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 1-10 of 10 results.

A188899 Third row of array in A187617.

Original entry on oeis.org

1, 5, 36, 281, 2245, 18061, 145601, 1174500, 9475901, 76455961, 616891945, 4977472781, 40161441636, 324048393905, 2614631600701, 21096536145301, 170220478472105, 1373448758774436, 11081871650713781, 89415697915538545, 721463601671126161, 5821234309893001301, 46969478172465070500, 378980086070257592201, 3057856106268358639861
Offset: 0

Views

Author

N. J. A. Sloane, Apr 13 2011

Keywords

Crossrefs

Bisection (odd part) of A005178. - Alois P. Heinz, Oct 28 2012

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);
    gt:=(m,n)->round(evalf(ft(m,n),300));
    tt:=[seq(gt(4,2*n),n=0..10)];
    # second Maple program:
    a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <-1|11|-25|11>>^n.
            <<1, 5, 36, 281>>)[1, 1]:
    seq(a(n), n=0..30);  # Alois P. Heinz, Oct 28 2012
  • Mathematica
    LinearRecurrence[{11, -25, 11, -1}, {1, 5, 36, 281}, 25] (* Jean-François Alcover, Jun 17 2018 *)
  • PARI
    x='x+O('x^200); Vec((1-x)*(x^2-5*x+1)/(x^4-11*x^3+25*x^2-11*x+1)) \\ Altug Alkan, Mar 23 2016

Formula

G.f.: (1-x)*(x^2-5*x+1)/(x^4-11*x^3+25*x^2-11*x+1). - Alois P. Heinz, Oct 28 2012

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

A187596 Array T(m,n) read by antidiagonals: number of domino tilings of the m X n grid (m>=0, n>=0).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Mar 11 2011

Keywords

Comments

A099390 supplemented by an initial row and column of 1's.
See A099390 (the main entry for this array) for further information.
If we work with the row index starting at 1 then every row of the array is a divisibility sequence, i.e., the terms satisfy the property that if n divides m then a(n) divide a(m) provided a(n) != 0. Row k satisfies a linear recurrence of order 2^floor(k/2) (Stanley, Ex. 36 p. 273). - Peter Bala, Apr 30 2014

Examples

			Array begins:
  1,  1,  1,  1,   1,    1,     1,     1,       1,      1,        1, ...
  1,  0,  1,  0,   1,    0,     1,     0,       1,      0,        1, ...
  1,  1,  2,  3,   5,    8,    13,    21,      34,     55,       89, ...
  1,  0,  3,  0,  11,    0,    41,     0,     153,      0,      571, ...
  1,  1,  5, 11,  36,   95,   281,   781,    2245,   6336,    18061, ...
  1,  0,  8,  0,  95,    0,  1183,     0,   14824,      0,   185921, ...
  1,  1, 13, 41, 281, 1183,  6728, 31529,  167089, 817991,  4213133, ...
  1,  0, 21,  0, 781,    0, 31529,     0, 1292697,      0, 53175517, ...
		

References

  • R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.

Crossrefs

Cf. A099390.
See A187616 for a triangular version, and A187617, A187618 for the sub-array T(2m,2n).
See also A049310, A053117.

Programs

  • Maple
    with(LinearAlgebra):
    T:= proc(m,n) option remember; local i, j, t, M;
          if m<=1 or n<=1 then 1 -irem(n*m, 2)
        elif irem(n*m, 2)=1 then 0
        elif mAlois P. Heinz, Apr 11 2011
  • Mathematica
    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}]; t[?OddQ, ?OddQ] = 0; Table[t[m-n, n] // FullSimplify, {m, 0, 13}, {n, 0, m}] // Flatten (* Jean-François Alcover, Jan 07 2014, after A099390 *)

