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.

A181140 The number of ways to color n balls in a row with 3 colors with no color runs having lengths greater than 4. This sequence is a special case of the general problem for coloring n balls in a row with p colors where each color has a given maximum run-length. In this example, the bounds are uniformly 4. It can be phrased in terms of tossing a p-faced die n times, requiring each face to have no runs longer than b.

Original entry on oeis.org

3, 9, 27, 81, 240, 714, 2124, 6318, 18792, 55896, 166260, 494532, 1470960, 4375296, 13014096, 38709768, 115140240, 342478800, 1018685808, 3030029232, 9012668160, 26807724000, 79738214400, 237177271584, 705471756288, 2098389932544
Offset: 1

Views

Author

William Sit (wyscc(AT)sci.ccny.cuny.edu), Oct 06 2010

Keywords

Comments

Generating function and recurrence for given p and uniform bound b are known.
a(n+b) = (p-1)(a(n) + ... + a(n+b-1)),
using b initial values a(1)=p, a(2)=p^2, ..., a(b)=p^(b).
The g.f. is p G/(1-(p-1)G) where G = t + t^2 + ... + t^b.

Examples

			The first nontrivial value is a(5)=240. These solutions are listed below: 11112,11113,11121,11122,11123,
11131,11132,11133,11211,11212,11213,11221,11222,11223,11231,11232,11233,11311,11312,11313,11321,11322,11323,
11331,11332,11333,12111,12112,12113,12121,12122,12123,12131,12132,12133,12211,12212,12213,12221,12222,12223,
12231,12232,12233,12311,12312,12313,12321,12322,12323,12331,12332,12333,13111,13112,13113,13121,13122,13123,
13131,13132,13133,13211,13212,13213,13221,13222,13223,13231,
13232,13233,13311,13312,13313,13321,13322,13323,13331,13332,13333,21111,21112,21113,21121,21122,21123,21131,
21132,21133,21211,21212,21213,21221,21222,21223,21231,21232,21233,21311,21312,21313,21321,21322,21323,21331,
21332,21333,22111,22112,22113,22121,22122,22123,22131,22132,22133,22211,22212,22213,22221,22223,22231,22232,
22233,22311,22312,22313,22321,22322,22323,22331,22332,22333,23111,23112,23113,23121,23122,23123,23131,23132,
23133,23211,23212,23213,23221,23222,23223,23231,23232,23233,23311,
23312,23313,23321,23322,23323,23331,23332,23333,31111,31112,31113,31121,31122,31123,31131,31132,31133,31211,
31212,31213,31221,31222,31223,31231,31232,31233,31311,31312,31313,31321,31322,31323,31331,31332,31333,32111,
32112,32113,32121,32122,32123,32131,32132,32133,32211,32212,32213,32221,32222,32223,32231,32232,32233,32311,
32312,32313,32321,32322,32323,32331,32332,32333,33111,33112,33113,33121,33122,33123,33131,33132,33133,33211,
33212,33213,33221,33222,33223,33231,33232,33233,33311,33312,33313,33321,33322,33323,33331,33332
		

Crossrefs

Programs

  • Mathematica
    (* next[p,z] computes the next member in a sequence and next[p,z] = a(n+b)= (p-1)( c(b)+ ... + c(n+b-1))
    where z is the preceding b items on the sequence starting with a(n) where b is the uniform bound on runs.
    The function sequence[p,z,n] computes the next n terms. *)
    next[p_,z_]:=(p-1) Apply[Plus,z]
    sequence[p_,z_,n_]:=Module[{y=z,seq=z, m=n, b=Length[z]}, While[m>0, seq = Join[seq,{next[p,y]}]; y = Take[seq, -b]; m-- ]; seq]
    (* sequence[3,{3,9,27,81},10] computes the next 10 terms after 3,9,27, 81. *)

Formula

For this sequence (p=3, b=4):
G.f.: 3t(t^3+t^2+t+1)/(1 - 2t(t^3+t^2+t+1));
a(n+4) = 2(a(n)+a(n+1)+a(n+2)+a(n+3)); a(1)=3, a(2)=9, a(3)=27, a(4)=81.