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.

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A005248 Bisection of Lucas numbers: a(n) = L(2*n) = A000032(2*n).

Original entry on oeis.org

2, 3, 7, 18, 47, 123, 322, 843, 2207, 5778, 15127, 39603, 103682, 271443, 710647, 1860498, 4870847, 12752043, 33385282, 87403803, 228826127, 599074578, 1568397607, 4106118243, 10749957122, 28143753123, 73681302247, 192900153618, 505019158607, 1322157322203
Offset: 0

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Author

Keywords

Comments

Drop initial 2; then iterates of A050411 do not diverge for these starting values. - David W. Wilson
All nonnegative integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = +4 together with b(n)=A001906(n), n>=0. - Wolfdieter Lang, Aug 31 2004
a(n+1) = B^(n)AB(1), n>=0, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 3=`10`, 7=`010`, 18=`0010`, 47=`00010`, ..., in Wythoff code. a(0) = 2 = B(1) in Wythoff code.
Output of Tesler's formula (as well as that of Lu and Wu) for the number of perfect matchings of an m X n Möbius band where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009
Numbers having two 1's in their base-phi representation. - Robert G. Wilson v, Sep 13 2010
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Feb 18 2013: (Start)
a(n) is also one half of the total number of round trips, each of length 2*n, on the graph P_4 (o-o-o-o) (the simple path with 4 points (vertices) and 3 lines (or edges)). See the array and triangle A198632 for the general case of the graph P_N (there N is n and the length is l=2*k).
O.g.f. for w(4,l) (with zeros for odd l): y*(d/dy)S(4,y)/S(4,y) with y=1/x and Chebyshev S-polynomials (coefficients A049310). See also A198632 for a rewritten form. One half of this o.g.f. for x -> sqrt(x) produces the g.f. (2-3x)/(1-3x+x^2) given below. (End)
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy - 5. - Michel Lagneau, Feb 01 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 7xy + y^2 + 45 = 0. - Colin Barker, Feb 16 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 18xy + y^2 + 320 = 0. - Colin Barker, Feb 16 2014
a(n) are the numbers such that a(n)^2-2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
All sequences of this form, b(n+1) = 3*b(n) - b(n-1), regardless of initial values, which includes this sequence, yield this sequence as follows: a(n) = (b(j+n) + b(j-n))/b(j), for any j, except where b(j) = 0. Also note formula below relating this a(n) to all sequences of the form G(n+1) = G(n) + G(n-1). - Richard R. Forberg, Nov 18 2014
A non-simple continued fraction expansion for F(2n*(k+1))/F(2nk) k>=1 is a(n) + (-1)/(a(n) + (-1)/(a(n) + ... + (-1)/a(n))) where a(n) appears exactly k times (F(n) denotes the n-th Fibonacci number). E.g., F(16)/F(12) equals 7 + (-1)/(7 + (-1)/7). Furthermore, these a(n) are exactly the positive integers k such that the non-simple infinite continued fraction k + (-1)/(k + (-1)/(k + (-1)/(k + ...))) belongs to Q(sqrt(5)). Compare to Benoit Cloitre and Thomas Baruchel's comments at A002878. - Greg Dresden, Aug 13 2019
For n >= 1, a(n) is the number of cyclic up-down words of length 2*n over an alphabet of size 3. - Sela Fried, Apr 08 2025

Examples

			G.f. = 2 + 3*x + 7*x^2 + 18*x^3 + 47*x^4 + 123*x^5 + 322*x^6 + 843*x^7 + ... - _Michael Somos_, Aug 11 2009
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Richard P. Stanley, Enumerative combinatorics, Vol. 2. Volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Crossrefs

Cf. A000032, A002878 (odd-indexed Lucas numbers), A001906 (Chebyshev S(n-1, 3)), a(n) = sqrt(4+5*A001906(n)^2), A228842.
a(n) = A005592(n)+1 = A004146(n)+2 = A065034(n)-1.
First differences of A002878. Pairwise sums of A001519. First row of array A103997.
Cf. A153415, A201157. Also Lucas(k*n): A000032 (k = 1), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

