A269599 - OEIS

%I #10 Dec 14 2023 03:12:20

%S 2,4,3,2,6,4,2,7,3,10,6,4,12,11,10,5,7,15,11,16,3,12,13,10,9,8,12,15,

%T 16,6,2,18,5,10,13,3,2,9,4,15,22,6,5,7,12,23,3,19,21,28,27,11,16,5,17,

%U 20,4,22,15

%N Irregular triangle giving T(n, k) = -(2*A269597(n, k))^(prime(n) -2) modulo prime(n) for n >= 2.

%C The length of row n >= 2 is (prime(n)-1)/2 = A005097(n-1).

%C The irregular companion triangle -(2*A269596(n, k))^(prime(n) -2) modulo prime(n) is given in A269598.

%C These numbers, called z_2 = z_2(x_2, prime(n)),  appear in a recurrence for the approximation sequence {x_n(prime(n), b, x_2)}  of the p-adic integer sqrt(-b) with entries congruent to x_2 modulo prime(n). The irregular triangle for the b values is given in  A269595(n, k) for n >= 2 (odd primes), and A269597(n, k) gives the corresponding x_2 values.

%C T(n, k) is the unique solution of the first order congruence 2*A269597(n, k)*z(n, k) + 1 == 0 (mod prime(n)), with 0 <= z(n, k) <= prime(n)-1, for  n >= 2.

%C For a(n), n >= 2, see column z_2 of the table of the paper given as a Wolfdieter Lang link.

%H Wolfdieter Lang, <a href="/A268922/a268922.pdf">Note on a Recurrence for Approximation Sequences of p-adic Square Roots</a>

%F T(n, k) = modp( -(2*A269597(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).

%e The irregular triangle T(n, k) begins (P(n) stands here for prime(n)):

%e n, P(n)\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14

%e 2,   3:   2

%e 3,   5:   4  3

%e 4,   7:   2  6 4

%e 5,  11:   2  7 3  10  6

%e 6:  13:   4 12 11 10  5  7

%e 7,  17:  15 11 16  3 12 13 10  9

%e 8,  19:   8 12 15 16  6  2 18  5 10

%e 9,  23:  13  3  2  9  4 15 22  6  5  7 12

%e 10, 29:  23  3 19 21 28 27 11 16  5 17 20  4 22 15

%e ...

%e T(5, 3) = 3  because 2*A269597(5, 3)*3 + 1 = 2*9*3 + 1 = 55 == 0 mod 11, hence modp(55, 11) = 0, and 3 is the unique nonnegative solution <= 10 of  2*A269597(5, 3)*z + 1 == 0 (mod 11).

%t 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[#, 1, -1], 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 Cf. A000040, A005097, A269597, A269598 (companion).

%K nonn,tabf,easy

%O 2,1

%A _Wolfdieter Lang_, Apr 03 2016