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%I #16 Oct 23 2021 12:38:56
%S 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,
%T 13,17,12,19,5,11,22,28,8,10,27,29,12,27,6,16,29,39,9,37,19,28,22,28,
%U 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
%N 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.
%C 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.
%C The length of row n of the triangle is A347833(n).
%C 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.
%D D. A. Buell, Binary Quadratic Forms, Springer, 1989.
%D A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, Sammlung Göschen Band 5131, Walter de Gruyter, 1973.
%e The irregular triangle T with A(n) = A347831(n) begins:
%e n A(n) \ k 1 2 3 4 ...
%e 1, 1: 0
%e 2, 2: 0 1
%e 3, 3: 1
%e 4, 4: 0 3
%e 5, 5: 2
%e 6, 6: 1 4
%e 7, 8: 3 4
%e 8, 10: 2 7
%e 9, 12: 4 7
%e 10, 15: 7
%e 11, 16: 3 12
%e 12, 17: 5 11
%e 13, 19: 8 10
%e 14, 20: 7 12
%e 15, 23: 6 16
%e 16, 24: 4 19
%e 17, 30: 7 22
%e 18, 31: 6 35
%e 19, 32: 12 19
%e 20, 34: 5 11 22 28
%e ...
%o (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
%o 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
%Y Cf. A000217, A347831, A347833.
%K nonn,tabf,easy
%O 1,6
%A _Wolfdieter Lang_, Sep 15 2021
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