Programs

  • Haskell
    a005248 n = a005248_list !! n
    a005248_list = zipWith (+) (tail a001519_list) a001519_list
    -- Reinhard Zumkeller, Jan 11 2012
  • Magma
    [Lucas(2*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011
    
  • Maple
    a:= n-> (<<2|3>>. <<3|1>, <-1|0>>^n)[1$2]: seq(a(n), n=0..30); # Alois P. Heinz, Jul 31 2008
    with(combinat): seq(5*fibonacci(n)^2+2*(-1)^n, n= 0..26);
  • Mathematica
    a[0] = 2; a[1] = 3; a[n_] := 3a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 27}] (* Robert G. Wilson v, Jan 30 2004 *)
    Fibonacci[1 + 2n] + 1/2 (-Fibonacci[2n] + LucasL[2n]) (* Tesler. Sarah-Marie Belcastro, Jul 04 2009 *)
    LinearRecurrence[{3, -1}, {2, 3}, 50] (* Sture Sjöstedt, Nov 27 2011 *)
    LucasL[Range[0,60,2]] (* Harvey P. Dale, Sep 30 2014 *)
  • PARI
    {a(n) = fibonacci(2*n + 1) + fibonacci(2*n - 1)}; /* Michael Somos, Jun 23 2002 */
    
  • PARI
    {a(n) = 2 * subst( poltchebi(n), x, 3/2)}; /* Michael Somos, Jun 28 2003 */
    
  • Sage
    [lucas_number2(n,3,1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008
    

