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%I #9 May 27 2021 17:14:53
%S 1,0,0,1,1,1,1,-1,2,1,2,-1,3,1,3,-1,1,2,1,-2,4,1,4,-1,3,2,3,-2,5,1,5,
%T -1,6,1,6,-1,5,2,5,-2,1,3,1,-3,2,3,2,-3,7,1,7,-1,4,3,3,-3,8,1,8,-1,7,
%U 2,7,-2,5,3,5,-3,1,4,1,-4,9,1,9,-1,3,4,3,-4,7,3,7,-3
%N Irregular triangle read by rows: row n gives the pairs of proper solutions (X, Y), with gcd(X, Y) = 1 and X >= 0, of the Diophantine equation X^2 + 5*Y^2 = A344231(n), for n >= 1.
%C The length of row n is r(n) = 2*A343240(b(n)), if A344231(n) = A343238(b(n)), for n >= 1. This sequence begins 2*(1, 1, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, ...).
%C The number of proper solutions (X, Y), with X > 0, is 1 for n = 1. X = 0 only for n = 2, and for n >= 2 only one half of the solutions are listed, namely those with X >= 0. There are also the solutions with (-X, -Y). Thus the total number of solutions for n >= 2 is actually r(n) given above.
%C For n >= 2 each distinct odd primes from {1, 3, 7, 9} (mod 20), i.e., from
%C A139513, that divides A344231(n) contributes a factor 2 to the number of solutions listed here. See A343238 and the corresponding A343240.
%F T(n, m) gives for m = 2^j-1, the nonnegative X(n, j) solution, and for m = 2*j the Y(n, j) solution of T(n, 2*j-1)^2 + 5*T(n, 2*j)^2 = A344231(n), for j = 1 ..r(n), for n >= 1. For n = 2 (X(2) = 0) the solution (0, -1) is not listed.
%e The irregular triangle T(n, m) begins (A(n) = A344231(n)):
%e n A(n) \ m 1 2 3 4 5 6 7 8 ...
%e 1, 1: 1 0
%e 2, 5: 0 1
%e 3, 6: 1 1 1 -1
%e 4, 9: 2 1 2 -1
%e 5, 14: 3 1 3 -1
%e 6, 21: 1 2 1 -2 4 1 4 -1
%e 7, 29: 3 2 3 -2
%e 8, 30: 5 1 5 -1
%e 9, 41: 6 1 6 -1
%e 10, 45: 5 2 5 -2
%e 11, 46: 1 3 1 -3
%e 12, 49: 2 3 2 -3
%e 13, 54: 7 1 7 -1
%e 14, 61: 4 3 3 -3
%e 15, 69: 8 1 8 -1 7 2 7 -2
%e 16, 70: 5 3 5 -3
%e 17, 81: 1 4 1 -4
%e 18, 86: 9 1 9 -1
%e 19, 89: 3 4 3 -4
%e 20, 94: 7 3 7 -3
%e ...
%e n = 2: Prime 5 is not a member of A139513, therefore only 1 solution appears here (see the remark above on the solution (0, -1)).
%e n = 4: Prime 3 a member A139513. Thus there are 2^1 = 2 solutions are listed. The solution (3, 0) does not appear; it is not proper.
%e n = 6: 21 = A344231(6) = A343238(12) = 3*7. hence A343240(12) = 2^2 = 4 and there are 4 pairs of proper solutions with X >= 0.
%Y Cf. A343238, A343240, A344231, A344234.
%K sign,tabf,easy
%O 1,9
%A _Wolfdieter Lang_, May 17 2021