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 1-4 of 4 results.

A219765 Triangle of coefficients of a polynomial sequence related to the coloring of labeled graphs.

Original entry on oeis.org

1, 0, 1, 0, -1, 2, 0, 5, -12, 8, 0, -79, 208, -192, 64, 0, 3377, -9520, 10240, -5120, 1024, 0, -362431, 1079744, -1282560, 778240, -245760, 32768, 0, 93473345, -291615296, 372893696, -255754240, 100925440, -22020096, 2097152, 0, -56272471039, 182582658048, -247557029888, 185511968768, -84263567360, 23488102400, -3758096384, 268435456
Offset: 0

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Author

Peter Bala, Apr 16 2013

Keywords

Comments

A simple graph G is a k-colorable graph if it is possible to assign one of k' <= k colors to each vertex of G so that no two adjacent vertices receive the same color. Such an assignment of colors is called a k-coloring function for the graph. The chromatic polynomial P(G,k) of a simple graph G gives the number of different k-colorings functions of the graph as a function of k. P(G,k) is a polynomial function of k.
The row polynomials R(n,x) of this triangle are defined to be the sum of the chromatic polynomials of all labeled simple graphs on n vertices: R(n,x) = sum {labeled graphs G on n nodes} P(G,x). An example is given below.

Examples

			Triangle begins:
  n\k.|..0.....1......2.......3......4.......5
  = = = = = = = = = = = = = = = = = = = = = = =
  .0..|..1
  .1..|..0.....1
  .2..|..0....-1......2
  .3..|..0.....5....-12.......8
  .4..|..0...-79....208....-192.....64
  .5..|..0..3377..-9520...10240..-5120....1024
  ...
Row 3 = [5, -12, 8]: There are 4 unlabeled graphs on 3 vertices, namely
(a)  o    o    o    (b)  o    o----o    (c)  o----o----o
(d)   0
     / \
    0---0
with chromatic polynomials x^3, x^2*(x-1), x*(x-1)^2 and (x-1)^3 - (x-1), respectively. Allowing for labeling, there is 1 labeled graph of type (a), 3 labeled graphs of type (b), 3 labeled graphs of type (c) and 1 labeled graph of type (d). Thus the sum of the chromatic polynomials over all labeled graphs on 3 nodes is x^3 + 3*x^2*(x-1) + 3*x*(x-1)^2  + (x-1)^3 - (x-1) = 5*x - 12*x^2 + 8*x^3.
Row 3 of A058843 is [1,12,8]. Thus R(3,x) = x + 12*x*(x-1) + 8*x*(x-1)*(x-2) = 5*x - 12*x^2 + 8*x^3.
		

Crossrefs

Cf. A003024 (alt. row sums), A006125 (main diagonal), A095351 (main subdiagonal), A134531 (column 1), A058843, A322280.

Programs

  • Maple
    R:= proc(n) option remember; `if`(n=0, 1, expand(x*add(
          binomial(n-1, k)*2^(k*(n-k))*subs(x=x-1, R(k)), k=0..n-1)))
        end:
    T:= n-> (p-> seq(coeff(p,x,i), i=0..degree(p)))(R(n)):
    seq(T(n), n=0..10);  # Alois P. Heinz, May 16 2024
  • Mathematica
    r[0] = 1; r[1] = x; r[n_] := r[n] = x*Sum[ Binomial[n-1, k]*2^(k*(n-k))*(r[k] /. x -> x-1), {k, 0, n-1}]; row[n_] := CoefficientList[ r[n], x]; Table[ row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 17 2013 *)

Formula

Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)) = 1 + x + x^2/(2!*2) + x^3/(3!*2^3) + x^4/(4!*2^6) + .... Then a generating function for the triangle is E(z)^x = Sum_{n >= 0} R(n,x)*z^n/(n!*2^C(n,2)) = 1 + x*z + (-x + 2*x^2)*z^2/(2!*2) + (5*x - 12*x^2 + 8*x^3)*z^3/(3!*2^3) + ....
The row polynomials satisfy the recurrence equation: R(0,x) = 1 and
R(n+1,x) = x*Sum_{k = 0..n} binomial(n,k)*2^(k*(n+1-k))*R(k,x-1).
In terms of the basis of falling factorial polynomials we have, for n >= 1, R(n,x) = Sum_{k = 1..n} A058843(n,k)*x*(x-1)*...*(x-k+1).
The polynomials R(n,x)/2^comb(n,2) form a sequence of binomial type (see Stanley). Setting D = d/dx, the associated delta operator begins D + D^2/(2!*2) + D^3/(3!*2^3) - D^4/(4!*2^4) + 23*D^5/(5!*2^5) + 559*D^6/(6!*2^6) - ... obtained by series reversion of the formal series D - D^2/(2!*2) + 5*D^3/(3!*2^3) - 79*D^4/(4!*2^4) + 3377*D^5/(5!*2^5) - ... coming from column 1 of the triangle.
Alternating row sums A003024. Column 1 = A134531. Main diagonal A006125. Main subdiagonal (-1)*A095351.
The row polynomials evaluated at k is A322280(n,k). - Geoffrey Critzer, Mar 02 2024
Let k>=1 and let D be a directed acyclic graph with vertices labeled on [n]. Then (-1)^n*R(n,-k) is the number of maps C:[n]->[k] such that for all vertices i,j in [n], if i is directed to j in D then C(i)>=C(j). Cf A003024 (k=1), A339934 (k=2). - Geoffrey Critzer, May 15 2024