Formula

a(n) = Fibonacci(2*n-1) + Fibonacci(2*n+1).
G.f.: (2-3*x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n, 3) - S(n-2, 3) = 2*T(n, 3/2) with S(n-1, 3) = A001906(n) and S(-2, x) = -1. U(n, x)=S(n, 2*x) and T(n, x) are Chebyshev's U- and T-polynomials.
a(n) = a(k)*a(n - k) - a(n - 2k) for all k, i.e., a(n) = 2*a(n) - a(n) = 3*a(n - 1) - a(n - 2) = 7*a(n - 2) - a(n - 4) = 18*a(n - 3) - a(n - 6) = 47*a(n - 4) - a(n - 8) etc., a(2n) = a(n)^2 - 2. - Henry Bottomley, May 08 2001
a(n) = A060924(n-1, 0) = 3*A001906(n) - 2*A001906(n-1), n >= 1. - Wolfdieter Lang, Apr 26 2001
a(n) ~ phi^(2*n) where phi=(1+sqrt(5))/2. - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(0)=2, a(1)=3, a(n) = 3*a(n-1) - a(n-2) = a(-n). - Michael Somos, Jun 28 2003
a(n) = phi^(2*n) + phi^(-2*n) where phi=(sqrt(5)+1)/2, the golden ratio. E.g., a(4)=47 because phi^(8) + phi^(-8) = 47. - Dennis P. Walsh, Jul 24 2003
With interpolated zeros, trace(A^n)/4, where A is the adjacency matrix of path graph P_4. Binomial transform is then A049680. - Paul Barry, Apr 24 2004
a(n) = (floor((3+sqrt(5))^n) + 1)/2^n. - Lekraj Beedassy, Oct 22 2004
a(n) = ((3-sqrt(5))^n + (3+sqrt(5))^n)/2^n (Note: substituting the number 1 for 3 in the last equation gives A000204, substituting 5 for 3 gives A020876). - Creighton Dement, Apr 19 2005
a(n) = (1/(n+1/2))*Sum_{k=0..n} B(2k)*L(2n+1-2k)*binomial(2n+1, 2k) where B(2k) is the (2k)-th Bernoulli number. - Benoit Cloitre, Nov 02 2005
a(n) = term (1,1) in the 1 X 2 matrix [2,3] . [3,1; -1,0]^n. - Alois P. Heinz, Jul 31 2008
a(n) = 2*cosh(2*n*psi), where psi=log((1+sqrt(5))/2). - Al Hakanson, Mar 21 2009
From Sarah-Marie Belcastro, Jul 04 2009: (Start)
a(n) - (a(n) - F(2n))/2 - F(2n+1) = 0. (Tesler)
Product_{r=1..n} (1 + 4*(sin((4r-1)*Pi/(4n)))^2). (Lu/Wu) (End)
a(n) = Fibonacci(2n+6) mod Fibonacci(2n+2), n > 1. - Gary Detlefs, Nov 22 2010
a(n) = 5*Fibonacci(n)^2 + 2*(-1)^n. - Gary Detlefs, Nov 22 2010
a(n) = A033888(n)/A001906(n), n > 0. - Gary Detlefs, Dec 26 2010
a(n) = 2^(2*n) * Sum_{k=1..2} (cos(k*Pi/5))^(2*n). - L. Edson Jeffery, Jan 21 2012
From Peter Bala, Jan 04 2013: (Start)
Let F(x) = Product_{n>=0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(3 - sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.31829 56058 81914 31334 ... = 2 + 1/(3 + 1/(7 + 1/(18 + ...))).
Also F(-alpha) = 0.64985 97768 07374 32950 has the continued fraction representation 1 - 1/(3 - 1/(7 - 1/(18 - ...))) and the simple continued fraction expansion 1/(1 + 1/((3-2) + 1/(1 + 1/((7-2) + 1/(1 + 1/((18-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((3^2-4) + 1/(1 + 1/((7^2-4) + 1/(1 + 1/((18^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + 1/2*F(alpha)/F(-alpha) = 1.5142923542... has the simple continued fraction expansion 1 + 1/((3 - 2) + 1/(1 + 1/((18 - 2) + 1/(1 + 1/(123 - 2) + 1/(1 + ...))))). (End)
G.f.: (W(0)+6)/(5*x), where W(k) = 5*x*k + x - 6 + 6*x*(5*k-9)/W(k+1) (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1. Compare with A001906, A002878 and A023039. - Peter Bala, Nov 29 2013
0 = a(n) * a(n+2) - a(n+1)^2 - 5 for all n in Z. - Michael Somos, Aug 24 2014
a(n) = (G(j+2n) + G(j-2n))/G(j), for n >= 0 and any j, positive or negative, except where G(j) = 0, and for any sequence of the form G(n+1) = G(n) + G(n-1) with any initial values for G(0), G(1), including non-integer values. G(n) includes Lucas, Fibonacci. Compare with A081067 for odd number offsets from j. - Richard R. Forberg, Nov 16 2014
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 5*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
From J. M. Bergot, Oct 28 2015: (Start)
For n>0, a(n) = F(n-1) * L(n) + F(2*n+1) - (-1)^n with F(k) = A000045(k).
For n>1, a(n) = F(n+1) * L(n) + F(2*n-1) - (-1)^n.
For n>2, a(n) = 5*F(2*n-3) + 2*L(n-3) * L(n) + 8*(-1)^n. (End)
For n>1, a(n) = L(n-2)*L(n+2) -7*(-1)^n. - J. M. Bergot, Feb 10 2016
a(n) = 6*F(n-1)*L(n-1) - F(2*n-6) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, Apr 21 2017
a(n) = F(2*n) + 2*F(n-1)*L(n) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, May 01 2017
E.g.f.: exp(4*x/(1+sqrt(5))^2) + exp((1/4)*(1+sqrt(5))^2*x). - Stefano Spezia, Aug 13 2019
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(2*n+2) - F(2*n-2) = A001906(n+1) - A001906(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^2 = [1, 1; 1, 2].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 1/a(n) ) = 1/5.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 3/(a(n) + 2/(a(n))) ) = 1/6.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 9/(a(n) + 4/(a(n) + 1/(a(n)))) ) = 1/9.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 3*x^2 + 8*x^3 + 21*x^4 + ... is the o.g.f. for A001906. (End)
a(n) = n + 2 + Sum_{k=1..n-1} k*a(n-k). - Yu Xiao, May 30 2020
Sum_{n>=1} 1/a(n) = A153415. - Amiram Eldar, Nov 11 2020
Sum_{n>=0} 1/(a(n) + 3) = (2*sqrt(5) + 1)/10 (André-Jeannin, 1991). - Amiram Eldar, Jan 23 2022
a(n) = 2*cosh(2*n*arccsch(2)) = 2*cosh(2*n*asinh(1/2)). - Peter Luschny, May 25 2022
a(n) = (5/2)*(Sum_{k=-n..n} binomial(2*n, n+5*k)) - (1/2)*4^n. - Greg Dresden, Jan 05 2023
a(n) = Sum_{k>=0} Lucas(2*n*k)/(Lucas(2*n)^(k+1)). - Diego Rattaggi, Jan 12 2025

