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A332416
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Positive integers r such that B(1,r) = B(2,r - 1) = ... = B(r,1) = 0, where B denotes the function mapping every pair of positive integers (m,n) into 1 if m * 2^(n + 2) + 1 is a prime number dividing F(n), where F(n) denotes the n-th Fermat number (i.e., F(n) = A000215(n)); and into 0 otherwise.
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2
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1, 3, 4, 5, 6, 7, 8, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 26, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57, 58, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 78, 79
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OFFSET
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1,2
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COMMENTS
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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).
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LINKS
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EXAMPLE
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3 is a term of this sequence, because B(1,3) = B(2,2) = B(3,1) = 0.
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MAPLE
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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:
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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