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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.

A005178 Number of domino tilings of 4 X (n-1) board.

Original entry on oeis.org

0, 1, 1, 5, 11, 36, 95, 281, 781, 2245, 6336, 18061, 51205, 145601, 413351, 1174500, 3335651, 9475901, 26915305, 76455961, 217172736, 616891945, 1752296281, 4977472781, 14138673395, 40161441636, 114079985111, 324048393905
Offset: 0

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Author

N. J. A. Sloane, David Singmaster, Frans J. Faase

Keywords

Comments

Or, number of perfect matchings in graph P_4 X P_{n-1}.
a(0) = 0, a(1) = 1 by convention.
It is easy to see that the g.f. for indecomposable tilings, i.e., those that cannot be split vertically into smaller tilings, is g = x + 4x^2 + 2x^3 + 3x^4 + 2x^5 + 3x^6 + 2x^7 + 3x^8 + ... = x + 4x^2 + x^3*(2+3x)/(1-x^2); then g.f. = 1/(1-g) = (1-x^2)/(1-x-5x^2-x^3+x^4). - Emeric Deutsch, Oct 16 2006
This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m). - T. D. Noe, Dec 22 2008
From Artur Jasinski, Dec 20 2008: (Start)
All numbers in this sequence are:
congruent to 0 mod 100 if n is congruent to 14 or 29 mod 30
congruent to 1 mod 100 if n is congruent to 0 or 1 or 12 or 16 or 27 or 28 mod 30
congruent to 5 mod 100 if n is congruent to 2 or 11 or 17 or 26 mod 30
congruent to 11 mod 100 if n is congruent to 3 or 25 mod 30
congruent to 36 mod 100 if n is congruent to 4 or 9 or 19 or 24 mod 30
congruent to 45 mod 100 if n is congruent to 8 or 20 mod 30
congruent to 51 mod 100 if n is congruent to 13 or 15 mod 30
congruent to 61 mod 100 if n is congruent to 10 or 18 mod 30
congruent to 81 mod 100 if n is congruent to 6 or 7 or 21 or 22 mod 30
congruent to 95 mod 100 if n is congruent to 5 or 23 mod 30
(End)
This is the case P1 = 1, P2 = -7, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 31 2014

Examples

			For n=2 the graph is
  o-o-o-o
and there is one perfect tiling:
  o-o o-o
For n=3 the graph is
  o-o-o-o
  | | | |
  o-o-o-o
and there are five perfect tilings:
  o o o o
  | | | |
  o o o o
two like:
  o o o-o
  | | ...
  o o o-o
and this
  o-o o-o
  .......
  o-o o-o
and this
  o o-o o
  | ... |
  o o-o o
a(n+1)=r(n)-r(n-2), r(n)=if n=0 then 1 else sum(sum(binomial(k,j)*sum(binomial(j,i-j)*5^(i-j)*binomial(k-j,n-i-3*(k-j))*(-1)^(n-i-3*(k-j)),i,j,n-k+j),j,0,k),k,1,n), n>1. - _Vladimir Kruchinin_, Sep 08 2010

References

  • F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics I, p. 292.

Links

Crossrefs

Row 4 of array A099390.
For all matchings see A033507.
Cf. A003775, A028468, A028469, A028470.
Cf. A003757. - T. D. Noe, Dec 22 2008
Bisection (odd part) gives A188899. - Alois P. Heinz, Oct 28 2012
Column k=2 of A250662.

Programs

  • Maple
    a[0]:=1: a[1]:=1: a[2]:=5: a[3]:=11: for n from 4 to 26 do a[n]:=a[n-1]+5*a[n-2]+a[n-3]-a[n-4] od: seq(a[n],n=0..26); # Emeric Deutsch, Oct 16 2006
A005178:=-(-1-4*z-z**2+z**3)/(1-z-5*z**2-z**3+z**4) # conjectured (correctly) by Simon Plouffe in his 1992 dissertation; gives sequence apart from an initial 1
  • Mathematica
    CoefficientList[Series[x(1-x^2)/(1-x-5x^2-x^3+x^4), {x,0,30}], x] (* T. D. Noe, Dec 22 2008 *)
LinearRecurrence[{1, 5, 1, -1}, {0, 1, 1, 5}, 28] (* Robert G. Wilson v, Aug 08 2011 *)
a[0] = 0; a[n_] := Product[2(2+Cos[2j Pi/5]+Cos[2k Pi/n]), {k, 1, (n-1)/2}, {j, 1, 2}] // Round;
Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Aug 20 2018 *)
  • Maxima
    r(n):=if n=0 then 1 else sum(sum(binomial(k,j)*sum(binomial(j,i-j)*5^(i-j)*binomial(k-j,n-i-3*(k-j))*(-1)^(n-i-3*(k-j)),i,j,n-k+j),j,0,k),k,1,n); a(n):=r(n)-r(n-2); /* Vladimir Kruchinin, Sep 08 2010 */

Formula

a(n) = a(n-1) + 5*a(n-2) + a(n-3) - a(n-4).
G.f.: x*(1 - x^2)/(1 - x - 5*x^2 - x^3 + x^4).
Limit_{n->oo} a(n)/a(n-1) = (1 + sqrt(29) + sqrt(14 + 2*sqrt(29)))/4 = 2.84053619409... - Philippe Deléham, Jun 12 2005
a(n) = (5*sqrt(29)/145)*(((1+sqrt(29)+sqrt(14+2*sqrt(29)))/4)^n+((1+sqrt(29)-sqrt(14+2*sqrt(29)))/4)^n-((1-sqrt(29)+sqrt(14-2*sqrt(29)))/4)^n-((1-sqrt(29)-sqrt(14-2*sqrt(29)))/4)^n). - Tim Monahan, Jul 30 2011
From Peter Bala, Mar 31 2014: (Start)
a(n) = ( T(n,alpha) - T(n,beta) )/(alpha - beta), where alpha = (1 + sqrt(29))/4 and beta = (1 - sqrt(29))/4 and T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 7/4; 1, 1/2].
a(n) = U(n-1,i*(1 + sqrt(5))/4)*U(n-1,i*(1 - sqrt(5))/4), where U(n,x) denotes the Chebyshev polynomial of the second kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials and 4th-order linear divisibility sequences. (End)
a(n) = A129113(n+2) - A129113(n). - R. J. Mathar, May 03 2021

Extensions

Amalgamated with (former) A003692, Dec 30 1995
Name changed and 0 prepended by T. D. Noe, Dec 22 2008
Edited by N. J. A. Sloane, Nov 15 2009

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