A005386 -id:A005386 - cp's OEIS frontend

A004253 a(n) = 5*a(n-1) - a(n-2), with a(1)=1, a(2)=4.

1, 4, 19, 91, 436, 2089, 10009, 47956, 229771, 1100899, 5274724, 25272721, 121088881, 580171684, 2779769539, 13318676011, 63813610516, 305749376569, 1464933272329, 7018916985076, 33629651653051, 161129341280179, 772017054747844, 3698955932459041, 17722762607547361

Offset: 1

Frans J. Faase, Per H. Lundow

Number of domino tilings in K_3 X P_2n (or in S_4 X P_2n).
Number of perfect matchings in graph C_{3} X P_{2n}.
Number of perfect matchings in S_4 X P_2n.
In general, Sum_{k=0..n} binomial(2*n-k, k)*j^(n-k) = (-1)^n * U(2*n, i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,5), where L is defined as in A108299; see also A030221 for L(n,-5). - Reinhard Zumkeller, Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4} which do not end in 0 (e.g., at n=2, we have 02, 03, 04, 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44). - Tanya Khovanova, Jan 10 2007
(sqrt(21)+5)/2 = 4.7912878... = exp(arccosh(5/2)) = 4 + 3/4 + 3/(4*19) + 3/(19*91) + 3/(91*436) + ... - Gary W. Adamson, Dec 18 2007
a(n+1) is the number of compositions of n when there are 4 types of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(3)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Right-shifted Binomial Transform of the left-shifted A030195. - R. J. Mathar, Oct 15 2012
Values of x (or y) in the solutions to x^2 - 5xy + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
From Wolfdieter Lang, Oct 15 2020: (Start)
All positive solutions of the Diophantine equation x^2 + y^2 - 5*x*y = -3 (see the preceding comment) are given by [x(n) = S(n, 5) - S(n-1, 5), y(n) = x(n-1)], for n =-oo..+oo, with the Chebyshev S-polynomials (A049310), with S(-1, 0) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.
This binary indefinite quadratic form has discriminant D = +21. There is only this family representing -3 properly with x and y positive, and there are no improper solutions.
See the  formula for a(n) = x(n-1), for n >= 1, in terms of S-polynomials below.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
From Wolfdieter Lang, Feb 08 2021: (Start)
All proper and improper solutions of the generalized Pell equation X^2 - 21*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n+1) from the preceding comment by X(n) = x(n) + x(n-1) = S(n-1, 5) - S(n-2, 5) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 5), for all integer numbers n. For positive integers X(n) = A003501(n) and Y(n) = A004254(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.
The two conjugated proper families of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer n. This follows from the mentioned paper by Robert K. Moniot. (End)
Equivalent definition: a(n) = ceiling(a(n-1)^2 / a(n-2)), with a(1)=1, a(2)=4, a(3)=19. The problem for USA Olympiad (see Andreescu and Gelca reference) asks to prove that a(n)-1 is always a multiple of 3. - Bernard Schott, Apr 13 2022

Titu Andreescu and Rǎzvan Gelca, Putnam and Beyond, New York, Springer, 2007, problem 311, pp. 104 and 466-467 (proposed for the USA Mathematical Olympiad by G. Heuer).
F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.
F. A. Haight, On a generalization of Pythagoras' theorem, pp. 73-77 of J. C. Butcher, editor, A Spectrum of Mathematics. Auckland University Press, 1971.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..200
F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Preliminary version of paper that appeared in Ars Combin. 49 (1998), 129-154.
F. Faase, Counting Hamiltonian cycles in product graphs
F. Faase, Results from the counting program
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Frank A. Haight, Letter to N. J. A. Sloane, Sep 06 1976
Frank A. Haight, On a generalization of Pythagoras' theorem, pp. 73-77 of J. C. Butcher, editor, A Spectrum of Mathematics. Auckland University Press, 1971. [Annotated scanned copy]
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 422
Tanya Khovanova, Recursive Sequences
Lisa Lokteva, Constructing Rational Homology 3-Spheres That Bound Rational Homology 4-Balls, arXiv:2208.14850 [math.GT], 2022.
Per Hakan Lundow, Computation of matching polynomials and the number of 1-factors in polygraphs, Research report, No 12, 1996, Department of Math., Umea University, Sweden.
Per Hakan Lundow, Enumeration of matchings in polygraphs, 1998.
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962, 2014
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
John Riordan, Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences. Note that the sequences are identified by their N-numbers, not their A-numbers.
James A. Sellers, Domino Tilings and Products of Fibonacci and Pell Numbers, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.2
F. M. van Lamoen, Square wreaths around hexagons, Forum Geometricorum, 6 (2006) 311-325.
Index entries for sequences related to dominoes
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A003501, A004254, A030221, A049310, A004254 (partial sums), A290902 (first differences).
Row 5 of array A094954.
Cf. similar sequences listed in A238379.

