

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 nth Fermat number (A000215).


2



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
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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..m1} 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)


LINKS

Table of n, a(n) for n=0..32.
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)^m1)/ (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

Cf. A000215 (Fermat numbers), A001146.
Sequence in context: A136225 A341724 A089460 * A278111 A223550 A178102
Adjacent sequences: A308692 A308693 A308694 * A308696 A308697 A308698


KEYWORD

nonn,more


AUTHOR

Lorenzo Sauras Altuzarra, Feb 11 2020


EXTENSIONS

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


STATUS

approved



