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A210844
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A pair of solutions of a congruence related to A141453.
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1
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3, 5, 9, 15, 33, 63, 255, 513, 16383, 131073, 262143, 1048575, 4294967295, 4611686018427387903, 1237940039285380274899124223, 324518553658426726783156020576255, 340282366920938463463374607431768211455
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OFFSET
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1,1
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COMMENTS
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See the comment on A141453. There r(a(n)) is the present a(n).
The next entry a(18) has 158 digits.
The sequence of exponents of 2 of the Fermat and Mersenne primes FM:=A141453 (including the prime 2) starts with k:=[0, 1, 2, 3, 4, 5, 7, 8, 13, 16, 17, 19, 31, 61, 89, 107, 127, 521,...], n>=1.
For the second k entry one can also take 2 instead of 1. Then a(2) should be replaced by 7.
a(n) and FM(n)*2^(k(n)+1) - a(n) are an incongruent pair of solutions of the congruence x^2 == 1 (mod FM(n)*2^(k(n)+1)), n>=1. For n>=3 there are all-together eight incongruent solutions. The trivial pair of positive solutions is always 1 and FM(n)*2^(k(n)+1) - 1. Two more pairs should therefore be found.
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LINKS
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FORMULA
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a(n) = sqrt(FM(n)*2^(k(n)+2) + 1), n>=1, with FM(n):=A141453(n) and the sequence k is given for n=1..18 in the comment section.
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EXAMPLE
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From Wolfdieter Lang, Apr 10 2012 (Start)
a(1)=3 because 3^2 = 9 == 1 (mod 2*2^(0+1)) = 1 (mod 4). The incongruent companion solution is 4 - 3 = 1. This is the trivial pair of solutions.
a(2)=5 because 5^2 = 25 == 1 (mod 3*2^(1+1)) = 1 (mod 12). The incongruent companion solution is 12 - 5 = 7, obtained also by taking k(2)=2. The trivial pair of solutions is (1,11).
1, 5, 7 and 11 are all the solutions of this congruence.
a(3)=9 because 9^2 = 81 == 1 (mod 5*2^(2+1)) = 1 (mod 40).
The companion solution is 40 - 9 = 31. The trivial pair is (1,39). The missing two pairs are (11,29) and (19,21), and all eight incongruent solutions are 1, 9, 11, 19, 21, 29, 31 and 39.
(End)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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