cp's OEIS Frontend

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.

%I A269595 #13 Oct 13 2017 00:21:22
%S A269595 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,
%T A269595 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,
%U A269595 22,23,24,25,28
%N 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.
%C A269595 The length of row 1 is 1 and of row n, n >= 2, is (prime(n)-1)/2, where prime(n) = A000040(n).
%F A269595 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.
%F A269595 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.
%e A269595 The irregular triangle T(n, k) begins (P(n) is here prime(n)):
%e A269595 n, P(n)\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14
%e A269595 1,   2:   1
%e A269595 2,   3:   2
%e A269595 3,   5:   1  4
%e A269595 4,   7:   1  2  4
%e A269595 5,  11:   1  3  4  5  9
%e A269595 6:  13:   1  3  4  9 10 12
%e A269595 7,  17:   1  2  4  8  9 13 15 16
%e A269595 8,  19:   1  4  5  6  7  9 11 16 17
%e A269595 9,  23:   1  2  3  4  6  8  9 12 13 16 18
%e A269595 10, 29:   1  4  5  6  7  9 13 16 20 22 23 24 25 28
%e A269595 ...
%t A269595 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 *)
%Y A269595 Cf. A000040, A063987.
%K A269595 nonn,tabf,easy
%O A269595 1,2
%A A269595 _Wolfdieter Lang_, Mar 06 2016