Extensions

Additional comments from Michael Somos, Jun 23 2001

A001835 a(n) = 4*a(n-1) - a(n-2), with a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 3, 11, 41, 153, 571, 2131, 7953, 29681, 110771, 413403, 1542841, 5757961, 21489003, 80198051, 299303201, 1117014753, 4168755811, 15558008491, 58063278153, 216695104121, 808717138331, 3018173449203, 11263976658481, 42037733184721, 156886956080403, 585510091136891
Offset: 0

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Author

Keywords

Comments

See A079935 for another version.
Number of ways of packing a 3 X 2*(n-1) rectangle with dominoes. - David Singmaster.
Equivalently, number of perfect matchings of the P_3 X P_{2(n-1)} lattice graph. - Emeric Deutsch, Dec 28 2004
The terms of this sequence are the positive square roots of the indices of the octagonal numbers (A046184) - Nicholas S. Horne (nairon(AT)loa.com), Dec 13 1999
Terms are the solutions to: 3*x^2 - 2 is a square. - Benoit Cloitre, Apr 07 2002
Gives solutions x > 0 of the equation floor(x*r*floor(x/r)) == floor(x/r*floor(x*r)) where r = 1 + sqrt(3). - Benoit Cloitre, Feb 19 2004
a(n) = L(n-1,4), where L is defined as in A108299; see also A001834 for L(n,-4). - Reinhard Zumkeller, Jun 01 2005
Values x + y, where (x, y) solves for x^2 - 3*y^2 = 1, i.e., a(n) = A001075(n) + A001353(n). - Lekraj Beedassy, Jul 21 2006
Number of 01-avoiding words of length n on alphabet {0,1,2,3} which do not end in 0. (E.g., for n = 2 we have 02, 03, 11, 12, 13, 21, 22, 23, 31, 32, 33.) - Tanya Khovanova, Jan 10 2007
sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571) + ... - Gary W. Adamson, Dec 18 2007
The lower principal convergents to 3^(1/2), beginning with 1/1, 5/3, 19/11, 71/41, comprise a strictly increasing sequence; numerators = A001834, denominators = A001835. - Clark Kimberling, Aug 27 2008
From Gary W. Adamson, Jun 21 2009: (Start)
A001835 and A001353 = bisection of denominators of continued fraction [1, 2, 1, 2, 1, 2, ...]; i.e., bisection of A002530.
a(n) = determinant of an n*n tridiagonal matrix with 1's in the super- and subdiagonals and (3, 4, 4, 4, ...) as the main diagonal.
Also, the product of the eigenvalues of such matrices: a(n) = Product_{k=1..(n-1)/2)} (4 + 2*cos(2*k*Pi/n).
(End)
Let M = a triangle with the even-indexed Fibonacci numbers (1, 3, 8, 21, ...) in every column, and the leftmost column shifted up one row. a(n) starting (1, 3, 11, ...) = lim_{n->oo} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n+1) is the number of compositions of n when there are 3 types of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
For n >= 2, a(n) equals the permanent of the (2*n-2) X (2*n-2) tridiagonal matrix with sqrt(2)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Primes in the sequence are apparently those in A096147. - R. J. Mathar, May 09 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 4xy + y^2 + 2 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14xy + y^2 + 32 = 0. - Colin Barker, Feb 10 2014
The (1,1) element of A^n where A = (1, 1, 1; 1, 2, 1; 1, 1, 2). - David Neil McGrath, Jul 23 2014
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001075. - René Gy, Feb 25 2018
a(n+1) is the number of spanning trees of the graph T_n, where T_n is a 2 X n grid with an additional vertex v adjacent to (1,1) and (2,1). - Kevin Long, May 04 2018
a(n)/A001353(n) is the resistance of an n-ladder graph whose edges are replaced by one-ohm resistors. The resistance in ohms is measured at two nodes at one end of the ladder. It approaches sqrt(3) - 1 for n -> oo. See A342568, A357113, and A357115 for related information. - Hugo Pfoertner, Sep 17 2022
a(n) is the number of ways to tile a 1 X (n-1) strip with three types of tiles: small isosceles right triangles (with small side length 1), 1 X 1 squares formed by joining two of those right triangles along the hypotenuse, and large isosceles right triangles (with large side length 2) formed by joining two of those right triangles along a short leg. As an example, here is one of the a(6)=571 ways to tile a 1 X 5 strip with these kinds of tiles:
| / \ |\ /| |
|/_\|\/_||. - Greg Dresden and Arjun Datta, Jun 30 2023
From Klaus Purath, May 11 2024: (Start)
For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 4xy + y^2 = -2 = A028872(-1). In general, the following applies to all sequences (t) satisfying t(i) = 4t(i-1) - t(i-2) with t(0) = 1 and two consecutive terms (x,y): x^2 - 4xy + y^2 = A028872(t(1)-2). This includes and interprets the Feb 04 2014 comments here and on A001075 by Colin Barker and the Dec 12 2012 comment on A001353 by Max Alekseyev. By analogy to this, for three consecutive terms (x,y,z) y^2 - xz = A028872(t(1)-2). This includes and interprets the Jul 10 2021 comment on A001353 by Bernd Mulansky.
If (t) is a sequence satisfying t(k) = 3t(k-1) + 3t(k-2) - t(k-3) or t(k) = 4t(k-1) - t(k-2) without regard to initial values and including this sequence itself, then a(n) = (t(k+2n+1) + t(k))/(t(k+n+1) + t(k+n)) always applies, as long as t(k+n+1) + t(k+n) != 0 for integer k and n >= 1. (End)
Binomial transform of 1, 0, 2, 4, 12, ... (A028860 without the initial -1) and reverse binomial transform of 1, 2, 6, 24, 108, ... (A094433 without the initial 1). - Klaus Purath, Sep 09 2024

