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A117898
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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.
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5
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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, 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, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1
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OFFSET
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0,4
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COMMENTS
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LINKS
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FORMULA
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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)).
T(n, k) = [k<=n]*2^abs(L(C(n,2)/3) - L(C(k,2)/3)).
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EXAMPLE
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Triangle begins
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, 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, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1;
1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1;
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MATHEMATICA
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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[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 *)
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PROG
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(Sage)
def T(n, k): return 2^abs(kronecker(binomial(n, 2), 3) - kronecker(binomial(k, 2), 3))
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 27 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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