Formula

From Peter Bala, Apr 30 2014: (Start)
T(n,k)^2 = absolute value of Product_{b=1..k} Product_{a=1..n} ( 2*cos(a*Pi/(n+1)) + 2*i*cos(b*Pi/(k+1)) ), where i = sqrt(-1). See Propp, Section 5.
Equivalently, working with both the row index n and column index k starting at 1 we have T(n,k)^2 = absolute value of Resultant (F(n,x), U(k-1,x/2)), where U(n,x) is a Chebyshev polynomial of the second kind and F(n,x) is a Fibonacci polynomial defined recursively by F(0,x) = 0, F(1,x) = 1 and F(n,x) = x*F(n-1,x) + F(n-2,x) for n >= 2. The divisibility properties of the array entries mentioned in the Comments are a consequence of this result. (End)

A239264 Number A(n,k) of domicule tilings of a k X n grid; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 3, 0, 1, 1, 1, 5, 5, 1, 1, 1, 0, 11, 0, 11, 0, 1, 1, 1, 21, 43, 43, 21, 1, 1, 1, 0, 43, 0, 280, 0, 43, 0, 1, 1, 1, 85, 451, 1563, 1563, 451, 85, 1, 1, 1, 0, 171, 0, 9415, 0, 9415, 0, 171, 0, 1, 1, 1, 341, 4945, 55553, 162409, 162409, 55553, 4945, 341, 1, 1
Offset: 0

Views

Author

Alois P. Heinz, Mar 13 2014

Keywords

Comments

A domicule is either a domino or it is formed by the union of two neighboring unit squares connected via their corners. In a tiling the connections of two domicules are allowed to cross each other.

Examples

			A(3,2) = 5:
  +-----+ +-----+ +-----+ +-----+ +-----+
  |o o-o| |o o o| |o o o| |o o o| |o-o o|
  ||    | ||  X | || | || | X  || |    ||
  |o o-o| |o o o| |o o o| |o o o| |o-o o|
  +-----+ +-----+ +-----+ +-----+ +-----+
A(4,3) = 43:
  +-------+ +-------+ +-------+ +-------+ +-------+
  |o o o o| |o o o-o| |o o-o o| |o o-o o| |o o-o o|
  ||  X  || | X     | | \   / | ||     || | \    ||
  |o o o o| |o o o o| |o o o o| |o o o o| |o o o o|
  |       | |     X | ||     || |   \ \ | ||    \ |
  |o-o o-o| |o-o o o| |o o-o o| |o-o o o| |o o-o o|
  +-------+ +-------+ +-------+ +-------+ +-------+ ...
Square array A(n,k) begins:
  1, 1,  1,   1,    1,      1,       1, ...
  1, 0,  1,   0,    1,      0,       1, ...
  1, 1,  3,   5,   11,     21,      43, ...
  1, 0,  5,   0,   43,      0,     451, ...
  1, 1, 11,  43,  280,   1563,    9415, ...
  1, 0, 21,   0, 1563,      0,  162409, ...
  1, 1, 43, 451, 9415, 162409, 3037561, ...
		

Crossrefs

Columns (or rows) k=0-10 give: A000012, A059841, A001045(n+1), A239265, A239266, A239267, A239268, A239269, A239270, A239271, A239272.
Bisection of main diagonal gives: A239273.

