A210844 A pair of solutions of a congruence related to A141453.

3, 5, 9, 15, 33, 63, 255, 513, 16383, 131073, 262143, 1048575, 4294967295, 4611686018427387903, 1237940039285380274899124223, 324518553658426726783156020576255, 340282366920938463463374607431768211455

Offset: 1

Author

Wolfdieter Lang, Mar 28 2012

Keywords

• nonn

Comments

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.

Examples

			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)
		

Crossrefs

Cf. A141453.

Formula

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.