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A344234 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 2*X^2 + 2*X*Y + 3*Y^2 = A344232(n), for n >= 1.

%I A344234 #16 Feb 15 2024 08:19:42
%S A344234 1,0,0,1,1,-1,1,1,2,-1,1,-2,2,1,3,-1,1,2,3,-2,1,-3,2,-3,3,1,4,-1,1,3,
%T A344234 4,-3,1,-4,3,2,3,-4,5,-2,4,1,5,-1,5,2,7,-2,4,3,7,-3,1,5,6,1,6,-5,7,-1,
%U A344234 3,4,7,-4,1,-6,5,-6
%N A344234 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 2*X^2 + 2*X*Y + 3*Y^2 = A344232(n), for n >= 1.
%C A344234 The length of row n is r(n) = 2*A343240(b(n)), if A344232(n) = A343238(b(n)), for n >= 1. This sequence begins 2*(1, 2, 2, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, ...).
%C A344234 See A344231 for references and links on parallel forms and half-reduced right neighbor forms (R-transformations), and also for the remark on the equivalent reduced form [2, -2, 3].
%C A344234 The number of proper solutions (X, Y), with X > 0, is 1 for n = 1 and 4. X = 0 only for n = 2, but the solution (0, -1) = (-0, -1) is not listed here.
%C A344234 For other n each distinct odd prime from {1, 3, 7, 9} (mod 20), i.e., from A139513, that divides A344232(n) contributes a factor of 2 to the listed  number of solutions. See A343238 and A343240 for the multiplicities.
%C A344234 Only solutions with nonnegative X are listed. There is also the corresponding solution (-X, -Y). Hence the total number of signed solution is twice the number considered here.
%F A344234 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 2*T(n, 2*j-1)^2 + 2*T(n, 2*j-1)*T(n, 2*j) + 3*T(n, 2*j)^2 =  A344232(n), for j = 1, 2, ..., r(n), for n >= 1. For n = 2 the solution (0, -1) is not listed here.
%e A344234 The irregular triangle T(n, m) begins (A(n) = A344232(n)):
%e A344234 n   A(n) \ m  1  2   3  4   5  6   7  8 ...
%e A344234 1,   2:       1  0
%e A344234 2,   3:       0  1   1 -1
%e A344234 3,   7:       1  1   2 -1
%e A344234 4,  10:       1 -2
%e A344234 5,  15:       2  1   3 -1
%e A344234 6,  18:       1  2   3 -2
%e A344234 7,  23:       1 -3   2 -3
%e A344234 8,  27:       3  1   4 -1
%e A344234 9,  35:       1  3   4 -3
%e A344234 10, 42:       1 -4   3  2   3 -4  5 -2
%e A344234 11, 43:       4  1   5 -1
%e A344234 12, 47:       2  3   5 -3
%e A344234 13, 58:       1  4   5 -4
%e A344234 14, 63:       2 -5   3 -5   5  1   6 -1
%e A344234 15, 67:       1 -5   4 -5
%e A344234 16, 82:       5  2   7 -2
%e A344234 17, 83:       4  3   7 -3
%e A344234 18, 87:       1  5   6  1   6 -5   7 -1
%e A344234 19, 90:       3  4   7 -4
%e A344234 20, 98:       1 -6   5 -6
%e A344234 ...
%e A344234 n = 2: The prime 3 is a member of A139513, hence 2^1 = 2 solutions are listed. There are also the corresponding (-X, -Y) solutions.
%e A344234 n = 4: 10 = A344232(4) = A343238(8) = 2*5, A343240(8) = 1, hence there is 1 pair of proper solution with X >= 0. This is because neither 2 nor 5  are primes from A139513. There is also the solution (-1, 2).
%e A344234 n = 6: Prime 3 is a member of A139513, not prime 2. This there are 2 solutions listed. The solution (3, 0) does not appear; it is not proper.
%e A344234 n = 10: 42 = A344232(10) = A343238(19) = 2*3*7,  A343240(19) = 2^2 = 4, hence there are 4 pairs of proper solution with X >= 0. 3 and 7 are primes from A139513.
%Y A344234 Cf. A343238, A343240, A344232.
%K A344234 sign,tabf,easy
%O A344234 1,9
%A A344234 _Wolfdieter Lang_, May 17 2021