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
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.
CROSSREFS
KEYWORD
sign,tabf,easy
AUTHOR
Wolfdieter Lang, May 17 2021
STATUS
approved