login
Numbers k such that ((k+1)^k-1)/k^2 is a prime.
4

%I #12 Nov 01 2024 21:11:43

%S 2,3,5,17,4357

%N Numbers k such that ((k+1)^k-1)/k^2 is a prime.

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

%C It seems that if p is in the sequence then the first three numbers k such that k^2 divides (p+1)^k-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,changed

%O 1,1

%A _Farideh Firoozbakht_ and _Alexander Adamchuk_, Mar 13 2007