a:=[1,4];; for n in [3..30] do a[n]:=5*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Oct 23 2019

[ n eq 1 select 1 else n eq 2 select 4 else 5*Self(n-1)-Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011

a[0]:=1: a[1]:=1: for n from 2 to 26 do a[n]:=5*a[n-1]-a[n-2] od: seq(a[n], n=1..22); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{5, -1}, {1, 4}, 22] (* Jean-François Alcover, Sep 27 2017 *)

Vec((1-x)/(1-5*x+x^2)+O(x^30)) \\ Charles R Greathouse IV, Jul 01 2013

[lucas_number1(n,5,1)-lucas_number1(n-1,5,1) for n in range(1, 23)] # Zerinvary Lajos, Nov 10 2009

G.f.: x*(1 - x) / (1 - 5*x + x^2).  Simon Plouffe in his 1992 dissertation.[offset 0]
For n>1, a(n) = A005386(n) + A005386(n-1). - Floor van Lamoen, Dec 13 2006
a(n) ~ (1/2 + 1/14*sqrt(21))*(1/2*(5 + sqrt(21)))^n. - Joe Keane (jgk(AT)jgk.org), May 16 2002[offset 0]
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then q(n, 3)=a(n). - Benoit Cloitre, Nov 10 2002 [offset 0]
For n>0, a(n)*a(n+3) = 15 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*3^k. - Paul Barry, Jul 26 2004[offset 0]
a(n) = (-1)^n*U(2n, i*sqrt(3)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005[offset 0]
[a(n), A004254(n)] = the 2 X 2 matrix [1,3; 1,4]^n * [1,0]. - Gary W. Adamson, Mar 19 2008
a(n) = ((sqrt(21)-3)*((5+sqrt(21))/2)^n + (sqrt(21)+3)*((5-sqrt(21))/2)^n)/2/sqrt(21). - Seiichi Kirikami, Sep 06 2011
a(n) = S(n-1, 5) - S(n-2, 5) = (-1)^n*S(2*n, i*sqrt(3)), n >= 1, with the Chebyshev S polynomials (A049310), and S(n-1, 5) = A004254(n), for n >= 0. See a Paul Barry formula (offset corrected). - Wolfdieter Lang, Oct 15 2020
From Peter Bala, Feb 10 2024: (Start)
a(n) = a(1-n).
a(n) = A004254(n) + A004254(1-n).
For n, j, k in Z, a(n)*a(n+j+k) - a(n+j)*a(n+k) = 3*A004254(j)*A004254(k). The case j = 1, k = 2 is given above.
a(n)^2 + a(n+1)^2 - 5*a(n)*a(n+1) = - 3.
More generally, a(n)^2 + a(n+k)^2 - (A004254(k+1) - A004254(k-1))*a(n)*a(n+k) = -3*A004254(k)^2. (End)
Sum_{n >= 2} 1/(a(n) - 1/a(n)) = 1/3 (telescoping series: for n >= 2,  3/(a(n) - 1/a(n)) = 1/A004254(n-1) - 1/A004254(n)). - Peter Bala, May 21 2025
E.g.f.: exp(5*x/2)*(7*cosh(sqrt(21)*x/2) - sqrt(21)*sinh(sqrt(21)*x/2))/7 - 1. - Stefano Spezia, Jul 02 2025

Additional comments from James Sellers and N. J. A. Sloane, May 03 2002

More terms from Ray Chandler, Nov 17 2003

A092936 Area of n-th triple of hexagons around a triangle.

