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