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%I #7 Mar 31 2012 13:20:34
%S 2,3,5,17,4357
%N Numbers n such that ((n+1)^n-1)/n^2 is a prime.
%C All terms are primes. Corresponding primes of the form ((n+1)^n-1)/n^2 are listed in A128466(n) = {2, 7, 311, 7563707819165039903, ...}.
%C It seems that if p is in the sequence then the first three numbers n such that n^2 divides (p+1)^n-1 are: 1, p & ((p+1)^p-1)/p. 2 is in the sequence and the first three terms of A127103 are : 1, 2 & ((2+1)^2-1)/2; 3 is in the sequence and the first three terms of A127104 are : 1, 3 & ((3+1)^3-1)/3; 5 is in the sequence and the first three terms of A127106 are : 1, 5 & ((5+1)^5-1)/5.
%C No other terms below 20000. - _Max Alekseyev_, Apr 25 2007
%e 4357 is in the sequence because (4358^4357-1)/4357^2 is prime.
%Y Cf. A128466, A037205, A060072, A060073, A058128, A128456, A127103, A127104, A127106, A128398.
%K hard,more,nonn
%O 1,1
%A _Farideh Firoozbakht_ and _Alexander Adamchuk_, Mar 13 2007
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