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 1-7 of 7 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

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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

A003696 Number of spanning trees in P_4 X P_n.

Original entry on oeis.org

1, 56, 2415, 100352, 4140081, 170537640, 7022359583, 289143013376, 11905151192865, 490179860527896, 20182531537581071, 830989874753525760, 34214941811800329425, 1408756312731277540744
Offset: 1

Views

Author

Keywords

Comments

Also number of domino tilings of the 7 X (2n-1) rectangle with upper left corner removed. - Alois P. Heinz, Apr 14 2011
A linear divisibility sequence of order 8; a(n) divides a(m) whenever n divides m. It is the product of a 2nd-order Lucas sequence and a 4th-order linear divisibility sequence. - Peter Bala, Apr 27 2014

References

  • F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.

Crossrefs

A row of A116469. - N. J. A. Sloane, May 27 2012
Bisection of A189004. - Alois P. Heinz, Sep 20 2012

Programs

  • Maple
    seq(resultant(simplify(ChebyshevU(3, (x-4)*(1/2))), simplify(ChebyshevU(n-1, (1/2)*x)), x), n = 1 .. 14); # Peter Bala, Apr 27 2014
  • Mathematica
    LinearRecurrence[{56, -672, 2632, -4094, 2632, -672, 56, -1}, {1, 56, 2415, 100352, 4140081, 170537640, 7022359583, 289143013376}, 20] (* Jean-François Alcover, Feb 28 2020 *)
  • PARI
    {a(n) = polresultant((x-4)*(x^2-8*x+14), polchebyshev(n-1, 2, x/2))}; /* Michael Somos, Oct 31 2022 */

Formula

a(1) = 1,
a(2) = 56,
a(3) = 2415,
a(4) = 100352,
a(5) = 4140081,
a(6) = 170537640,
a(7) = 7022359583,
a(8) = 289143013376 and
a(n) = 56a(n-1) - 672a(n-2) + 2632a(n-3) - 4094a(n-4) + 2632a(n-5) - 672a(n-6) + 56a(n-7) - a(n-8).
G.f.: x(x^6-49x^4+112x^3-49x^2+1) / (x^8-56x^7 +672x^6-2632x^5 +4094x^4 -2632x^3 +672x^2-56x+1). - Paul Raff, Mar 06 2009
From Peter Bala, Apr 27 2014: (Start)
a(n) = Resultant( U(3,(x-4)/2),U(n-1,x/2) ), where U(n,x) denotes the Chebyshev polynomial of the second kind. The polynomial U(3,(x-4)/2) = x^3 - 12*x^2 + 46*x - 56 (see A159764) has zeros z_1 = 4, z_2 = 4 + sqrt(2) and z_3 = 4 - sqrt(2). Hence a(n) = U(n-1,2)*U(n-1,1/2*(4 + sqrt(2)))*U(n-1,1/2*(4 - sqrt(2))).
a(n) = A001353(n)*A161158(n-1). (End)
a(n) = (9/3968)*(A028469(n+3)-A028469(n-4)) - (497/3968)*(A028469(n+2)-A028469(n-3)) + (5687/3968)*(A028469(n+1)-A028469(n-2)) - (19983/3968)*(A028469(n)-A028469(n-1)), n>3. - Sergey Perepechko, May 02 2016
a(n) = -a(-n) for all n in Z. - Michael Somos, Oct 31 2022

Extensions

Added recurrence from Faase's web page. - N. J. A. Sloane, Feb 03 2009

A028471 Number of perfect matchings (or domino tilings) in the graph P_9 X P_2n.

Original entry on oeis.org

1, 55, 6336, 817991, 108435745, 14479521761, 1937528668711, 259423766712000, 34741645659770711, 4652799879944138561, 623139489426439754945, 83456125990631342400791, 11177167872295392172767936, 1496943834332592837945956455, 200483802581126644843760725601
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    T[?OddQ, ?OddQ] = 0;
    T[m_, n_] := Product[2(2+Cos[2 j Pi/(m+1)]+Cos[2 k Pi/(n+1)]), {k, 1, n/2}, {j, 1, m/2}];
    a[n_] := T[2n, 9] // Round;
    Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 28 2022 *)
  • PARI
    {a(n) = sqrtint(polresultant(polchebyshev(2*n, 2, x/2), polchebyshev(9, 2, I*x/2)))} \\ Seiichi Manyama, Apr 13 2020

