A117904 -id:A117904 - cp's OEIS frontend

A117906 Inverse of number triangle A117904.

1, -1, 1, 0, 0, 1, 0, -1, 0, 1, 0, 0, 0, -1, 1, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, -1, 0, 1, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1

Offset: 0

Paul Barry, Apr 01 2006

Row sums are (1, 0, 1, 0, 0, 0, ...) with g.f. 1 + x^2.

Diagonal sums are A117907.

			Triangle begins
   1;
  -1,  1;
   0,  0,  1;
   0, -1,  0,  1;
   0,  0,  0, -1,  1;
   0,  0, -1,  0,  0,  1;
   0,  0,  0,  0, -1,  0,  1;
   0,  0,  0,  0,  0,  0, -1,  1;
   0,  0,  0,  0,  0, -1,  0,  0,  1;
   0,  0,  0,  0,  0,  0,  0, -1,  0,  1;
   0,  0,  0,  0,  0,  0,  0,  0,  0, -1,  1;
   0,  0,  0,  0,  0,  0,  0,  0, -1,  0,  0,  1;
   0,  0,  0,  0,  0,  0,  0,  0,  0,  0, -1,  0,  1;
   0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, -1, 1;
   0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, -1,  0, 0, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A117904, A117907.

M[n_, k_]:= M[n, k]= If[k>n, 0, If[Abs[JacobiSymbol[Binomial[n, 2], 3] - JacobiSymbol[Binomial[k, 2], 3]]==0, 1, 0]];
m:= m= With[{q=20}, Table[M[n, k], {n,0,q}, {k,0,q}]];
T[n_, k_]:= Inverse[m][[n+1, k+1]];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 20 2021 *)

G.f.: (1 -x*(1-y) +x^2*y^2 -x^3*y -x^5*y^2)/(1-x^3*y^3).

A117898 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.

Original entry on oeis.org

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

Offset: 0

Paul Barry, Apr 01 2006

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

			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;

G. C. Greubel, Table of n, a(n) for the first 101 rows, flattened

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

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 *)

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

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)).

A117908 Chequered (or checkered) triangle for odd prime p=3.

Original entry on oeis.org

1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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

Offset: 0

Paul Barry, Apr 01 2006

Row sums are A117909.
Diagonal sums are A117910.
For odd prime p, T(n,k;p) = [k<=n]*0^abs(L(C(n,p-1)/p) - 2*L(C(k,p-1)/p)) defines a checkered triangle for p.

			Triangle begins
  1;
  1, 1;
  0, 0, 0;
  1, 1, 0, 1;
  1, 1, 0, 1, 1;
  0, 0, 0, 0, 0, 0;
  1, 1, 0, 1, 1, 0, 1;
  1, 1, 0, 1, 1, 0, 1, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 0;
  1, 1, 0, 1, 1, 0, 1, 1, 0, 1;
  1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  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;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A117904, A117909, A117910.

A117908:= func< n,k | (n mod 3) lt 2 and (k mod 3) lt 2 select 1 else 0>;
[A117908(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Nov 18 2021

T[n_, k_]:= If[Abs[JacobiSymbol[Binomial[n, 2], 3] - 2*JacobiSymbol[Binomial[k, 2], 3]]==0, 1, 0];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 21 2021 *)

def A117908(n, k): return 1 if (n%3<2 and k%3<2) else 0
flatten([[A117908(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Oct 21 2021

G.f.: (1 +x*(1+y) +x^3*y)/((1-x^3)*(1-x^3*y^3)).
T(n,k) = [k<=n] * 0^abs(L(C(n,2)/3) - 2*L(C(k,2)/3)) where L(j/p) is the Legendre symbol of j and p.
T(n, k) = 1 if (n mod 3) < 2 and (k mod 3) < 2, otherwise 0. - Kevin Ryde, Oct 21 2021

A117905 Expansion of (1+2x+2x^2)/((1+x)*(1-x^3)^2).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 2, 4, 2, 2, 6, 2, 3, 7, 3, 3, 9, 3, 4, 10, 4, 4, 12, 4, 5, 13, 5, 5, 15, 5, 6, 16, 6, 6, 18, 6, 7, 19, 7, 7, 21, 7, 8, 22, 8, 8, 24, 8, 9, 25, 9, 9, 27, 9, 10, 28, 10, 10, 30, 10, 11, 31, 11, 11, 33, 11, 12, 34, 12, 12, 36

Offset: 0

Paul Barry, Apr 01 2006

Diagonal sums of number triangle A117904.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,0,2,2,0,-1,-1).

Cf. A014125, A117904, A144677.

R:=PowerSeriesRing(Integers(), 75); Coefficients(R!( (1+2*x+2*x^2)/((1+x)*(1-x^3)^2) )); // G. C. Greubel, Oct 18 2021

LinearRecurrence[{-1,0,2,2,0,-1,-1}, {1,1,1,1,3,1,2}, 75] (* G. C. Greubel, Oct 10 2021 *)

lista(n) = {my(x = 'x + 'x*O('x^n)); P = (1+2*x+2*x^2) / ((1-x^3)*(1+x-x^3-x^4)); Vec(P);}  \\ Michel Marcus, Mar 20 2013

def A117905_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1+2*x+2*x^2)/((1+x)*(1-x^3)^2) ).list()
A117905_list(75) # G. C. Greubel, Oct 18 2021

a(n) = -a(n-1) + 2*a(n-3) + 2*a(n-4) - a(n-6) - a(n-7).
a(n) = Sum_{k=0..floor(n/2)} 0^abs(L(C(n-k,2)/3) - L(C(k,2)/3)), where L(j/p) is the Legendre symbol of j and p.
From G. C. Greubel, Oct 18 2021: (Start)
a(n) = (1/36)*(10*n + 23 + (-1)^n*(9 + 16*u(n, 1/2) - 4*u(n-1, 1/2) - 12*Sum_{j=0..n} u(n-j, 1/2)*u(j, 1/2))), where u(n, x) = ChebyshevU(n, x).
a(n) = (1/36)*(39 + 30*n + 9*(-1)^n - 48*floor((n+2)/3) - 12*floor((n+1)/3) - 12*b(n)), where b(n) = binomial(n+3, 3) - 6*A014125(n-1) + 9*A144677(n-2). (End)

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A117906 Inverse of number triangle A117904.

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A117898 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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A117908 Chequered (or checkered) triangle for odd prime p=3.

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A117905 Expansion of (1+2*x+2*x^2)/((1+x)*(1-x^3)^2).

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A117905 Expansion of (1+2x+2x^2)/((1+x)*(1-x^3)^2).