|
%I #2 Mar 31 2012 13:20:35
%S 3,13,3553,4410001,117279379988209,2331742837337539921,
%T 1542124619897735720591412481,50763923011615077618322055249233,
%U 103811229730824186819640201001214367336729
%N Quotients A120492(p)/p, where p = Prime[n].
%C Prime p divides A120492(p), where A120492(n) = 1 + Sum[ Prime[k]^(n-1), {k,1,n}] = {2, 6, 39, 504, 17765, 552552, 30870007, ...}.
%F a(n) = A120492(Prime(n))/Prime(n). a(n) = (1 + Sum[ Prime[k]^(Prime[n]-1), {k,1,Prime[n]} ])/Prime[n].
%t Table[(1 + Sum[ Prime[k]^(Prime[n]-1), {k,1,Prime[n]} ])/Prime[n],{n,1,15}]
%Y Cf. A120492, A121166, A120329.
%K nonn
%O 1,1
%A _Alexander Adamchuk_, Feb 03 2007
|
