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

1,2

Comments

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

Links

Table of n, a(n) for n=1..60.

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.

Example

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
...

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 *)

Crossrefs

Cf. A000040, A063987.

Sequence in context: A030065 A362939 A328676 * A055176 A118267 A324755

Adjacent sequences: A269592 A269593 A269594 * A269596 A269597 A269598

Keyword

nonn,tabf,easy

Author

Wolfdieter Lang, Mar 06 2016

Status

approved