Original entry on oeis.org

1, 9, 100, 1089, 11881, 129600, 1413721, 15421329, 168220900, 1835008569, 20016873361, 218350598400, 2381839709041, 25981886201049, 283418908502500, 3091626107326449, 33724468272088441, 367877524885646400

Offset: 1

Peter J. C. Moses, Apr 18 2004

This is the unsigned member r=-9 of the family of Chebyshev sequences S_r(n) defined in A092184: ((-1)^(n+1))*a(n) = S_{-9}(n), n>=0.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/2,1/2)-fences, red half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal), green half-squares, and blue half-squares. A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,3/4)-fences, red (1/4,1/4)-fences, green (1/4,1/4)-fences, and blue (1/4,1/4)-fences. - Michael A. Allen, Dec 30 2022

			a(5) = 10*(1089+100)-9 = 11881. From A006190, a(5) = (3*33+10)^2 = 11881.

Muniru A Asiru, Table of n, a(n) for n = 1..200
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (10,10,-1).

Cf. A005386, A006190, A057076, A092184, A176019.

a:=[1,9,100];; for n in [4..18] do a[n]:=10*(a[n-1]+a[n-2])-a[n-3]; od; a; # Muniru A Asiru, Feb 20 2018

seq(fibonacci(n,3)^2,n=1..18); # Zerinvary Lajos, Apr 05 2008

CoefficientList[Series[(1-x)*x/(1-10*x-10*x^2+x^3), {x, 0, 20}], x]
(CoefficientList[Series[x/(1-3*x-x^2), {x, 0, 20}], x])^2
Table[Round[((3+Sqrt[13])^n)^2/(13*4^n)], {n, 0, 20}]
LinearRecurrence[{10, 10, -1}, {1, 9, 100}, 18] (* Georg Fischer, Feb 22 2019 *)

a(n) = A006190(n)^2.
a(n) = 10*(a(n-1)+a(n-2)) - a(n-3).
G.f.: (1-x)*x/(1-10*x-10*x^2+x^3).
a(n) = ((3-sqrt(13))^n-(3+sqrt(13))^n)^2/(13*4^n).
a(n) = 2*(T(n, 11/2)-(-1)^n)/13 with twice the Chebyshev polynomials of the first kind evaluated at x = 11/2: 2*T(n, 11/2) = A057076(n) = ((11+3*sqrt(13))^n + (11-3*sqrt(13))^n)/2^n. - Wolfdieter Lang, Oct 18 2004
From Michael A. Allen, Dec 30 2022: (Start)
a(n+1) = 11*a(n) - a(n-1) + 2*(-1)^n.
a(n+1) = (1 + (-1)^n)/2 + 9*Sum_{k=1..n} ( k*a(n+1-k) ). (End)
Product_{n>=2} (1 + (-1)^n/a(n)) = (3 + sqrt(13))/6 (A176019) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A003769 Number of perfect matchings (or domino tilings) in K_4 X P_n.

Original entry on oeis.org

3, 16, 75, 361, 1728, 8281, 39675, 190096, 910803, 4363921, 20908800, 100180081, 479991603, 2299777936, 11018898075, 52794712441, 252954664128, 1211978608201, 5806938376875, 27822713276176, 133306628004003, 638710426743841, 3060245505715200

Offset: 1

Frans J. Faase

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

Colin Barker, Table of n, a(n) for n = 1..1000
F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Preliminary version of paper that appeared in Ars Combin. 49 (1998), 129-154.
F. Faase, Counting Hamiltonian cycles in product graphs
F. Faase, Results from the counting program
Index entries for sequences related to dominoes
Index entries for linear recurrences with constant coefficients, signature (4,4,-1).

Essentially the same as A005386. First differences of A099025.

