A269597 Irregular triangle giving in row n the larger of the two roots of x^2 + b modulo prime(n) from {0, 1, ..., prime(n)-1} corresponding to b from row n of A269595.

1, 2, 3, 4, 5, 4, 6, 8, 7, 9, 6, 10, 8, 7, 10, 11, 9, 12, 13, 10, 9, 14, 12, 15, 11, 16, 13, 15, 12, 16, 11, 14, 10, 17, 18, 15, 19, 17, 14, 20, 13, 12, 21, 16, 18, 22, 17, 24, 16, 20, 15, 22, 25, 19, 26, 23, 21, 18, 27, 28

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) from {0, 1, ..., prime(n)-1} are given in A269596.

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 (prime(n)+1)/2, ..., prime(n) - 1.

For a(n), n >= 2, see column x_2 of the table in the Wolfdieter Lang link.

Examples

			The irregular triangle 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:   2
3,   5:   3  4
4,   7:   5  4  6
5,  11:   8  7  9  6 10
6:  13:   8  7 10 11  9 12
7,  17:  13 10  9 14 12 15 11 16
8,  19:  13 15 12 16 11 14 10 17 18
9,  23:  15 19 17 14 20 13 12 21 16 18 22
10, 29:  17 24 16 20 15 22 25 19 26 23 21 18 27 28
...
Row n=7, prime 17 has the permutation (in cycle notation) (9,13,12,14,15,11) (10) (16) of {9, 10, ..., 16}.
		

Links

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

Crossrefs

Cf. A000040, A005097, A269595, A269596 (x1).

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[#, 1, -1], 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 larger 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. Only representatives from the smallest nonnegative complete residue class modulo prime(n) are considered.

T(n, k) = prime(n) - A269596(n, k).