Formula

a(n) = 209*a(n-1) - 11936*a(n-2) + 274208*a(n-3) - 3112032*a(n-4) + 19456019*a(n-5) - 70651107*a(n-6) + 152325888*a(n-7) - 196664896*a(n-8) + 152325888*a(n-9) - 70651107*a(n-10) + 19456019*a(n-11) - 3112032*a(n-12) + 274208*a(n-13) - 11936*a(n-14) + 209*a(n-15) - a(n-16). - Jay Anderson (horndude77(AT)gmail.com), Apr 07 2007
G.f.: (1 - 154x + 6777x^2 - 123961x^3 + 1132714x^4 - 5684515x^5 + 16401668x^6 - 27757938x^7 + 27757938*x^8 - 16401668x^9 + 5684515x^10 - 1132714x^11 + 123961x^12 -6777x^13 + 154x^14 - x^15)/(1 - 209x + 11936x^2 - 274208x^3 + 3112032x^4 - 19456019x^5 + 70651107x^6 - 152325888x^7 + 196664896x^8 - 152325888x^9 + 70651107x^10 -19456019x^11 + 3112032x^12 - 274208x^13 + 11936x^14 - 209x^15 + x^16). - Sergey Perepechko, Nov 23 2012

Extensions

Edited by N. J. A. Sloane, Jul 03 2008 at the suggestion of R. J. Mathar

A189004 Number of domino tilings of the 7 X n grid with upper left corner removed iff n is odd.

Original entry on oeis.org

1, 1, 21, 56, 781, 2415, 31529, 100352, 1292697, 4140081, 53175517, 170537640, 2188978117, 7022359583, 90124167441, 289143013376, 3710708201969, 11905151192865, 152783289861989, 490179860527896, 6290652543875133
Offset: 0

Views

Author

Alois P. Heinz, Apr 15 2011

Keywords

Crossrefs

7th row of array A189006.
Bisection gives: A028469 (even part), A003696 (odd part).

Programs

  • Mathematica
    A[m_, n_] := A[m, n] = Module[{i, j, s, t, M}, Which[m == 0 || n == 0, 1, m < n, A[n, m], True, s = Mod[n*m, 2]; M[i_, j_] /; j < i := -M[j, i]; M[, ] = 0; For[i = 1, i <= n, i++, For[j = 1, j <= m, j++, t = (i - 1)*m + j - s; If[i > 1 || j > 1 || s == 0, If[j < m, M[t, t + 1] = 1]; If[i < n, M[t, t + m] = 1 - 2*Mod[j, 2]]]]]; Sqrt[Det[Array[M, {n*m - s, n*m - s}]] ]]];
    a[n_] := A[7, n];
    a /@ Range[0, 20] (* Jean-François Alcover, Feb 27 2020, after Alois P. Heinz in A189006 *)

Formula

G.f.: -(x^14-x^13-35*x^12+277*x^10 +49*x^9-727*x^8 -112*x^7+727*x^6 +49*x^5-277*x^4 +35*x^2-x-1) / (x^16-56*x^14 +672*x^12-2632*x^10 +4094*x^8-2632*x^6 +672*x^4-56*x^2+1).

A360799 Numbers m with m mod 3 = q, q != 2, such that the number of ones in its base-2 representation is even if q=0 and odd if q=1.

Original entry on oeis.org

0, 1, 3, 4, 6, 7, 9, 12, 13, 15, 16, 18, 19, 22, 24, 25, 27, 28, 30, 31, 33, 36, 37, 39, 45, 48, 49, 51, 52, 54, 55, 57, 60, 61, 63, 64, 66, 67, 70, 72, 73, 75, 76, 78, 79, 82, 88, 90, 91, 94, 96, 97, 99, 100, 102, 103, 105, 108, 109, 111, 112, 114, 115, 118, 120, 121
Offset: 0

