OFFSET
1,2
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.
LINKS
FORMULA
EXAMPLE
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}.
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 *)
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Wolfdieter Lang, Apr 03 2016
STATUS
approved