Vec(x*(3 + 4*x - x^2) / ((1 + x)*(1 - 5*x + x^2)) + O(x^40)) \\ Colin Barker, Dec 16 2017

a(n) = 4a(n-1) + 4a(n-2) - a(n-3), n>3.
a(n) = (1/7)*(6*A030221(n) - A054477(n) + 2(-1)^n).
G.f.: x*(3+4*x-x^2)/((1+x)*(1-5*x+x^2)). - R. J. Mathar, Dec 16 2008
a(n) = 2^(-1-n)*((-1)^n*2^(2+n) + (5-sqrt(21))^(1+n) + (5+sqrt(21))^(1+n)) / 7. - Colin Barker, Dec 16 2017

A099025 Expansion of 1 / ((1+x) * (1-5*x+x^2)).

Original entry on oeis.org

1, 4, 20, 95, 456, 2184, 10465, 50140, 240236, 1151039, 5514960, 26423760, 126603841, 606595444, 2906373380, 13925271455, 66719983896, 319674648024, 1531653256225, 7338591633100, 35161304909276, 168467932913279, 807178359657120, 3867423865372320

Offset: 0

Ralf Stephan, Sep 26 2004

			1 + 4*x + 20*x^2 + 95*x^3 + 456*x^4 + 2184*x^5 + 10465*x^6 + ...

Colin Barker, Table of n, a(n) for n = 0..1000
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Index entries for linear recurrences with constant coefficients, signature (4,4,-1).

First differences of A089927. First differences are in A003769 and A005386. Pairwise sums are in A004254.

I:=[1, 4, 20]; [n le 3 select I[n] else 4*Self(n-1) + 4*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 31 2017

CoefficientList[Series[1/((1+x)*(1-5*x+x^2)), {x,0,50}], x] (* or *) LinearRecurrence[{4,4,-1}, {1,4,20}, 30] (* G. C. Greubel, Dec 31 2017 *)

Vec(1/(1+x)/(1-5*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

{a(n) = (3 * (-1)^n + 38 * subst( poltchebi(n), x, 5/2) - 8 * subst( poltchebi(n-1), x, 5/2)) / 21} /* Michael Somos, Jan 25 2013 */

a(n) = (1/7)*[A030221(n+2) - A003501(n+2) + (-1)^n].
a(n) = 5*a(n-1) -a(n-2) +(-1)^n, a(0)=1, a(1)=4. - Vincenzo Librandi, Mar 22 2011
G.f.: 1 / ((1 + x) * (1 - 5*x + x^2)).
a(-3-n) = -a(n). - Michael Somos, Jan 25 2013
a(n) = (2^(-n)*(3*(-2)^n+(9-2*sqrt(21))*(5-sqrt(21))^n+(5+sqrt(21))^n*(9+2*sqrt(21))))/21. - Colin Barker, Nov 02 2016

A340561 Square array T(n,k), n >= 1, k >= 1, read by antidiagonals, where T(n,k) = sqrt( Product_{a=1..n-1} Product_{b=1..k-1} (4sin(aPi/n)^2 + 4cos(bPi/k)^2) ).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 3, 1, 1, 12, 16, 4, 1, 1, 29, 75, 45, 5, 1, 1, 70, 361, 384, 121, 6, 1, 1, 169, 1728, 3509, 1805, 320, 7, 1, 1, 408, 8281, 31500, 30976, 8100, 841, 8, 1, 1, 985, 39675, 284089, 508805, 261725, 35287, 2205, 9, 1

Offset: 1

Seiichi Manyama, Jan 11 2021

			Square array begins:
  1, 1,   1,    1,      1,       1, ...
  1, 2,   5,   12,     29,      70, ...
  1, 3,  16,   75,    361,    1728, ...
  1, 4,  45,  384,   3509,   31500, ...
  1, 5, 121, 1805,  30976,  508805, ...
  1, 6, 320, 8100, 261725, 7741440, ...

Columns 1..4 give A000012, A000027, A004146, A006235.
Rows 1..3 give A000012, A000129, A005386.
Main diagonal gives A340563.
T(n, 2*n) gives A252767.
Cf. A173958, A340476, A340560.