References

  • Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163-166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009).
  • Leonhard Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 375.
  • F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • 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 3 of array A099390.
Essentially the same as A079935.
First differences of A001353.
Partial sums of A052530.
Pairwise sums of A006253.
Bisection of A002530, A005246 and A048788.
First column of array A103997.
Cf. A001519, A003699, A082841, A101265, A125077, A001353, A001542, A096147 (subsequence of primes).

Programs

  • GAP
    a:=[1,1];; for n in [3..20] do a[n]:=4*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Haskell
    a001835 n = a001835_list !! n
    a001835_list =
       1 : 1 : zipWith (-) (map (4 *) $ tail a001835_list) a001835_list
    -- Reinhard Zumkeller, Aug 14 2011
    
  • Magma
    [n le 2 select 1 else 4*Self(n-1)-Self(n-2): n in [1..25]]; // Vincenzo Librandi, Sep 16 2016
    
  • Maple
    f:=n->((3+sqrt(3))^(2*n-1)+(3-sqrt(3))^(2*n-1))/6^n; [seq(simplify(expand(f(n))),n=0..20)]; # N. J. A. Sloane, Nov 10 2009
  • Mathematica
    CoefficientList[Series[(1-3x)/(1-4x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Jul 25 2011, after g.f. *)
    LinearRecurrence[{4,-1},{1,1},30] (* Harvey P. Dale, Jun 08 2013 *)
    Table[Round@Fibonacci[2n-1, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
    Table[(3*ChebyshevT[n, 2] - ChebyshevU[n, 2])/2, {n, 0, 20}] (* G. C. Greubel, Dec 23 2019 *)
  • PARI
    {a(n) = real( (2 + quadgen(12))^n * (1 - 1 / quadgen(12)) )} /* Michael Somos, Sep 19 2008 */
    
  • PARI
    {a(n) = subst( (polchebyshev(n) + polchebyshev(n-1)) / 3, x, 2)} /* Michael Somos, Sep 19 2008 */
    
  • Sage
    [lucas_number1(n,4,1)-lucas_number1(n-1,4,1) for n in range(25)] # Zerinvary Lajos, Apr 29 2009
    
  • Sage
    [(3*chebyshev_T(n,2) - chebyshev_U(n,2))/2 for n in (0..20)] # G. C. Greubel, Dec 23 2019
    

