A269596 Irregular triangle giving in row n the smaller of the two roots x1 of x^2 + b modulo prime(n) from {0, 1, ..., prime(n)-1} corresponding to b from row n of A269595.

1, 1, 2, 1, 2, 3, 1, 3, 4, 2, 5, 1, 5, 6, 3, 2, 4, 1, 4, 7, 8, 3, 5, 2, 6, 1, 6, 4, 7, 3, 8, 5, 9, 2, 1, 8, 4, 6, 9, 3, 10, 11, 2, 7, 5, 1, 12, 5, 13, 9, 14, 7, 4, 10, 3, 6, 8, 11, 2, 1

Offset: 1

Author

Wolfdieter Lang, Apr 03 2016

Keywords

• nonn
• tabf
• easy

Comments

The length of row n >= 2 is (prime(n)-1)/2 = A005097(n-1), and for row n = 1 it is 1.

The other roots of x^2 + b modulo prime(n) are given in A269597.

See A269595 for the irregular triangle with the quadratic residues -b modulo prime(n) = A000040(n), for n >= 1. For n=1 (prime 2) there is a double root x1 = x2 = 1 of x^2 + 1 (mod 2).

Each row n >= 2 consists of a certain permutation of 1, 2, ..., (prime(n)-1)/2.

For a(n), n >= 2, see column x_1 of the table in the Wolfdieter Lang link.

Examples

			The irregular triangle T(n, k) begins (P(n) stands here for prime(n)):
n, P(n)\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14
1,   2:   1
2,   3:   1
3,   5:   2  1
4,   7:   2  3  1
5,  11:   3  4  2  5  1
6:  13:   5  6  3  2  4  1
7,  17:   4  7  8  3  5  2  6  1
8,  19:   6  4  7  3  8  5  9  2  1
9,  23:   8  4  6  9  3 10 11  2  7  5  1
10, 29:  12  5 13  9 14  7  4 10  3  6  8 11  2  1
...
Row n=7 (prime 17) is the permutation (in cycle notation) (1,4,3,8)(2,7,6) of {1, 2, ..., 8}.
		

Links

• Wolfdieter Lang, Note on a Recurrence for Approximation Sequences of p-adic Square Roots

Crossrefs

Cf. A000040, A005097, A269595, A269597 (x2).

Programs

Mathematica

nn = 12; s = Table[Select[Range[Prime@ n - 1], JacobiSymbol[#, Prime@ n] == 1 &], {n, nn}]; t = Table[Prime@ n - s[[n, (Prime@ n - 1)/2 - k + 1]], {n, Length@ s}, {k, (Prime@ n - 1)/2}] /. {} -> {1};
Prepend[Table[SelectFirst[Range@ #, Function[x, Mod[x^2 + t[[n, k]], #] == 0]] &@ Prime@ n, {n, 2, Length@ t}, {k, (Prime@ n - 1)/2}], {1}] // Flatten (* Michael De Vlieger, Apr 04 2016, Version 10 *)

Formula

T(n, k) gives the smaller zero of x^2 + A269595(n, k) == 0 (mod prime(n)), n >= 1, for k=1 if n=1 and k = 1, 2, ..., (prime(n)-1)/2 = A005097(n-1) for n >= 2. Representatives are taken from the complete residue class {0, 1 ,..., prime(n)-1}.

T(n, k) = prime(n) - A269597(n, k).