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%I A269597 #9 Apr 05 2016 05:25:19
%S A269597 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,
%T A269597 12,16,11,14,10,17,18,15,19,17,14,20,13,12,21,16,18,22,17,24,16,20,15,
%U A269597 22,25,19,26,23,21,18,27,28
%N 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.
%C A269597 The length of row n >= 2 is (prime(n)-1)/2 = A005097(n-1), and for row n = 1 it is 1.
%C A269597 The other roots of x^2 + b modulo prime(n) from  {0, 1, ..., prime(n)-1} are given in A269596.
%C 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).
%C A269597 Each row n >= 2 consists of a certain permutation of  (prime(n)+1)/2, ..., prime(n) - 1.
%C A269597 For a(n), n >= 2, see column x_2 of the table in the Wolfdieter Lang link.
%H A269597 Wolfdieter Lang, <a href="/A268922/a268922.pdf">Note on a Recurrence for Approximation Sequences of p-adic Square Roots</a>
%F A269597 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.
%F A269597 T(n, k) = prime(n) - A269596(n, k).
%e A269597 The irregular triangle begins (P(n) stands here for prime(n)):
%e A269597 n, P(n)\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14
%e A269597 1,   2:   1
%e A269597 2,   3:   2
%e A269597 3,   5:   3  4
%e A269597 4,   7:   5  4  6
%e A269597 5,  11:   8  7  9  6 10
%e A269597 6:  13:   8  7 10 11  9 12
%e A269597 7,  17:  13 10  9 14 12 15 11 16
%e A269597 8,  19:  13 15 12 16 11 14 10 17 18
%e A269597 9,  23:  15 19 17 14 20 13 12 21 16 18 22
%e A269597 10, 29:  17 24 16 20 15 22 25 19 26 23 21 18 27 28
%e A269597 ...
%e A269597 Row n=7, prime 17 has the permutation (in cycle notation) (9,13,12,14,15,11) (10) (16) of {9, 10, ..., 16}.
%t A269597 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 *)
%Y A269597 Cf. A000040, A005097, A269595, A269596 (x1).
%K A269597 nonn,tabf,easy
%O A269597 1,2
%A A269597 _Wolfdieter Lang_, Apr 03 2016

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