cp's OEIS Frontend

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.

Showing 1-2 of 2 results.

A319093 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1, k) + T(n-1, k-1) - T(n-1, k-2) + 2*T(n-1, k-3) + T(n-1, k-4) for k = 0..4*n; T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 1, 1, -1, 2, 1, 1, 2, -1, 2, 7, -2, 2, 4, 1, 1, 3, 0, 1, 15, 3, -4, 24, 6, -1, 9, 6, 1, 1, 4, 2, 0, 23, 20, -14, 48, 55, -24, 46, 52, 2, 12, 20, 8, 1, 1, 5, 5, 0, 30, 51, -15, 60, 180, -25, 49, 280, 15, 30, 180, 72, 15, 45, 35, 10, 1, 1, 6, 9, 2, 36, 96, 11, 54, 387, 116, -51, 774, 376, -162, 804, 532
Offset: 0

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Author

Shara Lalo, Oct 01 2018

Keywords

Comments

Row n gives the coefficients in the expansion of (1 + x - x^2 + 2*x^3 + x^4)^n, where n is a nonnegative integer. The row sum at row n is s(n) = 4^n. In the center-justified triangle, the sum of numbers along "first layer" skew diagonals pointing top-right are the coefficients in the expansion of 1/(1 - x - x^2 + x^3 - 2*x^4 - x^5) and the sum of numbers along "first layer" skew diagonals pointing top-left are the coefficients in the expansion of 1/(1 - x - 2*x^2 + x^3 - x^4 - x^5), see links. The central coefficients are given in A319096.

Examples

			Triangle begins:
1;
1, 1, -1, 2,  1;
1, 2, -1, 2,  7, -2,  2,   4,   1;
1, 3,  0, 1, 15,  3, -4,  24,   6,  -1,  9,   6,  1;
1, 4,  2, 0, 23, 20, -14, 48,  55, -24, 46,  52,  2, 12,  20,  8,  1;
1, 5,  5, 0, 30, 51, -15, 60, 180, -25, 49, 280, 15, 30, 180, 72, 15, 45, 35, 10, 1;
...
		

References

  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

Crossrefs

Cf. A319096.

Programs

  • Mathematica
    Clear[t, n, k]; t[n_, k_] := t[n, k] = Sum[((-1)^(i + q - 2*j)*2^(j - 2*i)*n!)/((n - k + q)!*(k + j - 2*q)!*(i + q - 2*j)!*(j - 2*i)!*i!), {i, 0, k}, {j, 2*i, k}, {q, 3*i, k}]; Flatten[Table[t[n, k], {n, 0, 7}, {k, 0, 4*n}]]
    Clear[t, n, k]; t[0, 0] = 1; t[n_, k_] :=  t[n, k] =   If[n < 0 || k < 0, 0,    t[n - 1, k] + t[n - 1, k - 1] - t[n - 1, k - 2] + 2 t[n - 1, k - 3] + t[n - 1, k - 4]]; Table[t[n, k], {n, 0, 6}, {k, 0, 4*n}  ]  // Flatten
  • PARI
    row(n) = Vecrev((1 + x - x^2 + 2*x^3 + x^4)^n);
    tabl(nn) = for (n=0, nn, print(row(n))); \\ Michel Marcus, Oct 15 2018

Formula

T(n,k) = Sum_{i=0..k} Sum_{j=2*i..k} Sum_{q=3*i..k}(f) for k = 0..4*n; f= (-1)^(i + q - 2*j)*2^(j - 2*i)*n!)/((n - k + q)!*(k + j - 2*q)!*(i + q - 2*j)!*(j - 2*i)!*i!); f=0 for (n - k + q)<0 or (k + j - 2*q)<0 or (i + q - 2*j) <0 or (j - 2*i) <0. A novel formula proven by Shara Lalo and Zagros Lalo. Also see formula in Links section.
G.f.: 1/(1 - t*x - t*x^2 + t*x^3 - 2*t*x^4 - t*x^5).

A321614 Number of nonequivalent ways to place 2n nonattacking kings on a 4 X 2n chessboard under all symmetry operations of the rectangle.

Original entry on oeis.org

1, 4, 23, 106, 473, 1939, 7618, 28703, 105112, 375597, 1316944, 4544124, 15474559, 52108212, 173799309, 574908646, 1888125243, 6162032375, 19998659760, 64584817367, 207655073310, 665017743665
Offset: 0

Views

Author

Anton Nikonov, Dec 19 2018

Keywords

Comments

A maximum of 2n nonattacking kings can be placed on a 4 X 2n chessboard.
Number of nonequivalent ways of placing 2n 2 X 2 tiles in an 5 X (2n+1) rectangle under all symmetry operations of the rectangle. - Andrew Howroyd, Dec 21 2018

Crossrefs

Formula

a(n) = A231145(2*n+1, 2n).
Conjectures from Colin Barker, Dec 22 2018: (Start)
G.f.: (1 - 2*x)*(1 - 6*x + 17*x^2 - 18*x^3 - 2*x^4 + 7*x^5 + 6*x^6 - 3*x^7) / ((1 - x)^2*(1 - 3*x)^2*(1 - 3*x + x^2)*(1 - x - x^2)*(1 - 3*x^2)).
a(n) = 12*a(n-1) - 54*a(n-2) + 98*a(n-3) + 17*a(n-4) - 346*a(n-5) + 505*a(n-6) - 210*a(n-7) - 120*a(n-8) + 126*a(n-9) - 27*a(n-10) for n>9.
(End)
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