Programs

  • Maple
    b:= proc(n, l) option remember; local d, f, k;
          d:= nops(l)/2; f:=false;
          if n=0 then 1
        elif l[1..d]=[f$d] then b(n-1, [l[d+1..2*d][], true$d])
        else for k to d while not l[k] do od;
             `if`(k1 and l[k+d+1],
                                  b(n, subsop(k=f, k+d+1=f, l)), 0)+
             `if`(k>1 and n>1 and l[k+d-1],
                                  b(n, subsop(k=f, k+d-1=f, l)), 0)+
             `if`(n>1 and l[k+d], b(n, subsop(k=f, k+d=f, l)), 0)+
             `if`(k `if`(irem(n*k, 2)>0, 0,
        `if`(k>n, A(k, n), b(n, [true$(k*2)]))):
    seq(seq(A(n, d-n), n=0..d), d=0..14);
  • Mathematica
    b[n_, l_List] := b[n, l] = Module[{d = Length[l]/2, f = False, k}, Which [n == 0, 1, l[[1 ;; d]] == Array[f&, d], b[n-1, Join[l[[d+1 ;; 2*d]], Array[True&, d]]], True, For[k=1, !l[[k]], k++]; If[k1 && l[[k+d+1]], b[n, ReplacePart[l, {k -> f, k+d+1 -> f}]], 0] + If[k>1 && n>1 && l[[k+d-1]], b[n, ReplacePart[l, {k -> f, k+d-1 -> f}]], 0] + If[n>1 && l[[k+d]], b[n, ReplacePart[l, {k -> f, k+d -> f}]], 0] + If[k f, k+1 -> f}]], 0]]]; A[n_, k_] := If[Mod[n*k, 2]>0, 0, If[k>n, A[k, n], b[n, Array[True&, k*2]]]]; Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 14}] // Flatten (* Jean-François Alcover, Feb 02 2015, after Alois P. Heinz *)

A187616 Triangle T(m,n) read by rows: number of domino tilings of the m X n grid (0 <= m <= n).

Original entry on oeis.org

1, 1, 0, 1, 1, 2, 1, 0, 3, 0, 1, 1, 5, 11, 36, 1, 0, 8, 0, 95, 0, 1, 1, 13, 41, 281, 1183, 6728, 1, 0, 21, 0, 781, 0, 31529, 0, 1, 1, 34, 153, 2245, 14824, 167089, 1292697, 12988816, 1, 0, 55, 0, 6336, 0, 817991, 0, 108435745, 0, 1, 1, 89, 571, 18061, 185921, 4213133, 53175517, 1031151241, 14479521761, 258584046368
Offset: 0

Views

Author

N. J. A. Sloane, Mar 11 2011

Keywords

Comments

A099390 is the main entry for this problem.
Triangle read by rows: the square array in A187596 with entries above main diagonal deleted.

Examples

			Triangle begins:
1
1 0
1 1  2
1 0  3   0
1 1  5  11   36
1 0  8   0   95     0
1 1 13  41  281  1183   6728
1 0 21   0  781     0  31529       0
1 1 34 153 2245 14824 167089 1292697 12988816
...
		

Crossrefs

Cf. A099390, A187596. See A099390 for sequences appearing in the rows and columns. See also A187617, A187618.

Programs

  • Maple
    with(LinearAlgebra):
    T:= proc(m,n) option remember; local i, j, t, M;
          if m<=1 or n<=1 then 1 -irem(n*m, 2)
        elif irem(n*m, 2)=1 then 0
        else M:= Matrix(n*m, shape =skewsymmetric);
             for i to n do
               for j to m do
                 t:= (i-1)*m+j;
                 if jAlois P. Heinz, Apr 11 2011
  • Mathematica
    T[m_, n_] := T[m, n] = Module[{i, j, t, M}, Which[m <= 1 || n <= 1, 1 - Mod[n*m, 2], Mod[n*m, 2] == 1, 0, True, 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; If[j < m, M[t, t+1] = 1]; If[i < n, M[t, t+m] = 1 - 2*Mod[j, 2]]]]; Sqrt[Det[Table[M[i, j], {i, 1, n*m}, {j, 1, n*m}]]]]]; Table[Table[T[m, n], {n, 0, m}], {m, 0, 10}] // Flatten (* Jean-François Alcover, Jan 07 2014, translated from Maple *)