Formula

G.f.: (1 - 3*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
a(1-n) = a(n).
a(n) = ((3 + sqrt(3))^(2*n - 1) + (3 - sqrt(3))^(2*n - 1))/6^n. - Dean Hickerson, Dec 01 2002
a(n) = (8 + a(n-1)*a(n-2))/a(n-3). - Michael Somos, Aug 01 2001
a(n+1) = Sum_{k=0..n} 2^k * binomial(n + k, n - k), n >= 0. - Len Smiley, Dec 09 2001
Limit_{n->oo} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 10 2002
a(n) = 2*A061278(n-1) + 1 for n > 0. - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n - i, i); then q(n, 2) = a(n+1). - Benoit Cloitre, Nov 10 2002
a(n+1) = Sum_{k=0..n} ((-1)^k)*((2*n+1)/(2*n + 1 - k))*binomial(2*n + 1 - k, k)*6^(n - k) (from standard T(n,x)/x, n >= 1, Chebyshev sum formula). The Smiley and Cloitre sum representation is that of the S(2*n, i*sqrt(2))*(-1)^n Chebyshev polynomial. - Wolfdieter Lang, Nov 29 2002
a(n) = S(n-1, 4) - S(n-2, 4) = T(2*n-1, sqrt(3/2))/sqrt(3/2) = S(2*(n-1), i*sqrt(2))*(-1)^(n - 1), with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x) = 0, S(-2, x) = -1, S(n, 4) = A001353(n+1), T(-1, x) = x.
a(n+1) = sqrt((A001834(n)^2 + 2)/3), n >= 0 (see Cloitre comment).
Sequence satisfies -2 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = (1/6)*(3*(2 - sqrt(3))^n + sqrt(3)*(2 - sqrt(3))^n + 3*(2 + sqrt(3))^n - sqrt(3)*(2 + sqrt(3))^n) (Mathematica's solution to the recurrence relation). - Sarah-Marie Belcastro, Jul 04 2009
If p[1] = 3, p[i] = 2, (i > 1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n+1) = det A. - Milan Janjic, Apr 29 2010
a(n) = (a(n-1)^2 + 2)/a(n-2). - Irene Sermon, Oct 28 2013
a(n) = A001353(n+1) - 3*A001353(n). - R. J. Mathar, Oct 30 2015
a(n) = a(n-1) + 2*A001353(n-1). - Kevin Long, May 04 2018
From Franck Maminirina Ramaharo, Nov 11 2018: (Start)
a(n) = (-1)^n*(A125905(n) + 3*A125905(n-1)), n > 0.
E.g.f.: exp^(2*x)*(3*cosh(sqrt(3)*x) - sqrt(3)*sinh(sqrt(3)*x))/3. (End)
From Peter Bala, Feb 12 2024: (Start)
For n in Z, a(n) = A001353(n) + A001353(1-n).
For n, j, k in Z, a(n)*a(n+j+k) - a(n+j)*a(n+k) = 2*A001353(j)*A001353(k). The case j = 1, k = 2 is given above. (End)

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

A103999 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2M x 2N Klein bottle.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 16, 34, 1, 1, 54, 196, 198, 1, 1, 196, 1666, 2704, 1154, 1, 1, 726, 16384, 64152, 37636, 6726, 1, 1, 2704, 171394, 1844164, 2549186, 524176, 39202, 1, 1, 10086, 1844164, 57523158, 220581904, 101757654, 7300804, 228486, 1
Offset: 0

Views

Author

Ralf Stephan, Feb 26 2005

Keywords

Examples

			Array begins:
  1,   1,     1,        1,           1,             1,                1, ...
  1,   6,    34,      198,        1154,          6726,            39202, ...
  1,  16,   196,     2704,       37636,        524176,          7300804, ...
  1,  54,  1666,    64152,     2549186,     101757654,       4064620168, ...
  1, 196, 16384,  1844164,   220581904,   26743369156,    3252222705664, ...
  1, 726,171394, 57523158, 21050622914, 7902001927776, 2988827208115522, ...
		

Crossrefs

Rows include A003499, A067902+2. Columns include A003500+2.
Main diagonal gives A340557.

