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.

A248093 Triangle read by rows: TR(n,k) is the number of unordered vertex pairs at distance k of the hexagonal triangle T_n, defined in the He et al. reference (1<=k<=2n+1).

Original entry on oeis.org

1, 0, 6, 6, 6, 3, 13, 15, 21, 21, 15, 6, 22, 27, 42, 48, 45, 36, 24, 9, 33, 42, 69, 84, 87, 81, 69, 51, 33, 12, 46, 60, 102, 129, 141, 141, 132, 114, 93, 66, 42, 15, 61, 81, 141, 183, 207, 216, 213, 198, 177, 147, 117, 81, 51, 18, 78, 105, 186, 246, 285
Offset: 0

Views

Author

Emeric Deutsch, Nov 14 2014

Keywords

Comments

Number of entries in row n is 2*n+2.
The entries in row n are the coefficients of the Hosoya polynomial of T_n.
TR(n,0) = A028872(n+2) = number of vertices of T_n.
TR(n,1) = A140091(n) = number of edges of T_n.
sum(j*TR(n,j), j=0..2n+1) = A033544(n) = the Wiener index of T_n.
(1/2)*sum(j*(j+1)TR(n,j), j=0..2n+1) = A248094(n) = the hyper-Wiener index of T_n.
sum((-1)^j*TR(n,j), j=0..2n+1) = A002061(n). - Peter Luschny, Nov 15 2014

Examples

			Row n=1 is 6, 6, 6, 3; indeed, T_1 is a hexagon ABCDEF; it has 6 distances equal to 0 (= number of vertices), 6 distances equal to 1 (= number of edges), 6 distances equal to 2 (AC, BD, CE, DA, EA, FB), and 3 distances equal to 3 (AD, BE, CF).
Triangle starts:
1, 0;
6, 6, 6, 3;
13, 15, 21, 21, 15, 6;
22, 27, 42, 48, 45, 36, 24, 9;
33, 42, 69, 84, 87, 81, 69, 51, 33, 12;
		

Crossrefs

Programs

  • Maple
    G := (1+(3+6*t+4*t^2+3*t^3)*z-(1+t+2*t^2)*(2+t-2*t^2)*z^2+t^2*(1-3*t^2)*z^3+t^4*z^4)/((1-z)^3*(1-t^2*z)^2): Gser := simplify(series(G, z = 0, 25)): for n from 0 to 22 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 12 do seq(coeff(P[n], t, j), j = 0 .. 2*n+1) end do; # yields sequence in triangular form

Formula

G.f.: (1 + (3 + 6*t + 4*t^2 + 3*t^3)*z - (1 + t + 2*t^2)*(2 + t - 2*t^2)*z^2 +t^2*(1 - 3*t^2)*z^3 + t^4*z^4)/((1-z)^3*(1 - t^2*z^2)^2); follows from Theorem 3.6 of the He et al. reference.