This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A230077 #22 May 07 2015 18:46:47
%S A230077 1,1,4,1,4,2,1,4,9,5,3,1,4,9,3,12,10,1,4,9,16,8,2,15,13,1,4,9,16,6,17,
%T A230077 11,7,5,1,4,9,16,2,13,3,18,12,8,6,1,4,9,16,25,7,20,6,23,13,5,28,24,22,
%U A230077 1,4,9,16,25,5,18,2,19,7,28,20,14,10,8
%N A230077 Table a(n,m) giving in row n all k from {1, 2, ..., prime(n)-1} for which the Legendre symbol (k/prime(n)) = +1, for odd prime(n).
%C A230077 The length of row n is r(n) = (prime(n) - 1)/2, with prime(n) = A000040(n), n >= 2.
%C A230077 If k from {1, 2, ..., p-1} appears in row n then the Legendre symbol (k/prime(n)) = +1 otherwise it is -1.
%C A230077 The Legendre symbol (k/p), p an odd prime and gcd(k,p) = 1, is +1 if there exists an integer x with x^2 == k (mod p) and -1 otherwise. It is sufficient to consider k from {1, 2, ..., p-1} (gcd(0,p) = p, not 1) because (k/p) = ((k + l*p)/p) for integer l. Because (p - x)^2 == x^2 (mod p), it is also sufficient to consider only x^2 from {1^2, 2^2, ..., ((p-1)/2)^2} which are pairwise incongruent modulo p. See the Hardy-Wright reference. p. 68-69.
%C A230077 For odd primes p one has for the Legendre symbol ((p-1)/p) = (-1/p) = (-1)^(r(n)) (see above for the row length r(n), and theorem 82, p. 69 of Hardy-Wright), and this is +1 for prime p == 1 (mod 4) and -1 for p == 3 (mod 4). Therefore k = p-1 appears in row n iff p = prime(n) is from A002144 = 5, 13, 17, 29, 37, 41,...
%C A230077 For n>=4 (prime(n)>=7) at least one of the integers 2, 3, or 6 appears in every row. - _Geoffrey Critzer_, May 01 2015
%D A230077 G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. Fifth ed., Clarendon Press, 2003.
%H A230077 Alois P. Heinz, <a href="/A230077/b230077.txt">Rows n = 2..100, flattened</a>
%F A230077 a(n,m) = m^2 (mod prime(n)), n >= 2, where prime(n) = A000040(n), m = 1, 2, ..., (prime(n) - 1)/2.
%e A230077 The irregular table a(n,m) begins (here p(n)=prime(n)):
%e A230077 n, p(n)\m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
%e A230077 2, 3: 1
%e A230077 3, 5: 1 4
%e A230077 4, 7: 1 4 2
%e A230077 5, 11: 1 4 9 5 3
%e A230077 6, 13: 1 4 9 3 12 10
%e A230077 7, 17: 1 4 9 16 8 2 15 13
%e A230077 8, 19: 1 4 9 16 6 17 11 7 5
%e A230077 9, 23: 1 4 9 16 2 13 3 18 12 8 6
%e A230077 10, 29: 1 4 9 16 25 7 20 6 23 13 5 28 24 22
%e A230077 11, 31 1 4 9 16 25 5 18 2 19 7 28 20 14 10 8
%e A230077 ...
%e A230077 Row n=12, p(n)=37: 1, 4, 9, 16, 25, 36, 12, 27, 7, 26, 10, 33, 21, 11, 3, 34, 30, 28.
%e A230077 Row n=13, p(n)=41: 1, 4, 9, 16, 25, 36, 8, 23, 40, 18, 39, 21, 5, 32, 20, 10, 2, 37, 33, 31.
%e A230077 (2/p) = +1 for n=4, p(4) = 7; p(7) = 17, p(9) = 23, p(11) = 31, p(13) = 41, ... This leads to A001132 (primes 1 or 7 (mod 8)).
%e A230077 4 = 5 - 1 appears in row n=3 for p(3)=5 because 5 is from A002144. 10 cannot appear in row 5 for p(5)=11 because 11 == 3 (mod 4), hence 11 is not in A002144.
%p A230077 T:= n-> (p-> seq(irem(m^2, p), m=1..(p-1)/2))(ithprime(n)):
%p A230077 seq(T(n), n=2..12); # _Alois P. Heinz_, May 07 2015
%t A230077 Table[Table[Mod[a^2, p], {a, 1, (p - 1)/2}], {p,
%t A230077 Prime[Range[2, 20]]}] // Grid (* _Geoffrey Critzer_, Apr 30 2015 *)
%Y A230077 Cf. A000040, A002144, A063987.
%K A230077 nonn,tabf,easy
%O A230077 2,3
%A A230077 _Wolfdieter Lang_, Oct 25 2013