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%I #10 Jul 13 2019 14:13:06
%S 0,2,3,7,4,14,5,23,12,18,6,34,7,47,25,33,17,43,8,62,29,49,9,79,42,52,
%T 22,78,10,98,36,84,11,119,63,75,52,93,40,108,27,123,12,142,74,104,13,
%U 167,88,102,61,137,47,157,14,194,80,128,32,178
%N Irregular triangle with the representative solutions of the Diophantine equation x^2 + x - 1 congruent to 0 modulo N(n), with N(n) = A089270(n), for n >= 1.
%C The length of row n is 1 for n = 1 and n = 2, and for n >= 3 it is 2^{r1 + r4} with the number r1 and r4 of distinct primes congruent to 1 and 4 modulo 5, respectively, in the prime number factorization of N(n). E.g., n = 29, N = 209 = 11*19, has r1 = 1 and r4 = 1, with four solutions. The next rows with four solutions are n = 41, 43, 59,..., with N = 319, 341, 451, ... ; for n = 643, 688, 896, ..., with N = 6061, 6479, 8569, ..., there are eight solutions.
%C For N(1) = 1 every integer solves this Diophantine equation, and the representative solution is 0.
%C For N(2) = 5 there is only one representative solution, namely 2.
%C For n >= 3 the representative solutions come in nonnegtive power of 2 pairs (x1, x2) with x2 = N - 1 - x1.
%C See the link in A089270 to the W. Lang paper, section 3, and Table 6.
%e The irregular triangle T(n, k) begins (pairs (x, N - 1 - x) in brackets):
%e n, N \ k 1 2 3 4 ...
%e ----------------------------------
%e 1, 1: 0
%e 2, 5: 2
%e 3, 11: (3 7)
%e 4, 19: (4 14)
%e 5, 29: (5 23)
%e 6, 31: (12 18)
%e 7, 41: (6 34)
%e 8, 55: (7 47)
%e 9, 59: (25 33)
%e 10, 61: (17 43)
%e 11, 71: (8 62)
%e 12, 79: (29 49)
%e 13, 89: (9 79)
%e 14, 95: (42 52)
%e 15, 101: (22 78)
%e 16, 109: (10 98)
%e 17, 121: (36 84)
%e 18, 131: (11 119)
%e 19, 139: (63 75)
%e 20, 145: (52 93)
%e ....
%e 29, 209: (14 194) (80 128)
%e ...
%e 41, 319: (139 179) (150 168)
%e ...
%e 43, 341: (18 322) (80 260)
%e ...
%e 59, 451: (47 403) (157 293)
%e ...
%Y Cf. A089270, A324599 (x^2 - 5 == 0 (mod N)).
%K nonn,tabf
%O 1,2
%A _Wolfdieter Lang_, Jul 08 2019
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