A189006 Array A(m,n) read by antidiagonals: number of domino tilings of the m X n grid with upper left corner removed iff m*n is odd, (m>=0, n>=0).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 5, 4, 5, 1, 1, 1, 1, 8, 11, 11, 8, 1, 1, 1, 1, 13, 15, 36, 15, 13, 1, 1, 1, 1, 21, 41, 95, 95, 41, 21, 1, 1, 1, 1, 34, 56, 281, 192, 281, 56, 34, 1, 1, 1, 1, 55, 153, 781, 1183, 1183, 781, 153, 55, 1, 1, 1, 1, 89, 209, 2245, 2415, 6728, 2415, 2245, 209, 89, 1, 1
Offset: 0

Views

Author

Alois P. Heinz, Apr 15 2011

Keywords

Examples

			A(3,3) = 4, because there are 4 domino tilings of the 3 X 3 grid with upper left corner removed:
  . .___. . .___. . .___. . .___.
  ._|___| ._|___| ._| | | ._|___|
  | |___| | | | | | |_|_| |___| |
  |_|___| |_|_|_| |_|___| |___|_|
Array begins:
  1, 1,  1,  1,   1,    1,    1, ...
  1, 1,  1,  1,   1,    1,    1, ...
  1, 1,  2,  3,   5,    8,   13, ...
  1, 1,  3,  4,  11,   15,   41, ...
  1, 1,  5, 11,  36,   95,  281, ...
  1, 1,  8, 15,  95,  192, 1183, ...
  1, 1, 13, 41, 281, 1183, 6728, ...
		

Crossrefs

Rows m=0+1, 2-12 give: A000012, A000045(n+1), A002530(n+1), A005178(n+1), A189003, A028468, A189004, A028470, A189005, A028472, A210724, A028474.
Main diagonal gives: A189002.

Programs

  • Maple
    with(LinearAlgebra):
    A:= proc(m, n) option remember; local i, j, s, t, M;
          if m=0 or n=0 then 1
        elif m1 or j>1 or s=0 then
                   if j
    				
  • Mathematica
    A[1, 1] = 1; 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}]]]]]; Table[Table[A[m, d-m], {m, 0, d}], {d, 0, 15}] // Flatten (* Jean-François Alcover, Dec 26 2013, translated from Maple *)

A187618 Triangle T(m,n) read by rows: number of domino tilings of the 2m X 2n grid (0 <= m <= n).

Original entry on oeis.org

1, 1, 2, 1, 5, 36, 1, 13, 281, 6728, 1, 34, 2245, 167089, 12988816, 1, 89, 18061, 4213133, 1031151241, 258584046368, 1, 233, 145601, 106912793, 82741005829, 65743732590821, 53060477521960000, 1, 610, 1174500, 2720246633, 6675498237130, 16848161392724969, 43242613716069407953, 112202208776036178000000
Offset: 0

Views

Author

N. J. A. Sloane, Mar 12 2011

Keywords

Comments

A099390 is the main entry for this problem.

Examples

			Triangle begins:
1
1       2
1       5       36
1       13      281     6728
1       34      2245    167089  12988816
1       89      ...
		

Crossrefs

Extensions

More terms from Nathaniel Johnston, Mar 22 2011

A348566 Triangle read by rows: T(m, n) is the number of symmetric recurrent sandpiles on an m X n grid (m >= 0, 0 <= n <= m).

Original entry on oeis.org

1, 1, 4, 1, 3, 2, 1, 14, 7, 128, 1, 11, 5, 71, 36, 1, 52, 18, 1358, 539, 43264, 1, 41, 13, 769, 281, 17753, 6728, 1, 194, 47, 14852, 4271, 1452866, 434657, 151519232, 1, 153, 34, 8449, 2245, 603126, 167089, 46069729, 12988816, 1, 724, 123, 163534, 34276, 49704772, 10894561, 16236962114, 3625549353, 5475450241024
Offset: 0

Views

Author

Andrey Zabolotskiy, Oct 22 2021

Keywords

Comments

Terms of this triangle count recurrent sandpiles on rectangular grids that have vertical and horizontal symmetries. Terms of A348567 count recurrent sandpiles on square grids that also have diagonal symmetries.