A380336 Triangular array read by rows. T(n,k) is the number of ways to choose a size k subset S of [n] and form a labeled acyclic digraph on S. Then form another labeled acyclic digraph on [n]-S. For each pair u in S and v in [n]-S add the directed edge u->v or not, n>=0, 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 3, 4, 3, 25, 36, 36, 25, 543, 800, 864, 800, 543, 29281, 43440, 48000, 48000, 43440, 29281, 3781503, 5621952, 6255360, 6400000, 6255360, 5621952, 3781503, 1138779265, 1694113344, 1888975872, 1946112000, 1946112000, 1888975872, 1694113344, 1138779265
Offset: 0

Views

Author

Geoffrey Critzer, Jan 21 2025

Keywords

Examples

			Triangle T(n,k) begins:
     1;
     1,     1;
     3,     4,     3;
    25,    36,    36,    25;
   543,   800,   864,   800,   543;
 29281, 43440, 48000, 48000, 43440, 29281;
 ...
		

Crossrefs

Cf. A339934 (row sums), A003024 (column k=0 and main diagonal).

Programs

  • Mathematica
    nn = 6; B[n_] := n! 2^Binomial[n, 2]; e[z_] := Sum[z^n/B[n], {n, 0, nn}]; Map[Select[#, # > 0 &] &,Table[B[n], {n, 0, nn}] CoefficientList[Series[1/e[-u z]*1/e[-z], {z, 0, nn}], {z, u}]] // Grid

Formula

Sum_{n>=0} T(n,k)*y^k*x^n/(2^binomial(n,2)*n!) = 1/E(-y*x)*1/E(-x) where E(x) = Sum_{n>=0} x^n/(2^binomial(n,2)*n!).
T(n,k) = binomial(n,k)*A003024(k)*A003024(n-k)*2^(k*(n-k)). - Alois P. Heinz, Jan 22 2025

A340798 Square array read by descending antidiagonals. Let G be a simple labeled graph on n nodes. T(n,k) is the number of ways to give G an acyclic orientation and a coloring function C:V(G) -> {1,2,...,k} so that u->v implies C(u) >= C(v) for all u,v in V(G), n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 3, 10, 25, 0, 1, 4, 21, 122, 543, 0, 1, 5, 36, 339, 3550, 29281, 0, 1, 6, 55, 724, 12477, 241442, 3781503, 0, 1, 7, 78, 1325, 32316, 1035843, 37717630, 1138779265, 0
Offset: 0

Views

Author

Geoffrey Critzer, Jan 21 2021

Keywords

Examples

			Array begins
  1,     1,      1,       1,       1,       1, ...
  0,     1,      2,       3,       4,       5, ...
  0,     3,     10,      21,      36,      55, ...
  0,    25,    122,     339,     724,    1325, ...
  0,   543,   3550,   12477,   32316,   69595, ...
  0, 29281, 241442, 1035843, 3180484, 7934885, ...
  ...
		

Crossrefs

Cf. A003024 (column k=1), A339934 (column k=2), A322280, A219765.

Programs

  • Mathematica
    nn = 6; e[x_] := Sum[x^n/(n! 2^Binomial[n, 2]), {n, 0, nn}];
    Prepend[Table[Table[n! 2^Binomial[n, 2], {n, 0, nn}] CoefficientList[
          Series[1/e[-x]^k, {x, 0, nn}], x], {k, 1, nn}],PadRight[{1}, nn + 1]] // Transpose // Grid

Formula

Let E(x) = Sum_{n>=0} x^n/(n!*2^binomial(n,2)). Then Sum_{n>=0} T(n,k)*x^n/(n!*2^binomial(n,2)) = 1/E(-x)^k.
T(n,k) = (-1)^n p_n(-k) where p_n is the n-th polynomial described in A219765.

A342113 Number of surjective compatible pairs (C,O), where O is an acyclic orientation of simple labeled graph G on n nodes and C:V(G) -> {1,2,...}.

Original entry on oeis.org

1, 1, 7, 145, 7999, 1103041, 365051647, 281898887425, 497570152386559, 1976049386530790401, 17439288184770966867967, 338596445913833207323643905, 14343481992486219718322674565119
Offset: 0

Views

Author

Geoffrey Critzer, Feb 28 2021

Keywords

Comments

A pair (C,O) is a surjective compatible pair if O is an acyclic orientation of the edges of a simple labeled graph G on n nodes, and C is a surjective function from V(G)->{1,2,...k} for some positive integer k such that for all u,v in V(G) if u->v under the orientation then C(u)>= C(v).

Crossrefs

Cf. A339934.

Programs

  • Mathematica
    nn = 12; b[n_] := q^Binomial[n, 2] n! /. q -> 2; e[z_] := Sum[z^n/b[n], {n, 0, nn}];Table[b[n], {n, 0, nn}] CoefficientList[ Series[1/(1 - (1/e[-z] - 1)), {z, 0, nn}], z]

Formula

Let E(x) = Sum_{n>=0}x^n/(n! 2^Binomial(n,2)). Then Sum_{n>=0}a_n x^n/(n! 2^Binomial(n,2)) = 1/(2 - E(-x)^-1).
Showing 1-4 of 4 results.