OFFSET
1,2
COMMENTS
Note that A332414 is a subsequence of this sequence.
Prime q = m*2^(n + 2) + 1 does not divide ((F(n + 2) - 1)^m - 1)/F(n) if and only if q divides F(n). Direct implication is Theorem 2.24 of my article (see the links). Proof of the reciprocal implication (by Wang): A001146(n) = 2^(2^n) == - 1 (mod q), so ((F(n + 2) - 1)^m - 1)/F(n) = Sum_{i = 0..4*m-1} (-1)^(i+1)*(2^(2^n))^i == -4*m (mod q).
LINKS
Lorenzo Sauras Altuzarra, Some arithmetical problems that are obtained by analyzing proofs and infinite graphs, arXiv:2002.03075 [math.NT], 2020.
EXAMPLE
3 is a term of this sequence, because B(1,3) = B(2,2) = B(3,1) = 0.
MAPLE
A332416:=proc(n)
local c, i, k, q, r, v:
c:=0:
i:=0:
r:=1:
while c < n do
for k from 0 to r-1 do
q:=(k+1)*2^(r-k+2)+1:
if not isprime(q) or (2^(2^(r-k)) + 1) mod q != 0 then
i:=i+1:
fi:
od:
if i = r then
v:=r:
c:=c+1:
fi:
i:=0:
r:=r+1:
od:
return v:
end proc:
MATHEMATICA
Select[Range@ 29, NoneTrue[Transpose@ {#, Reverse@ #} &@ Range@ #, And[PrimeQ[#4], Mod[((#3 - 1)^#1 - 1)/(2^(2^#2) + 1), #4] != 0] & @@ {#1, #2, 2^(2^(#2 + 2)) + 1, #1*2^(#2 + 2) + 1} & @@ # &] &] (* Michael De Vlieger, Feb 14 2020 *)
PROG
(PARI) isB(m, t) = ispseudoprime(q=4*m*2^t+1) && Mod(2, q)^(2^t)==-1;
isok(r) = sum(i=1, r, isB(i, r-i+1)) == 0; \\ Jinyuan Wang, Feb 18 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Lorenzo Sauras Altuzarra, Feb 12 2020
EXTENSIONS
a(25)-a(68) from Jinyuan Wang, Feb 18 2020
STATUS
approved