A269598 -id:A269598 - cp's OEIS frontend

A269599 Irregular triangle giving T(n, k) = -(2*A269597(n, k))^(prime(n) -2) modulo prime(n) for n >= 2.

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, 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, 20, 4, 22, 15

Offset: 2

Wolfdieter Lang, Apr 03 2016

The length of row n >= 2 is (prime(n)-1)/2 = A005097(n-1).
The irregular companion triangle -(2*A269596(n, k))^(prime(n) -2) modulo prime(n) is given in A269598.
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.
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.
For a(n), n >= 2, see column z_2 of the table of the paper given as a Wolfdieter Lang link.

			The irregular triangle T(n, k) 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
2,   3:   2
3,   5:   4  3
4,   7:   2  6 4
5,  11:   2  7 3  10  6
6:  13:   4 12 11 10  5  7
7,  17:  15 11 16  3 12 13 10  9
8,  19:   8 12 15 16  6  2 18  5 10
9,  23:  13  3  2  9  4 15 22  6  5  7 12
10, 29:  23  3 19 21 28 27 11 16  5 17 20  4 22 15
...
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).

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

Cf. A000040, A005097, A269597, A269598 (companion).

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 *)

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).

cp's OEIS Frontend

A269599 Irregular triangle giving T(n, k) = -(2*A269597(n, k))^(prime(n) -2) modulo prime(n) for n >= 2.

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