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%I #19 Jan 06 2024 14:32:10
%S 0,1,1,2,0,1,5,3,4,2,7,5,3,11,5,10,7,11,4,10,11,17,8,15,7,20,13,16,5,
%T 25,10,25,6,35,11,17,25,31,9,34,20,25,15,31,18,29,17,32,7,47,13,45,19,
%U 42,11,25,38,52,14,53,8,31,38,61,25,45,20,61,35,47,24,59,9,77,13,16,71,74
%N Irregular triangle read by rows giving the solutions x for x^2 == -5 (mod A343238(n)), for x from {0, 1, 2, ..., A343238(n)-1}, for n >= 1.
%C The length of row n is A343240(n).
%H Andrew Howroyd, <a href="/A343239/b343239.txt">Table of n, a(n) for n = 1..1468</a> (first 500 rows)
%F T(n, k) gives the solutions x from {0, 1, ..., A343238(n)-1} of the congruence x^2 + 5 == 0 (mod A343238(n)), for n >= 1.
%e The irregular triangle T with A(n) = A343238(n) begins:
%e n A(n) \ k 1 2 3 4 ...
%e ---------------------------
%e 1, 1: 0
%e 2, 2: 1
%e 3, 3: 1 2
%e 4, 5: 0
%e 5, 6: 1 5
%e 6, 7: 3 4
%e 7, 9: 2 7
%e 8, 10: 5
%e 9, 14: 3 11
%e 10, 15: 5 10
%e 11, 18: 7 11
%e 12, 21: 4 10 11 17
%e 13, 23: 8 15
%e 14, 27: 7 20
%e 15, 29: 13 16
%e 16, 30: 5 25
%e 17, 35: 10 25
%e 18, 41: 6 35
%e 19, 42: 11 17 25 31
%e 20, 43: 9 34
%e ...
%o (PARI) isok(k) = issquare(Mod(-5, k)); \\ A343238
%o lista(nn) = my(list = List()); for (n=1, nn, if (issquare(Mod(-5, n)), my(row = select(x->(Mod(x,n)^2 + 5 == 0), [0..n-1])); listput(list, row))); Vec(list); \\ _Michel Marcus_, Sep 17 2023
%Y Cf. A343238, A343240.
%K nonn,tabf,easy
%O 1,4
%A _Wolfdieter Lang_, May 16 2021