A343239 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.

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, 25, 10, 25, 6, 35, 11, 17, 25, 31, 9, 34, 20, 25, 15, 31, 18, 29, 17, 32, 7, 47, 13, 45, 19, 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

Offset: 1

Author

Wolfdieter Lang, May 16 2021

Keywords

• nonn
• tabf
• easy

Comments

The length of row n is A343240(n).

Examples

			The irregular triangle T with A(n) = A343238(n) begins:
   n  A(n) \ k  1  2  3  4 ...
  ---------------------------
   1,  1:       0
   2,  2:       1
   3,  3:       1  2
   4,  5:       0
   5,  6:       1  5
   6,  7:       3  4
   7,  9:       2  7
   8, 10:       5
   9, 14:       3 11
  10, 15:       5 10
  11, 18:       7 11
  12, 21:       4 10 11 17
  13, 23:       8 15
  14, 27:       7 20
  15, 29:      13 16
  16, 30:       5 25
  17, 35:      10 25
  18, 41:       6 35
  19, 42:      11 17 25 31
  20, 43:       9 34
  ...
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1468 (first 500 rows)

Crossrefs

Cf. A343238, A343240.

Programs

PARI

isok(k) = issquare(Mod(-5, k)); \\ A343238
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

Formula

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.