Examples

			The triangle begins:
  1
  1  4
  1  3  2
  1 14  7  128
  1 11  5   71  36
  1 52 18 1358 539 43264
  1 41 13  769 281 17753 6728
...
See Fig. 9 of the paper by Florescu et al. for the T(4, 4) = 36 symmetric recurrent sandpiles on a 4x4 grid.
		

Crossrefs

Formula

T(2m, 2n) = A187617(m, n) = A187618(m, n). [Florescu et al., Theorem 15]
T(2m, 2n-1) = T(2n-1, 2m) = A103997(m, n). [Florescu et al., Theorem 18]
T(2m-1, 2n-1) = Product_{h=1..m, k=1..n} 4*(z(h, m) + z(k, n)) where z(k, n) = cos(Pi*(2k-1)/(4n)). [Florescu et al., Theorem 23]
A256045(m, n) divides T(m, n), T(m, n) divides A116469(m+1, n+1).
This triangle can obviously be extended to n > m as T(m, n) = T(n, m).

A340475 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Product_{a=1..n} Product_{b=1..k} (4*sin(a*Pi/(2*n+1))^2 + 4*sin(b*Pi/(2*k+1))^2).

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 29, 29, 1, 1, 139, 500, 139, 1, 1, 666, 8329, 8329, 666, 1, 1, 3191, 138301, 463736, 138301, 3191, 1, 1, 15289, 2295701, 25543057, 25543057, 2295701, 15289, 1, 1, 73254, 38105729, 1404312491, 4614756624, 1404312491, 38105729, 73254, 1
Offset: 0

Views

Author

Seiichi Manyama, Jan 09 2021

Keywords

Examples

			Square array begins:
  1,   1,      1,        1,          1, ...
  1,   6,     29,      139,        666, ...
  1,  29,    500,     8329,     138301, ...
  1, 139,   8329,   463736,   25543057, ...
  1, 666, 138301, 25543057, 4614756624, ...
		

Crossrefs

Rows and columns 0..1 give A000012, A030221.
Main diagonal gives A127605.

Programs

  • PARI
    default(realprecision, 120);
    {T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*sin(a*Pi/(2*n+1))^2+4*sin(b*Pi/(2*k+1))^2)))}

Formula

T(n,k) = T(k,n).

A340476 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = Product_{a=1..n} Product_{b=1..k} (4*sin(a*Pi/(2*n+1))^2 + 4*cos(b*Pi/(2*k+1))^2).

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 19, 11, 1, 1, 91, 176, 29, 1, 1, 436, 2911, 1471, 76, 1, 1, 2089, 48301, 79808, 11989, 199, 1, 1, 10009, 801701, 4375897, 2091817, 97021, 521, 1, 1, 47956, 13307111, 240378643, 372713728, 53924597, 783511, 1364, 1
Offset: 0

Views

Author

Seiichi Manyama, Jan 09 2021

Keywords

Examples

			Square array begins:
  1,  1,     1,       1,         1, ...
  1,  4,    19,      91,       436, ...
  1, 11,   176,    2911,     48301, ...
  1, 29,  1471,   79808,   4375897, ...
  1, 76, 11989, 2091817, 372713728, ...
		

Crossrefs

Column k=0..1 give A000012, A002878.
Main diagonal gives A127606.

Programs

  • PARI
    default(realprecision, 120);
    {T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*sin(a*Pi/(2*n+1))^2+4*cos(b*Pi/(2*k+1))^2)))}
    
  • PARI
    {T(n, k) = sqrtint(4^k*polresultant(polchebyshev(2*n+1, 1, I*x/2), polchebyshev(2*k, 2, x/2)))}

Formula

T(n,k) = 2^k * sqrt(Resultant(T_{2*n+1}(i*x/2), U_{2*k}(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).
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