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.

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, 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, 3, 4, 7, -4, 1, -6, 5, -6

Offset: 1

Author

Wolfdieter Lang, May 17 2021

Keywords

• sign
• tabf
• easy

Comments

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, ...).

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].

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.

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.

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.

Examples

			The irregular triangle T(n, m) begins (A(n) = A344232(n)):
n   A(n) \ m  1  2   3  4   5  6   7  8 ...
1,   2:       1  0
2,   3:       0  1   1 -1
3,   7:       1  1   2 -1
4,  10:       1 -2
5,  15:       2  1   3 -1
6,  18:       1  2   3 -2
7,  23:       1 -3   2 -3
8,  27:       3  1   4 -1
9,  35:       1  3   4 -3
10, 42:       1 -4   3  2   3 -4  5 -2
11, 43:       4  1   5 -1
12, 47:       2  3   5 -3
13, 58:       1  4   5 -4
14, 63:       2 -5   3 -5   5  1   6 -1
15, 67:       1 -5   4 -5
16, 82:       5  2   7 -2
17, 83:       4  3   7 -3
18, 87:       1  5   6  1   6 -5   7 -1
19, 90:       3  4   7 -4
20, 98:       1 -6   5 -6
...
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.
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).
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.
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.
		

Crossrefs

Cf. A343238, A343240, A344232.

Formula

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.