n = 1..100</a>
%e a(0) = 0 because 0 = 0 * b and 0' = 0' * b' = 0;
%e a(1) = 1 because 1 = 1 * 1 and 1' = 1' * 1' = 0;
%e a(2) = 256 because 256 = 16 * 16 and 256' = 16' * 16' = 32 * 32 = 1024;
%e a(3) = 512 because 512 = 8 * 64 and 512' = 8' * 64' = 12 * 192 = 2304.
%p with(numtheory): P:=proc(q) local a,b,c,j,k,n,p;
%p for n from 1 to q do j:=sort([op(divisors(n))]);
%p for k from 2 to trunc((nops(j)+1)/2) do
%p a:=j[k]*add(op(2,p)/op(1,p), p=ifactors(j[k])[2]);
%p b:=(n/j[k])*add(op(2,p)/op(1,p), p=ifactors(n/j[k])[2]);
%p c:=n*add(op(2,p)/op(1,p), p=ifactors(n)[2]);
%p if c=a*b then print(n); break; fi; od; od; end: P(10^6);
%Y Cf. A003415, A046311, A165558.
%K nonn
%O 1,3
%A _Paolo P. Lava_, Oct 24 2017
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