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

A094083 Numerators of ratio of sides of n-th triple of rectangles of unit area sum around a triangle.

Original entry on oeis.org

1, 1, 1, 4, 9, 64, 25, 256, 1225, 16384, 3969, 65536, 53361, 1048576, 184041, 4194304, 41409225, 1073741824, 147744025, 4294967296, 2133423721, 68719476736, 7775536041, 274877906944, 457028729521, 17592186044416, 1690195005625

Offset: 1

Peter J. C. Moses, Apr 30 2004

Page 13 of the link shows the type of configuration. When n is odd, the numerators 1,1,9,25,1225,3969,.. are A038534 and (A001790)^2, and the denominators 1,4,64,256,16384,65536,.. are A056982, A038533/2, and (A046161)^2. When n is even, the numerators 1,4,64,256,16384,65536,.. are A056982, A038533/2, and (A046161)^2, and the denominators 3,27,675,3675,297675,1440747,.. are 3*(A001803)^2. The limit of a(n+1)/a(n) as n(odd) tends to infinity = Pi^2/12, A072691. The limit of a(n+2)/a(n) as n tends to infinity = 1. a(n), for large odd n, tends to 2/(Pi*n). a(n), for large even n, tends to Pi/(6*n). The expansion of 2*x*EllipticK(x)/Pi gives the odd fractions. The expansion of 1/3*x*HypergeometricPFQ({1,1,1},{3/2,3/2},x) gives the even fractions.

			a(5) = a(5-2)*((5-2)/(5-1))^2 = 1/4*(3/4)^2 = 9/64

Paul Yiu, EuclideanGeometry Notes

Cf. A005386, A092936, A094084.

a[n_]:=If[OddQ[n], ((n/2-1)!)^2/(Pi*((n/2-1/2)!)^2), Pi*((n/2-1)!)^2/(12*((n/2-1/2)!)^2)] a[n_]:=If[OddQ[n], (2^(1-n)*(n-2)!!^2)/((n-1)/2)!^2, (2^(n-2)*((n-2)/2)!^2)/(3*(n-1)!!^2)] a[n_]:=((12+Pi^2+E^(I*n*Pi)*(Pi^2-12))*((n/2-1)!)^2)/(24*Pi*((n/2-1/2)!)^2) (CoefficientList[Series[(I*x*(6+Sqrt[3]*Pi)-2*x*Sqrt[3]*Log[x+Sqrt[x^2-1]])/(6*Sqrt[x^2-1]), {x, 0, 20}], x])^2

a(n)=a(n-2)*((n-2)/(n-1))^2, a(1)=1, a(2)=1/3. a(n)=((n/2-1)!)^2/(Pi*((n/2-1/2)!)^2) for n odd. a(n)=(2^(1-n)*(n-2)!!^2)/((n-1)/2)!^2 for n odd. a(n)=Pi*((n/2-1)!)^2/(12*((n/2-1/2)!)^2) for n even. a(n)=(2^(n-2)*((n-2)/2)!^2)/(3*(n-1)!!^2) for n even.

cp's OEIS Frontend

A004253 a(n) = 5*a(n-1) - a(n-2), with a(1)=1, a(2)=4.

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A092936 Area of n-th triple of hexagons around a triangle.

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A003769 Number of perfect matchings (or domino tilings) in K_4 X P_n.

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A099025 Expansion of 1 / ((1+x) * (1-5*x+x^2)).

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A340561 Square array T(n,k), n >= 1, k >= 1, read by antidiagonals, where T(n,k) = sqrt( Product_{a=1..n-1} Product_{b=1..k-1} (4*sin(a*Pi/n)^2 + 4*cos(b*Pi/k)^2) ).

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A094083 Numerators of ratio of sides of n-th triple of rectangles of unit area sum around a triangle.

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A094084 Denominators associated with A094083.

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A340561 Square array T(n,k), n >= 1, k >= 1, read by antidiagonals, where T(n,k) = sqrt( Product_{a=1..n-1} Product_{b=1..k-1} (4sin(aPi/n)^2 + 4cos(bPi/k)^2) ).