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 1-3 of 3 results.

A099390 Array T(m,n) read by antidiagonals: number of domino tilings (or dimer tilings) of the m X n grid (or m X n rectangle), for m>=1, n>=1.

Original entry on oeis.org

0, 1, 1, 0, 2, 0, 1, 3, 3, 1, 0, 5, 0, 5, 0, 1, 8, 11, 11, 8, 1, 0, 13, 0, 36, 0, 13, 0, 1, 21, 41, 95, 95, 41, 21, 1, 0, 34, 0, 281, 0, 281, 0, 34, 0, 1, 55, 153, 781, 1183, 1183, 781, 153, 55, 1, 0, 89, 0, 2245, 0, 6728, 0, 2245, 0, 89, 0, 1, 144, 571, 6336, 14824, 31529, 31529, 14824, 6336, 571, 144, 1
Offset: 1

Views

Author

Ralf Stephan, Oct 16 2004

Keywords

Comments

There are many versions of this array (or triangle) in the OEIS. This is the main entry, which ideally collects together all the references to the literature and to other versions in the OEIS. But see A004003 for further information. - N. J. A. Sloane, Mar 14 2015

Examples

			0,  1,  0,   1,    0,    1, ...
1,  2,  3,   5,    8,   13, ...
0,  3,  0,  11,    0,   41, ...
1,  5, 11,  36,   95,  281, ...
0,  8,  0,  95,    0, 1183, ...
1, 13, 41, 281, 1183, 6728, ...
		

References

  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 406-412.
  • P. E. John, H. Sachs, and H. Zernitz, Problem 5. Domino covers in square chessboards, Zastosowania Matematyki (Applicationes Mathematicae) XIX 3-4 (1987), 635-641.
  • R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 2nd ed., pp. 547 and 570.
  • Darko Veljan, Kombinatorika: s teorijom grafova (Croatian) (Combinatorics with Graph Theory) mentions the value 12988816 = 2^4*901^2 for the 8 X 8 case on page 4.

Crossrefs

See A187596 for another version (with m >= 0, n >= 0). See A187616 for a triangular version. See also A187617, A187618.
See also A004003 for more literature on the dimer problem.
Main diagonal is A004003.

Programs

  • Maple
    (Maple code for the even-numbered rows from N. J. A. Sloane, Mar 15 2015. This is not totally satisfactory since it uses floating point. However, it is useful for getting the initial values quickly.)
    Digits:=100;
    p:=evalf(Pi);
    z:=proc(h,d) global p; evalf(cos( h*p/(2*d+1) )); end;
    T:=proc(m,n) global z; round(mul( mul( 4*z(h,m)^2+4*z(k,n)^2, k=1..n), h=1..m)); end;
    [seq(T(1,n),n=0..10)]; # A001519
    [seq(T(2,n),n=0..10)]; # A188899
    [seq(T(3,n),n=0..10)]; # A256044
    [seq(T(n,n),n=0..10)]; # A004003
  • Mathematica
    T[?OddQ, ?OddQ] = 0;
    T[m_, n_] := Product[2*(2+Cos[2j*Pi/(m+1)]+Cos[2k*Pi/(n+1)]), {k, 1, n/2}, {j, 1, m/2}];
    Flatten[Table[Round[T[m-n+1, n]], {m, 1, 12}, {n, 1, m}]] (* Jean-François Alcover, Nov 25 2011, updated May 28 2022 *)
  • PARI
    {T(n, k) = sqrtint(abs(polresultant(polchebyshev(n, 2, x/2), polchebyshev(k, 2, I*x/2))))} \\ Seiichi Manyama, Apr 13 2020

Formula

T(m, n) = Product_{j=1..ceiling(m/2)} Product_{k=1..ceiling(n/2)} (4*cos(j*Pi/(m+1))^2 + 4*cos(k*Pi/(n+1))^2).

Extensions

Old link fixed and new link added by Frans J. Faase, Feb 04 2009
Entry edited by N. J. A. Sloane, Mar 15 2015

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

Views

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.

Crossrefs

Row 4 of array A099390.
For all matchings see A033507.
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

A187617 Array T(m,n) read by antidiagonals: number of domino tilings of the 2m X 2n grid (m>=0, n>=0).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 5, 1, 1, 13, 36, 13, 1, 1, 34, 281, 281, 34, 1, 1, 89, 2245, 6728, 2245, 89, 1, 1, 233, 18061, 167089, 167089, 18061, 233, 1, 1, 610, 145601, 4213133, 12988816, 4213133, 145601, 610, 1
Offset: 0

Views

Author

N. J. A. Sloane, Mar 11 2011

Keywords

Comments

A099390 is the main entry for this problem.
The even-indexed rows and columns of the square array in A187596.
Row (and column) 2 is given by A122367. - Nathaniel Johnston, Mar 22 2011

Examples

			The array begins:
  1,  1,     1,       1,          1,            1, ...
  1,  2,     5,      13,         34,           89, ...
  1,  5,    36,     281,       2245,        18061, ...
  1, 13,   281,    6728,     167089,      4213133, ...
  1, 34,  2245,  167089,   12988816,   1031151241, ...
  1, 89, 18061, 4213133, 1031151241, 258584046368, ...
		

Crossrefs

A187618 is the triangle version.
Main diagonal is A004003. Second and third rows give A001519, A188899.

Programs

  • Maple
    ft:=(m,n)->
    2^(m*n/2)*mul( mul(
    (cos(Pi*i/(n+1))^2+cos(Pi*j/(m+1))^2), j=1..m/2), i=1..n/2);
    T:=(m,n)->round(evalf(ft(m,n),300));
  • Mathematica
    T[m_, n_] := Product[2(2 + Cos[(2j Pi)/(2m+1)] + Cos[(2k Pi)/(2n+1)]), {j, 1, m}, {k, 1, n}];
    Table[T[m-n, n] // Round, {m, 0, 8}, {n, 0, m}] // Flatten (* Jean-François Alcover, Aug 05 2018 *)
  • PARI
    default(realprecision, 120);
    {T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*cos(a*Pi/(2*n+1))^2+4*cos(b*Pi/(2*k+1))^2)))} \\ Seiichi Manyama, Jan 09 2021

Extensions

More terms from Nathaniel Johnston, Mar 22 2011
Showing 1-3 of 3 results.