OFFSET
0,1
COMMENTS
Presumably all terms are prime numbers, if they exist.
EXAMPLE
a(0) = 2 because ((2^0+1)^2-1)/2^0 = 3 (prime);
a(1) = 3 because ((2^1+1)^3-1)/2^1 = 13 (prime);
a(2) = 3 because ((2^2+1)^3-1)/2^2 = 43 (prime);
a(4) = 3 because ((2^4+1)^3-1)/2^4 = 307 (prime);
a(5) = 3 because ((2^5+1)^3-1)/2^5 = 1123 (prime).
MATHEMATICA
a[n_]:=Module[{k=1}, If[n==3, -1, While[!PrimeQ[((2^n+1)^k-1)/(2^n)], k++]; k]]; Array[a, 13, 0] (* James C. McMahon, Dec 03 2025 *)
PROG
(Magma) [2, 3, 3, -1] cat [Min([k: k in [1..100] | ((2^n+1)^k-1) mod 2^n eq 0 and IsPrime(((2^n+1)^k-1) div 2^n)]): n in [4..12]];
(PARI) a(n) = if(n==3, return(-1)); my(k=1); while (!isprime(((2^n+1)^k - 1)/2^n), k++); k; \\ Michel Marcus, Nov 26 2025
CROSSREFS
KEYWORD
sign,more
AUTHOR
Juri-Stepan Gerasimov, Nov 25 2025
STATUS
approved
