A324599 - OEIS

%I #11 Jul 13 2019 14:16:01

%S 0,0,4,7,9,10,11,18,6,25,13,28,15,40,8,51,26,35,17,54,20,59,19,70,10,

%T 85,45,56,21,88,48,73,23,108,12,127,40,105,68,81,55,96,25,130,30,149,

%U 27,154,14,177,76,123,95,110,29,180,48,161,65,146

%N Irregular triangle with the representative solutions of the Diophantine equation x^2 - 5 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.

%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 0.

%C For n >= 3 the solutions come in a nonnegative power of 2 pairs, each of the type (x1, x2) with x2 = N - x1.

%C See the link in A089270 to the W. Lang paper, section 3, and Table 7.

%e The irregular triangle T(n, k) begins (pairs (x, N - x) in brackets):

%e n,    N \ k   1   2     3   4  ...

%e ----------------------------------

%e 1,    1:      0

%e 2,    5:      0

%e 3,   11:     (4   7)

%e 4,   19:     (9  10)

%e 5,   29:    (11  18)

%e 6,   31:     (6  25)

%e 7,   41:    (13  28)

%e 8,   55:    (15  40)

%e 9,   59:     (8  51)

%e 10,  61:    (26  35)

%e 11,  71:    (17  54)

%e 12,  79:    (20  59)

%e 13,  89:    (19  70)

%e 14,  95:    (10  85)

%e 15, 101:    (45  56)

%e 16, 109:    (21  88)

%e 17, 121:    (48  73)

%e 18, 131:    (23 108)

%e 19, 139:    (12 127)

%e 20, 145:    (40 105)

%e ....

%e 29, 209:    (29 180)  (48 161)

%e ...

%e 41, 319:    (18 301)  (40 279)

%e ...

%e 43, 341:    (37 304) (161 180)

%e ...

%e 59, 451:    (95 356) (136 315)

%Y Cf. A089270, A324598 (x^2 + x - 1 == 0 (mod N)).

%K nonn,tabf

%O 1,3

%A _Wolfdieter Lang_, Jul 08 2019