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A324598
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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.
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1
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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, 22, 78, 10, 98, 36, 84, 11, 119, 63, 75, 52, 93, 40, 108, 27, 123, 12, 142, 74, 104, 13, 167, 88, 102, 61, 137, 47, 157, 14, 194, 80, 128, 32, 178
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OFFSET
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1,2
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COMMENTS
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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.
For N(1) = 1 every integer solves this Diophantine equation, and the representative solution is 0.
For N(2) = 5 there is only one representative solution, namely 2.
For n >= 3 the representative solutions come in nonnegtive power of 2 pairs (x1, x2) with x2 = N - 1 - x1.
See the link in A089270 to the W. Lang paper, section 3, and Table 6.
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LINKS
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EXAMPLE
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The irregular triangle T(n, k) begins (pairs (x, N - 1 - x) in brackets):
n, N \ k 1 2 3 4 ...
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1, 1: 0
2, 5: 2
3, 11: (3 7)
4, 19: (4 14)
5, 29: (5 23)
6, 31: (12 18)
7, 41: (6 34)
8, 55: (7 47)
9, 59: (25 33)
10, 61: (17 43)
11, 71: (8 62)
12, 79: (29 49)
13, 89: (9 79)
14, 95: (42 52)
15, 101: (22 78)
16, 109: (10 98)
17, 121: (36 84)
18, 131: (11 119)
19, 139: (63 75)
20, 145: (52 93)
....
29, 209: (14 194) (80 128)
...
41, 319: (139 179) (150 168)
...
43, 341: (18 322) (80 260)
...
59, 451: (47 403) (157 293)
...
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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