cp's OEIS Frontend

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).

%I A308695 #77 Nov 17 2022 18:41:39
%S A308695 1,2,1,8,4,2,1,128,64,32,16,8,4,2,1,6300,3150,26,13,579,1069378,
%T A308695 534689,10,5,387304,193652,96826,48413,141015,298082,149041,2958,1479,
%U A308695 51418638746,25709319373,20,10,5,6,3
%N 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).
%C A308695 Note that some terms are obtained by dividing the previous one by 2.
%C A308695 From _Jinyuan Wang_, Feb 18 2020: (Start)
%C A308695 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).
%C A308695 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)
%C A308695 a(40) <= 74327396788321657. - _Max Alekseyev_, Nov 17 2022
%H A308695 Lorenzo Sauras Altuzarra, <a href="https://arxiv.org/abs/2002.03075">Some arithmetical problems that are obtained by analyzing proofs and infinite graphs</a>, arXiv:2002.03075 [math.NT], 2020.
%e A308695 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).
%p A308695 A308695:=proc(n)
%p A308695    local m:
%p A308695    m:=1:
%p A308695    while not isprime(m*2^(n+2)+1) or (2^(2^(n+2))-1) mod (m*2^(n+2)+1) != 0 do
%p A308695       m:=m+1:
%p A308695    od:
%p A308695    return m:
%p A308695 end proc:
%t A308695 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 *)
%o A308695 (PARI) F(n) = 2^(2^n) + 1;
%o A308695 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
%o A308695 (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
%Y A308695 Cf. A000215 (Fermat numbers), A001146.
%K A308695 nonn
%O A308695 0,2
%A A308695 _Lorenzo Sauras Altuzarra_, Feb 11 2020
%E A308695 a(15)-a(32) from _Jinyuan Wang_, Feb 18 2020
%E A308695 a(33)-a(39) from _Max Alekseyev_, Nov 17 2022