OFFSET
1,6
COMMENTS
T(n, k) gives the solutions x from {0, 1, ..., A347831(n) - 1} of the congruence (x + 1)*x + 4 == 0 (mod A347831(n)), for n >= 1. No other positive modulus has a solution.
The length of row n of the triangle is A347833(n).
The present congruence 2*T(x) + 4 == 0 (mod k), for k >= 1, with the triangular numbers T(n) = A000217(n), is equivalent to the congruence s^2 + 15 == 0 (mod 4*k) where s = 2*x + 1. Each of these two congruences has a solution for k >= 1 if and only if k is prepresented by some positive definite binary quadratic form of discriminant disc = -15. See e.g., Buell Proposition 41, p. 50, or Scholz-Schoeneberg Satz 74, p. 105.
REFERENCES
D. A. Buell, Binary Quadratic Forms, Springer, 1989.
A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, Sammlung Göschen Band 5131, Walter de Gruyter, 1973.
EXAMPLE
The irregular triangle T with A(n) = A347831(n) begins:
n A(n) \ k 1 2 3 4 ...
1, 1: 0
2, 2: 0 1
3, 3: 1
4, 4: 0 3
5, 5: 2
6, 6: 1 4
7, 8: 3 4
8, 10: 2 7
9, 12: 4 7
10, 15: 7
11, 16: 3 12
12, 17: 5 11
13, 19: 8 10
14, 20: 7 12
15, 23: 6 16
16, 24: 4 19
17, 30: 7 22
18, 31: 6 35
19, 32: 12 19
20, 34: 5 11 22 28
...
PROG
(PARI) isok(m) = {my(f=factor(m)); for (k=1, #f~, my(p=f[k, 1]); if ((p==3) || (p==5), if (f[k, 2] > 1, return (0)), if (kronecker(p, 15) != 1, return(0))); ); return (1); } \\ A347831
tabf(nn) = {for (n=1, nn, if (isok(n), for (x=0, n-1, if (Mod(x*(x+1), n) == -4, print1(x, ", ")); ); ); ); } \\ Michel Marcus, Oct 23 2021
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Wolfdieter Lang, Sep 15 2021
STATUS
approved