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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. 4
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 (list; graph; refs; listen; history; text; internal format)
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
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
Sequence in context: A030065 A362939 A328676 * A055176 A118267 A324755
KEYWORD
nonn,tabf,easy
AUTHOR
Wolfdieter Lang, Mar 06 2016
STATUS
approved

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Last modified April 12 19:05 EDT 2024. Contains 371636 sequences. (Running on oeis4.)