A269595 Irregular triangle in which n-th row the gives quadratic residues prime(n)- m modulo prime(n), for m from {1, 2, ..., prime(n)-1}, in increasing order.

1, 2, 1, 4, 3, 5, 6, 2, 6, 7, 8, 10, 1, 3, 4, 9, 10, 12, 1, 2, 4, 8, 9, 13, 15, 16, 2, 3, 8, 10, 12, 13, 14, 15, 18, 5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22, 1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28

Offset: 1

Author

Wolfdieter Lang, Mar 06 2016

Keywords

• nonn
• tabf
• easy

Comments

The length of row 1 is 1 and of row n, n >= 2, is (prime(n)-1)/2, where prime(n) = A000040(n).

Examples

			The irregular triangle T(n, k) begins (P(n) is here prime(n)):
n, P(n)\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14
1,   2:   1
2,   3:   2
3,   5:   1  4
4,   7:   1  2  4
5,  11:   1  3  4  5  9
6:  13:   1  3  4  9 10 12
7,  17:   1  2  4  8  9 13 15 16
8,  19:   1  4  5  6  7  9 11 16 17
9,  23:   1  2  3  4  6  8  9 12 13 16 18
10, 29:   1  4  5  6  7  9 13 16 20 22 23 24 25 28
...
		

Crossrefs

Cf. A000040, A063987.

Programs

Mathematica

t = Table[Select[Range[Prime@ n - 1], JacobiSymbol[#, Prime@ n] == 1 &], {n, 10}]; Table[Prime@ n - t[[n, (Prime@ n - 1)/2 - k + 1]], {n, Length@ t}, {k, (Prime@ n - 1)/2}] /. {} -> 1 // Flatten (* Michael De Vlieger, Mar 31 2016, after Jean-François Alcover at A063987 *)

Formula

For n = 1, prime(1) = 2: 1, and for odd primes n >= 2: the increasing values of m from {1, 2, ..., p-1} with the Legendre symbol (-m/prime(n)) = + 1.

T(n, k) = prime(n) - A063987(n,(prime(n)-1)/2-k+1). k=1..(prime(n)-1)/2, for n >= 2, and T(1, 1) = 1.