Views

Author

Gerhard Kirchner, Feb 24 2023

Keywords

Comments

For q=0, the terms in A180963 are excluded.
The terms of the sequence occur, with some exceptions, while tiling a wall (odd width w) with 1 X 2 dominos. The current tiling status can be described by a number x with 0 <= x < 2^w. In the base-2 representation, 1 stands for an overstanding unit square, see example.
Statement:
The tiling always starts with q=1 and an odd number of ones (type 1) and is followed by a term with q=0 and an even number of ones (type 2) and so on, alternately.
Proof:
Start, provisionally, with w upright dominos. The corresponding term is x = (11..1) = 2^w-1 with x mod 3 = 1 (type 1). Another first profile can be generated by replacing a pair of adjacent upright dominos with one flat domino. In the base-2 representation, this is the subtraction (11..11..1) - (00..11..0) = (11..00..1). The subtrahend is 3*2^j with 0 <= j < w. Therefore, the modified term also is type 1. This way, any first profile can be found and it is type 1.
In the next provisional step, an upright domino is placed on each not overstanding unit square. If p1 is the first profile, then the second is p2 = 2^w - 1 - p1 with p2 mod 3 = 0. Moreover, the transition from p1 to p2 exchanges the ones and zeros such that p2 is type 2. Again, replacing adjacent upright dominos by one flat domino does not change the type of the profile. The next profile is type 1 and so on. QED. Condition to be satisfied by a tiling profile: The continued removal of 00 and 11 (reduction) leads to (0) or (1). Example: a(10)=18=(10010) -> (110) -> (0). The first exceptions are a(314) = 682 = (01010101010), a(611) = 1365 = (10101010101) and a(988) = 2218 = (0100010101010). Note that the reduction of 2218 is 682.

Examples

			 5 X 4 wall is tiled bottom-up with 1 X 2 dominos:
                                      _    ___ ___ _
                 _ _          _ _ ___| |  |_ _|___| |
        _       | | |_ ___   | | |_ _|_|  | | |_ _|_|
    ___| |___   |_|_| |___|  |_|_| |___|  |_|_| |___|
   |___|_|___|  |___|_|___|  |___|_|___|  |___|_|___|
    0 0 1 0 0    1 1 0 0 0    0 0 0 0 1    0 0 0 0 0
     4 = a(3)   24 = a(14)     1 = a(1)     0 = a(0)
		

Crossrefs

Programs

  • Maxima
    block(kmax: 100, a:[],
     even_ones(x):= block(su:0,
      while x>0 do(p: mod(x,2), x:(x-p)/2, su:su+p),
       return(mod(su,2))),
    for k from 0 thru kmax do(r:mod(k,3),
     if r<2 and r=even_ones(k) then a:append(a,[k])),a);
    
  • PARI
    isok(m) = my(k=m%3); if (hammingweight(m) % 2, k==1, k==0); \\ Michel Marcus, Feb 27 2023

A360800 Numbers Sum_{i=1..2r+1} 2^k(i) such that k(1) is even and, for r > 0 and i < 2r+1, the difference k(i+1)-k(i) is > 0 and odd.

Original entry on oeis.org

1, 4, 7, 16, 19, 25, 28, 31, 64, 67, 73, 76, 79, 97, 100, 103, 112, 115, 121, 124, 127, 256, 259, 265, 268, 271, 289, 292, 295, 304, 307, 313, 316, 319, 385, 388, 391, 400, 403, 409, 412, 415, 448, 451, 457, 460, 463, 481, 484, 487, 496, 499, 505, 508, 511, 1024
Offset: 1

Views

Author

Gerhard Kirchner, Feb 24 2023

Keywords

Comments

This is a subsequence of A360799. Another description of the terms: in the base-2 representation, the number of ones is odd and all zeros are grouped in blocks of even length.
That is why the terms less than 2^(2j+1) describe start profiles for tiling a (2j+1) X m wall with 1 X 2 dominos, see examples and A360799.

Examples

			A 5 X m wall is tiled bottom-up with dominos, start profiles:
            _        _            _ _ _    _     _ _    _ _ _ _ _
    ___ ___| |   ___| |___    ___| | | |  | |___| | |  | | | | | |
   |___|___|_|  |___|_|___|  |___|_|_|_|  |_|___|_|_|  |_|_|_|_|_|
    0 0 0 0 1    0 0 1 0 0    0 0 1 1 1    1 0 0 1 1    1 1 1 1 1
    1 = a(1)     4 = a(2)     7 = a(3)     19 = a(5)    31 = a(7)
    also the mirror images of 1 (16), 19 (25) and 7 (28).
		

Crossrefs

Programs

  • Maxima
    block(kmax: 100, a:[],
     oddsum(y):= block(su1:0, su2:0, pold:0, ok: true,
      while y>0 and ok do(p:mod(y,2), y:(y-p)/2,
       if p=1 then(if pold=0 and su2=1 then ok:false, su1:1-su1, su2:0)
       elseif p=0 then su2:1-su2, pold:p), return(is(ok and su1=1))),
    for k from 1 thru kmax do if oddsum(k) then a:append(a,[k]),a);
Showing 1-7 of 7 results.