OFFSET
0,2
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
LINKS
Lorenzo Sauras Altuzarra, Some arithmetical problems that are obtained by analyzing proofs and infinite graphs, arXiv:2002.03075 [math.NT], 2020.
EXAMPLE
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).
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Lorenzo Sauras Altuzarra, Feb 11 2020
EXTENSIONS
a(15)-a(32) from Jinyuan Wang, Feb 18 2020
a(33)-a(39) from Max Alekseyev, Nov 17 2022
STATUS
approved