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.

A068921 Number of ways to tile a 2 X n room with 1 X 2 Tatami mats. At most 3 Tatami mats may meet at a point.

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201, 39865, 58425, 85626, 125491, 183916, 269542, 395033, 578949, 848491, 1243524, 1822473, 2670964, 3914488, 5736961
Offset: 0

Views

Author

Dean Hickerson, Mar 11 2002

Keywords

Crossrefs

Cf. A068927 for incongruent tilings, A068920 for more info.
Cf. A000930, A078012, first column of A272471.

Programs

  • Mathematica
    LinearRecurrence[{1, 0, 1}, {1, 1, 2}, 42] (* Robert G. Wilson v, Jul 12 2014 *)
  • PARI
    my(x='x+O('x^50)); Vec((1+x^2)/(1-x-x^3)) \\ G. C. Greubel, Apr 26 2017

Formula

For n >= 3, a(n) = a(n-1) + a(n-3).
a(n) = A000930(n+1).
From Frank Ruskey, Jun 07 2009: (Start)
G.f.: (1+x^2)/(1-x-x^3).
a(n) = Sum_{j=0..floor(n/2)} binomial(n-2j+1, j). (End)
G.f.: Q(0)*( 1+x^2 )/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + x^2)/( x*(4*k+3 + x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 08 2013

A068926 Table read by antidiagonals: ti(r,s) is the number of incongruent ways to tile an r X s room with 1 X 2 Tatami mats. At most 3 Tatami mats may meet at a point.

Original entry on oeis.org

0, 1, 1, 0, 1, 0, 1, 2, 2, 1, 0, 3, 0, 3, 0, 1, 4, 2, 2, 4, 1, 0, 6, 0, 1, 0, 6, 0, 1, 8, 2, 2, 2, 2, 8, 1, 0, 12, 0, 2, 0, 2, 0, 12, 0, 1, 16, 4, 2, 1, 1, 2, 4, 16, 1, 0, 24, 0, 3, 0, 1, 0, 3, 0, 24, 0, 1, 33, 5, 3, 1, 1, 1, 1, 3, 5, 33, 1, 0, 49, 0, 4, 0, 1, 0, 1, 0, 4, 0, 49, 0
Offset: 0

Views

Author

Dean Hickerson, Mar 11 2002

Keywords

Examples

			Table begins:
  0, 1, 0, 1, 0, 1, 0, ...
  1, 1, 2, 3, 4, 6, 8, ...
  0, 2, 0, 2, 0, 2, 0, ...
  1, 3, 2, 1, 2, 2, 2, ...
  0, 4, 0, 2, 0, 1, 0, ...
  1, 6, 2, 2, 1, 1, 1, ...
  0, 8, 0, 2, 0, 1, 0, ...
  ...
		

Crossrefs

Cf. A068920 (total number of tilings), A052270 (count by area).
Cf. A068927 (row 2), A068928 (row 3), A068929 (row 4), A068930 (row 5), A068931 (row 6).

