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A347832
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Irregular triangle T, read by rows, giving the solutions x for x*(x + 1) == -4 (mod A347831 (n)), for x from {0, 1, 2, ..., A347831(n)-1}, for n >= 1.
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2
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0, 0, 1, 1, 0, 3, 2, 1, 4, 3, 4, 2, 7, 4, 7, 7, 3, 12, 5, 11, 8, 10, 7, 12, 6, 16, 4, 19, 7, 22, 13, 17, 12, 19, 5, 11, 22, 28, 8, 10, 27, 29, 12, 27, 6, 16, 29, 39, 9, 37, 19, 28, 22, 28, 20, 32, 10, 46, 7, 52, 15, 45, 13, 17, 44, 48, 19, 44, 11, 28, 39, 56, 16, 52, 8, 27, 48, 67, 35, 43, 12, 67, 31, 51
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OFFSET
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1,6
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COMMENTS
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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.
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REFERENCES
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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.
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LINKS
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EXAMPLE
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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
...
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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STATUS
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approved
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