A344233 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.

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, -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, 2, 7, -2, 5, 3, 5, -3, 1, 4, 1, -4, 9, 1, 9, -1, 3, 4, 3, -4, 7, 3, 7, -3

Offset

1,9

Comments

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, ...).

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.

For n >= 2 each distinct odd primes from {1, 3, 7, 9} (mod 20), i.e., from

A139513, that divides A344231(n) contributes a factor 2 to the number of solutions listed here. See A343238 and the corresponding A343240.

Links

Table of n, a(n) for n=1..84.

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 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.

Example

The irregular triangle T(n, m) begins (A(n) = A344231(n)):
n   A(n) \ m  1  2   3  4   5  6   7  8 ...
1,   1:       1  0
2,   5:       0  1
3,   6:       1  1   1 -1
4,   9:       2  1   2 -1
5,  14:       3  1   3 -1
6,  21:       1  2   1 -2   4  1   4 -1
7,  29:       3  2   3 -2
8,  30:       5  1   5 -1
9,  41:       6  1   6 -1
10, 45:       5  2   5 -2
11, 46:       1  3   1 -3
12, 49:       2  3   2 -3
13, 54:       7  1   7 -1
14, 61:       4  3   3 -3
15, 69:       8  1   8 -1   7  2   7 -2
16, 70:       5  3   5 -3
17, 81:       1  4   1 -4
18, 86:       9  1   9 -1
19, 89:       3  4   3 -4
20, 94:       7  3   7 -3
...
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)).
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.
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.

Crossrefs

Cf. A343238, A343240, A344231, A344234.

Sequence in context: A322317 A263259 A137163 * A072625 A268187 A232440

Adjacent sequences: A344230 A344231 A344232 * A344234 A344235 A344236

Keyword

sign,tabf,easy

Author

Wolfdieter Lang, May 17 2021

Status

approved