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a(1) = 1; a(n+1) = (product{k|n} a(k)) (sum{j|n} 1/a(j)), where both the product and sum are over the positive divisors of n.
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%I #13 Mar 28 2017 07:25:29

%S 1,1,2,3,7,8,42,43,304,914,13717,13718,1948004,1948005,165580467,

%T 3808350755,1157738629649,1157738629650,14800530641450464,

%U 14800530641450465,703750431470328179479,90080055228202006973396,2471346315185722061315132977

%N a(1) = 1; a(n+1) = (product{k|n} a(k)) (sum{j|n} 1/a(j)), where both the product and sum are over the positive divisors of n.

%H Alois P. Heinz, <a href="/A068342/b068342.txt">Table of n, a(n) for n = 1..68</a>

%e a(7) = a(1) a(2) a(3) a(6) (1/a(1) + 1/a(2) + 1/a(3) + 1/a(6)) = 1 *1 *2 *8 *(1 + 1 + 1/2 + 1/8) = 42.

%p with(numtheory):

%p a:= proc(n) option remember; `if`(n=1, 1, (l->

%p mul(a(i), i=l)*add(1/a(i), i=l))(divisors(n-1)))

%p end:

%p seq(a(n), n=1..25); # _Alois P. Heinz_, May 22 2015

%t a[1] = 1;

%t a[n_] := a[n] = With[{k = a /@ Divisors[n-1]}, (Times @@ k)*Total[1/k]];

%t Array[a, 25] (* _Jean-François Alcover_, Mar 28 2017 *)

%K nonn

%O 1,3

%A _Leroy Quet_, Feb 27 2002