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A117904 Number triangle [k<=n]*0^abs(L(C(n,2)/3) - L(C(k,2)/3)) where L(j/p) is the Legendre symbol of j and p.

%I A117904 #8 Oct 21 2021 01:30:14
%S A117904 1,1,1,0,0,1,1,1,0,1,1,1,0,1,1,0,0,1,0,0,1,1,1,0,1,1,0,1,1,1,0,1,1,0,
%T A117904 1,1,0,0,1,0,0,1,0,0,1,1,1,0,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,0,1,1,0,0,
%U A117904 1,0,0,1,0,0,1,0,0,1,1,1,0,1,1,0,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,0,1,1,0,1,1
%N A117904 Number triangle [k<=n]*0^abs(L(C(n,2)/3) - L(C(k,2)/3)) where L(j/p) is the Legendre symbol of j and p.
%C A117904 Row sums are A009947(n+2).
%C A117904 Diagonal sums are A117905.
%C A117904 Inverse is A117906.
%C A117904 Equals A117898 mod 2.
%H A117904 G. C. Greubel, <a href="/A117904/b117904.txt">Rows n = 0..50 of the triangle, flattened</a>
%F A117904 G.f.: (1 +x*(1+y) +x^2*y^2 +x^3*y)/((1-x^3)*(1-x^3*y^3)).
%F A117904 T(n, k) = [k<=n] * 2^abs(L(C(n,2)/3) - L(C(k,2)/3)) mod 2.
%e A117904 Triangle begins
%e A117904   1;
%e A117904   1, 1;
%e A117904   0, 0, 1;
%e A117904   1, 1, 0, 1;
%e A117904   1, 1, 0, 1, 1;
%e A117904   0, 0, 1, 0, 0, 1;
%e A117904   1, 1, 0, 1, 1, 0, 1;
%e A117904   1, 1, 0, 1, 1, 0, 1, 1;
%e A117904   0, 0, 1, 0, 0, 1, 0, 0, 1;
%e A117904   1, 1, 0, 1, 1, 0, 1, 1, 0, 1;
%e A117904   1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1;
%t A117904 T[n_, k_]:= If[Abs[JacobiSymbol[Binomial[n, 2], 3] - JacobiSymbol[Binomial[k, 2], 3]]==0, 1, 0];
%t A117904 Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Oct 20 2021 *)
%o A117904 (Sage)
%o A117904 def A117904(n,k): return 1 if abs(jacobi_symbol(binomial(n,2), 3) - jacobi_symbol(binomial(k,2), 3))==0 else 0
%o A117904 flatten([[A117904(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Oct 20 2021
%Y A117904 Cf. A009947, A117898, A117905, A117906.
%K A117904 easy,nonn,tabl
%O A117904 0,1
%A A117904 _Paul Barry_, Apr 01 2006