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Number triangle 2^abs(L(C(n,2)/3) - L(C(k,2)/3))*[k<=n] where L(j/p) is the Legendre symbol of j and p.
5

%I #9 Oct 01 2021 13:10:42

%S 1,1,1,2,2,1,1,1,2,1,1,1,2,1,1,2,2,1,2,2,1,1,1,2,1,1,2,1,1,1,2,1,1,2,

%T 1,1,2,2,1,2,2,1,2,2,1,1,1,2,1,1,2,1,1,2,1,1,1,2,1,1,2,1,1,2,1,1,2,2,

%U 1,2,2,1,2,2,1,2,2,1,1,1,2,1,1,2,1,1,2,1,1,2,1

%N Number triangle 2^abs(L(C(n,2)/3) - L(C(k,2)/3))*[k<=n] where L(j/p) is the Legendre symbol of j and p.

%C Row sums are A117899. Diagonal sums are A117900. Inverse is A117901. A117898 mod 2 is A117904.

%H G. C. Greubel, <a href="/A117898/b117898.txt">Table of n, a(n) for the first 101 rows, flattened</a>

%F G.f.: (1 +x*(1+y) +x^2*(2+2*y+y^2) +x^3*y(1+2*y) +2*x^4*y^2)/((1-x^3)*(1-x^3*y^3)).

%F T(n, k) = [k<=n]*2^abs(L(C(n,2)/3) - L(C(k,2)/3)).

%e Triangle begins

%e 1;

%e 1, 1;

%e 2, 2, 1;

%e 1, 1, 2, 1;

%e 1, 1, 2, 1, 1;

%e 2, 2, 1, 2, 2, 1;

%e 1, 1, 2, 1, 1, 2, 1;

%e 1, 1, 2, 1, 1, 2, 1, 1;

%e 2, 2, 1, 2, 2, 1, 2, 2, 1;

%e 1, 1, 2, 1, 1, 2, 1, 1, 2, 1;

%e 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1;

%e 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1;

%e 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1;

%t Flatten[CoefficientList[CoefficientList[Series[(1 +x(1+y) +x^2(2+2y+y^2) +x^3*y(1 +2y) +2x^4*y^2)/((1-x^3)(1-x^3*y^3)), {x,0,15}, {y,0,15}], x], y]] (* _G. C. Greubel_, May 03 2017 *)

%t T[n_, k_]:= 2^Abs[JacobiSymbol[Binomial[n, 2], 3] - JacobiSymbol[Binomial[k, 2], 3]]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* _G. C. Greubel_, Sep 27 2021 *)

%o (Sage)

%o def T(n, k): return 2^abs(kronecker(binomial(n,2), 3) - kronecker(binomial(k,2), 3))

%o flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Sep 27 2021

%Y Cf. A117898, A117899, A117900, A117901, A117904.

%K easy,nonn,tabl

%O 0,4

%A _Paul Barry_, Apr 01 2006