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A269598 Irregular triangle giving T(n, k) = -(2*A269596(n, k))^(prime(n)-2) modulo prime(n) for n >= 2.

%I A269598 #15 Dec 14 2023 03:12:30
%S A269598 1,1,2,5,1,3,9,4,8,1,5,9,1,2,3,8,6,2,6,1,14,5,4,7,8,11,7,4,3,13,17,1,
%T A269598 14,9,10,20,21,14,19,8,1,17,18,16,11,6,26,10,8,1,2,18,13,24,12,9,25,7,
%U A269598 14
%N A269598 Irregular triangle giving T(n, k) = -(2*A269596(n, k))^(prime(n)-2) modulo prime(n) for n >= 2.
%C A269598 The length of row n >= 2 is (prime(n)-1)/2 = A005097(n-1).
%C A269598 The irregular companion triangle -(2*A269597(n, k))^(prime(n)-2) modulo prime(n) is given in A269599.
%C A269598 These numbers, called z_1 = z_1(x_1,prime(n)), appear in a recurrence for the approximation sequence {x_n(prime(n), b, x1)} of the p-adic integer sqrt(-b) with entries congruent to x1 modulo prime(n). The irregular triangle for the b values is given in  A269595(n, k) for n >= 2 (odd primes), and A269596(n, k) gives the corresponding x1 values.
%C A269598 T(n, k) is the unique solution of the first order congruence 2*A269596(n, k)*z(n, k) + 1 == 0 (mod prime(n)), with 0 <= z(n, k) <= prime(n)-1, for  n >= 2.
%C A269598 For a(n), n >= 2, see column z_1 of the table of the paper given as a Wolfdieter Lang link.
%H A269598 Wolfdieter Lang, <a href="/A268922/a268922.pdf">Note on a Recurrence for Approximation Sequences of p-adic Square Roots</a>
%F A269598 T(n, k) = modp(-(2*A269596(n, k))^(prime(n) -2), prime(n)), for n >= 2 and k=1, 2, ...., (prime(n)-1)/2, with modp(a, p) giving the number a' from {0, 1, ...,  p-1} with a' == a (mod p).
%F A269598 T(n, k) = prime(n) - A269599(n, k).
%e A269598 The irregular triangle T(n, k) begins (P(n) stands here for prime(n)):
%e A269598 n, P(n)\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14
%e A269598 2,   3:   1
%e A269598 3,   5:   1  2
%e A269598 4,   7:   5  1  3
%e A269598 5,  11:   9  4  8  1  5
%e A269598 6:  13:   9  1  2  3  8  6
%e A269598 7,  17:   2  6  1 14  5  4  7  8
%e A269598 8,  19:  11  7  4  3 13 17  1 14  9
%e A269598 9,  23:  10 20 21 14 19  8  1 17 18 16 11
%e A269598 10, 29:   6 26 10  8  1  2 18 13 24 12  9 25  7 14
%e A269598 ...
%e A269598 T(5, 3) = 8  because 2*A269596(5, 3)*8 + 1 = 2*2*8 + 1 = 33 == 0 mod 11, hence modp(33, 11) = 0 , and 8 is the unique nonnegative solution <= 10 of 2*A269596(5, 3)*z + 1 == 0 (mod 11).
%t A269598 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}; u = 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}]; Table[SelectFirst[Range@ #, Function[z, Mod[-(2 u[[n, k]] z + 1), #] == 0]] &@ Prime@ n, {n, 2, Length@ u}, {k, (Prime@ n - 1)/2}] // Flatten (* _Michael De Vlieger_, Apr 04 2016, Version 10 *)
%Y A269598 Cf. A000040, A005097, A269596, A269599 (companion).
%K A269598 nonn,tabf,easy
%O A269598 2,3
%A A269598 _Wolfdieter Lang_, Apr 03 2016