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(Product k^k) * (Sum 1/k^k) where both the sum and product are over those positive integers k that divide n.
2

%I #12 Mar 23 2017 04:14:45

%S 1,5,28,1284,3126,6485292,823544,21541946368,10847773719,

%T 156290000012500,285311670612,14847746691430172786688,302875106592254,

%U 45756121633729931379676,38327538543365478600000,397378771500072999738379599872,827240261886336764178

%N (Product k^k) * (Sum 1/k^k) where both the sum and product are over those positive integers k that divide n.

%H Alois P. Heinz, <a href="/A057642/b057642.txt">Table of n, a(n) for n = 1..200</a>

%e The positive divisors of 4 are 1, 2, 4. So a(4) = 1^1 *2^2 *4^4 *(1/1^1 +1/2^2 +1/4^4) = 1284.

%p with(numtheory):

%p a:= n-> (l-> mul(k^k, k=l)*add(1/k^k, k=l))(divisors(n)):

%p seq(a(n), n=1..20); # _Alois P. Heinz_, May 22 2015

%t a[n_] := Function[l, Product[k^k, {k, l}] * Sum[1/k^k, {k, l}]] @ Divisors[n]; Array[a, 20] (* _Jean-François Alcover_, Mar 23 2017 *)

%K nonn

%O 1,2

%A _Leroy Quet_, Oct 11 2000