Programs

  • Mathematica
    (* See link above for Mathematica programs. *)
    c[r_, s_] := Which[s<0, 0, r==1, 1 - Mod[s, 2], r == 2, c1[2, s] + c2[2, s] + Boole[s == 0], OddQ[r], c[r, s] = c[r, s - r + 1] + c[r, s - r - 1] + Boole[s == 0], EvenQ[r], c[r, s] = c1[r, s] + c2[r, s] + Boole[s == 0]];
    c1[r_, s_] := Which[s <= 0, 0, r == 2, c[2, s - 1], EvenQ[r], c2[r, s - 1] + Boole[s == 1]];
    c2[r_, s_] := Which[s <= 0, 0, r == 2, c2[2, s] = c1[2, s - 2] + Boole[s == 2], EvenQ[r], c2[r, s] = c1[r, s - r + 2] + c1[r, s - r] + Boole[s == r - 2] + Boole[s == r]];
    cs[r_, s_] := Which[s < 0, 0, r == 1, c[r, s], r == 2, cs[2, s] = c1s[r, s] + c2s[r, s] + Boole[s == 0], OddQ[r], cs[r, s] = cs[r, s - 2 r + 2] + cs[r, s - 2 r - 2] + Boole[s == 0] + Boole[s == r - 1] + Boole[s == r + 1], EvenQ[r], cs[r, s] = c1s[r, s] + c2s[r, s] + Boole[s == 0]];
    c1s[r_, s_] := Which[s <= 0, 0, r == 2, cs[r, s - 2] + Boole[s == 1], EvenQ[r], c2s[r, s - 2] + Boole[s == 1]];
    c2s[r_, s_] := Which[s <= 0, 0, r == 2, c2s[2, s] = c1s[2, s - 4] + Boole[s == 2], EvenQ[r], c2s[r, s] = c1s[r, s - 2 r + 4] + c1s[r, s - 2 r] + Boole[s == r - 2] + Boole[s == r]];
    ti[r_, s_] := Which[r > s, ti[s, r], r == s, 1 - Mod[r, 2], True, (c[r, s] + cs[r, s])/2];
    A068926[n_] := Module[{x}, x = Floor[(Sqrt[8 n + 1] - 1)/2]; ti[n + 1 - x (x + 1)/2, (x + 1) (x + 2)/2 - n]];
    Table[A068926[n], {n, 0, 100}] (* Jean-François Alcover, May 12 2017, copied and adapted from Dean Hickerson's programs *)

A102543 Antidiagonal sums of the antidiagonals of Losanitsch's triangle.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 6, 8, 12, 16, 24, 33, 49, 69, 102, 145, 214, 307, 452, 653, 960, 1393, 2046, 2978, 4371, 6376, 9354, 13665, 20041, 29307, 42972, 62884, 92191, 134974, 197858, 289772, 424746, 622198, 911970, 1336121, 1958319, 2869417, 4205538, 6162579, 9031996, 13235661, 19398240
Offset: 0

Views

Author

Gerald McGarvey, Feb 24 2005

Keywords

Comments

The Ca1 and Ca2 sums, see A180662 for their definitions, of Losanitsch's triangle A034851 equal this sequence. - Johannes W. Meijer, Jul 14 2011
For n >= 5, a(n+1)-1 is the number of non-isomorphic snake polyominoes with n cells that can be inscribed in a rectangle of height 2. - Christian Barrientos and Sarah Minion, Jul 29 2018

Crossrefs

Programs

  • Maple
    A102543 := proc(n): (A000930(n)+x(n)+x(n-1)+x(n-3))/2 end: A000930:=proc(n): sum(binomial(n-2*i, i), i=0..n/3) end: x:=proc(n): if type(n, even) then A000930(n/2) else 0 fi: end: seq(A102543(n), n=0..38); # Johannes W. Meijer, Jul 14 2011
  • Mathematica
    CoefficientList[Series[(1 - x^2 - x^4 - x^6)/((x^3 + x - 1)*(x^6 + x^2 - 1)), {x, 0, 50}], x] (* G. C. Greubel, Apr 27 2017 *)
    LinearRecurrence[{1,1,0,0,-1,1,-1,0,-1},{1,1,1,2,2,3,4,6,8},50] (* Harvey P. Dale, Dec 14 2023 *)
  • PARI
    x='x+O('x^50); Vec((1 - x^2 - x^4 - x^6)/((x^3 + x - 1)*(x^6 + x^2 - 1))) \\ G. C. Greubel, Apr 27 2017

Formula

a(n) = A068927(n-1), n>3.
From Johannes W. Meijer, Jul 14 2011: (Start)
G.f.: (-1/2)*(1/(x^3+x-1)+(1+x+x^3)/(x^6+x^2-1))= ( 1-x^2-x^4-x^6 ) / ( (x^3+x-1)*(x^6+x^2-1) ).
a(n) = (A000930(n)+x(n)+x(n-1)+x(n-3))/2 with x(2*n) = A000930(n) and x(2*n+1) = 0. (End)
Showing 1-3 of 3 results.