A308695 a(n) is the minimum positive integer m such that m * 2^(n + 2) + 1 is a prime number which does not divide ((F(n + 2) - 1)^m - 1)/(F(n + 2) - 2), where F(n) is the n-th Fermat number (A000215).

1, 2, 1, 8, 4, 2, 1, 128, 64, 32, 16, 8, 4, 2, 1, 6300, 3150, 26, 13, 579, 1069378, 534689, 10, 5, 387304, 193652, 96826, 48413, 141015, 298082, 149041, 2958, 1479, 51418638746, 25709319373, 20, 10, 5, 6, 3

Offset: 0

Author

Lorenzo Sauras Altuzarra, Feb 11 2020

Keywords

• nonn

Comments

Note that some terms are obtained by dividing the previous one by 2.

From Jinyuan Wang, Feb 18 2020: (Start)

a(n) is the least m such that q = m*2^(n + 2) + 1 is a prime factor of F(n + 2) - 2. Proof: if r(n + 2)/s(n + 2) = ((F(n + 2) - 1)^m - 1)/(F(n + 2) - 2) is not divisible by q, then q divides s(n + 2) because r(n + 2) is always divisible by q (by Fermat's little theorem). Also note that if F(n + 2) - 1 == 1 (mod q), then r(n + 2)/s(n + 2) = Sum_{i = 0..m-1} A001146(n + 2)^i == m (mod q). In conclusion, prime q = m*2^(n + 2) + 1 does not divide r(n + 2)/s(n + 2) if and only if q divides F(n + 2) - 2 = Product_{i = 0..n + 1} F(i).

a(n) always exists because prime factors of F(n) are of the form k*2^(n + 2) + 1. a(n) is not greater than the smallest such k. (End)

a(40) <= 74327396788321657. - Max Alekseyev, Nov 17 2022

Examples

			2 is the minimum positive integer m such that m * 2^(1 + 2) + 1 is a prime number (note that 2 * 2^(1 + 2) + 1 = 17) which does not divide ((F(1 + 2) - 1)^m - 1)/(F(1 + 2) - 2) (note that ((F(1 + 2) - 1)^2 - 1)/(F(1 + 2) - 2) = 257, which is a prime number).
		

Links

• Lorenzo Sauras Altuzarra, Some arithmetical problems that are obtained by analyzing proofs and infinite graphs, arXiv:2002.03075 [math.NT], 2020.

Crossrefs

Cf. A000215 (Fermat numbers), A001146.

Programs

Maple

A308695:=proc(n)
   local m:
   m:=1:
   while not isprime(m*2^(n+2)+1) or (2^(2^(n+2))-1) mod (m*2^(n+2)+1) != 0 do
      m:=m+1:
   od:
   return m:
end proc:

Mathematica

Array[Block[{m = 1}, While[Nand[PrimeQ[#4], Mod[((#3 - 1)^#1 - 1)/(#3 - 2), #4] != 0] & @@ {m, #, 2^(2^(# + 2)) + 1, m*2^(# + 2) + 1}, m++]; m] &, 14] (* Michael De Vlieger, Feb 14 2020 *)

PARI

F(n) = 2^(2^n) + 1;
a(n) = {my(m=1); while (!isprime(p=(m*2^(n+2)+1)) || !((((F(n+2)-1)^m-1)/ (F(n+2)-2)) % p), m++); m;} \\ Michel Marcus, Feb 14 2020

PARI

a(n) = {my(d=4*2^n, q=1); for(m=1, oo, q+=d; if(ispseudoprime(q) && Mod(2, q)^d==1, return(m))); } \\ Jinyuan Wang, Feb 18 2020

Extensions

a(15)-a(32) from Jinyuan Wang, Feb 18 2020

a(33)-a(39) from Max Alekseyev, Nov 17 2022