1,1

All terms are primes. Corresponding primes of the form ((n+1)^n-1)/n^2 are listed in A128466(n) = {2, 7, 311, 7563707819165039903, ...}.

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.

No other terms below 20000. - Max Alekseyev (maxal(AT)cs.ucsd.edu), Apr 25 2007

4357 is in the sequence because (4358^4357-1)/4357^2 is prime.

Cf. A128466 = Primes of the form ((n+1)^n-1)/n^2. Cf. A060071, A060072, A060073 = (n^(n-1)-1)/(n-1)^2. Cf. A058128. Cf. A128456 = least prime factor of ((p+1)^p - 1)/p^2, where p = Prime[n].

Cf. A127103, A127104, A127106, A128398.

Sequence in context: A099936 A092506 A127063 this_sequence A004249 A121510 A132346

Adjacent sequences: A127834 A127835 A127836 this_sequence A127838 A127839 A127840

hard,more,nonn

Farideh Firoozbakht (mymontain(AT)yahoo.com) and Alexander Adamchuk (alex(AT)kolmogorov.com), Mar 13 2007