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 1-4 of 4 results.

A103997 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2*M X 2*N Moebius strip.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 11, 7, 1, 1, 41, 71, 18, 1, 1, 153, 769, 539, 47, 1, 1, 571, 8449, 17753, 4271, 123, 1, 1, 2131, 93127, 603126, 434657, 34276, 322, 1, 1, 7953, 1027207, 20721019, 46069729, 10894561, 276119, 843, 1, 1, 29681, 11332097, 714790675, 4974089647, 3625549353, 275770321, 2226851, 2207, 1
Offset: 0

Views

Author

Ralf Stephan, Feb 26 2005

Keywords

Examples

			Array begins:
  1,   1,     1,        1,          1,             1,               1,
  1,   3,     7,       18,         47,           123,             322,
  1,  11,    71,      539,       4271,         34276,          276119,
  1,  41,   769,    17753,     434657,      10894561,       275770321,
  1, 153,  8449,   603126,   46069729,    3625549353,    289625349454,
  1, 571, 93127, 20721019, 4974089647, 1234496016491, 312007855309063,
  ...
		

Crossrefs

Rows include A005248, A103998.
Columns 1..7 give A001835(n+1), A334135, A334179, A334180, A334181, A334182, A334183.
Main diagonal gives A334124.

Programs

  • Mathematica
    T[M_, N_] := Product[4Sin[(4n-1)Pi/(4N)]^2 + 4Cos[m Pi/(2M+1)]^2, {n, 1, N}, {m, 1, M}];
    Table[T[M - N, N] // Round, {M, 0, 9}, {N, 0, M}] // Flatten (* Jean-François Alcover, Dec 03 2018 *)

Formula

T(M, N) = Product_{m=1..M} (Product_{n=1..N} 4*sin(Pi*(4*n-1)/(4*N))^2 + 4*cos(Pi*m/(2*M + 1))^2).
For k > 0, T(n,k) = 2^n * sqrt(Resultant(U_{2*n}(x/2), T_{2*k}(i*x/2))), where T_n(x) is a Chebyshev polynomial of the first kind, U_n(x) is a Chebyshev polynomial of the second kind and i = sqrt(-1). - Seiichi Manyama, Apr 15 2020

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

A256045 Triangle read by rows: order of all-2s configuration on the n X k sandpile grid graph.

Original entry on oeis.org

2, 3, 1, 7, 7, 8, 11, 5, 71, 3, 26, 9, 679, 77, 52, 41, 13, 769, 281, 17753, 29, 97, 47, 3713, 4271, 726433, 434657, 272, 153, 17, 8449, 2245, 33507, 167089, 46069729, 901, 362, 123, 81767, 8569, 24852386, 265721, 8118481057, 190818387, 73124, 571, 89, 93127, 18061, 20721019, 4213133, 4974089647, 1031151241, 1234496016491, 89893
Offset: 1

Views

Author

N. J. A. Sloane, Mar 15 2015

Keywords

Examples

			Triangle begins:
[2]
[3, 1]
[7, 7, 8]
[11, 5, 71, 3]
[26, 9, 679, 77, 52]
[41, 13, 769, 281, 17753, 29]
[97, 47, 3713, 4271, 726433, 434657, 272]
[153, 17, 8449, 2245, 33507, 167089, 46069729, 901]
[362, 123, 81767, 8569, 24852386, 265721, 8118481057, 190818387, 73124]
[571, 89, 93127, 18061, 20721019, 4213133, 4974089647, 1031151241, 1234496016491, 89893]
...
		

Crossrefs

Main diagonal gives A256046, A256043, and A256047.

Formula

From Andrey Zabolotskiy, Oct 22 2021: (Start)
It seems that T(k, 1) = A005246(k+2).
For the formula for T(k, 2), see the last theorem of Morar and Perkinson in Perkinson's slides. In particular, T(2*k, 2) = A195549(k).
T(n, k) divides A348566(n, k). (End)

Extensions

Column 1 added by Andrey Zabolotskiy, Oct 22 2021

A348567 Number of fully symmetric recurrent sandpiles on an n X n grid.

Original entry on oeis.org

1, 4, 2, 32, 12, 832, 232, 69632, 14416, 18719744, 2876576
Offset: 0

Views

Author

Andrey Zabolotskiy, Oct 22 2021

Keywords

Examples

			a(0) = 1, the empty sandpile on an empty grid.
a(1) = 4, all 4 sandpiles on a 1 X 1 grid are fully symmetric.
a(2) = 2 fully symmetric recurrent sandpiles on an 2 X 2 grid are the all-2 and all-3 sandpiles.
Among A348566(4, 4) = 36 sandpiles with horizontal and vertical symmetry, only a(4) = 12 also have diagonal symmetries:
  3333  1331  1331  3223  3223  3333  2222  0330  2332  2332  0330  2222
  3333  3223  3333  2222  2332  3223  2222  3223  3223  3333  3333  2332
  3333  3223  3333  2222  2332  3223  2222  3223  3223  3333  3333  2332
  3333  1331  1331  3223  3223  3333  2222  0330  2332  2332  0330  2222
		

Crossrefs

Formula

A256046(n) divides a(n), a(n) divides A348566(n, n).
Showing 1-4 of 4 results.