Programs

  • Mathematica
    T[m_, n_] := Product[4 Sin[(4k-1) Pi/(4n)]^2 + 4 Cos[j Pi/(2m+1)]^2, {j, 1, m}, {k, 1, n}] // Round;
    Table[T[m-n, n], {m, 0, 9}, {n, 0, m}] // Flatten (* Jean-François Alcover, Aug 20 2018 *)
  • PARI
    default(realprecision, 120);
    {T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*sin((4*a-1)*Pi/(4*n))^2+4*sin((2*b-1)*Pi/(2*k))^2)))} \\ Seiichi Manyama, Jan 11 2021

Formula

T(M, N) = Product_{m=1..M} Product_{n=1..N} ( 4sin(Pi*(4n-1)/(4N))^2 + 4sin(Pi*(2m-1)/(2M))^2 ).

A341741 Square array T(n,k), n >= 1, k >= 1, read by antidiagonals downwards: number of perfect matchings in the graph C_{2n} x C_k.

Original entry on oeis.org

2, 8, 2, 14, 36, 2, 36, 50, 200, 2, 82, 272, 224, 1156, 2, 200, 722, 3108, 1058, 6728, 2, 478, 3108, 9922, 39952, 5054, 39204, 2, 1156, 10082, 90176, 155682, 537636, 24200, 228488, 2, 2786, 39952, 401998, 3113860, 2540032, 7379216, 115934, 1331716, 2
Offset: 1

Views

Author

Seiichi Manyama, Feb 18 2021

Keywords

Comments

Dimer tilings of 2n x k toroidal grid.

Examples

			Square array begins:
  2,     8,    14,      36,       82,        200, ...
  2,    36,    50,     272,      722,       3108, ...
  2,   200,   224,    3108,     9922,      90176, ...
  2,  1156,  1058,   39952,   155682,    3113860, ...
  2,  6728,  5054,  537636,  2540032,  114557000, ...
  2, 39204, 24200, 7379216, 41934482, 4357599552, ...
		

Crossrefs

Columns 1..12 give A007395, A162484(2*n), A231087, A220864(2*n), A231485, A232804(2*n), A230033, A253678(2*n), A281583, A281679(2*n), A308761, A309018(2*n).
T(n,2*n) gives A335586.

Formula

T(n,k) = A341533(n,k)/2 + A341738(n,k) + 2 * ((k+1) mod 2) * A341739(n,ceiling(k/2)).
T(n, 2k) = T(k, 2n).
If k is odd, T(n,k) = A341533(n,k) = 2*A341738(n,k).

Extensions

New name from Andrey Zabolotskiy, Dec 26 2021

A334178 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) = 2^n * sqrt(Resultant(U_{2*n}(x/2), T_{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).

Original entry on oeis.org

1, 1, 2, 1, 1, 4, 1, 3, 1, 8, 1, 4, 11, 1, 16, 1, 7, 19, 41, 1, 32, 1, 11, 71, 91, 153, 1, 64, 1, 18, 176, 769, 436, 571, 1, 128, 1, 29, 539, 2911, 8449, 2089, 2131, 1, 256, 1, 47, 1471, 17753, 48301, 93127, 10009, 7953, 1, 512, 1, 76, 4271, 79808, 603126, 801701, 1027207, 47956, 29681, 1, 1024
Offset: 0

Views

Author

Seiichi Manyama, Apr 17 2020

Keywords

Examples

			Square array begins:
   1, 1,    1,     1,       1,        1,         1, ...
   2, 1,    3,     4,       7,       11,        18, ...
   4, 1,   11,    19,      71,      176,       539, ...
   8, 1,   41,    91,     769,     2911,     17753, ...
  16, 1,  153,   436,    8449,    48301,    603126, ...
  32, 1,  571,  2089,   93127,   801701,  20721019, ...
  64, 1, 2131, 10009, 1027207, 13307111, 714790675, ...
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := 2^n * Sqrt[Resultant[ChebyshevU[2*n, x/2], ChebyshevT[k, I*x/2], x]]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 04 2021 *)
  • PARI
    {T(n, k) = sqrtint(4^n*polresultant(polchebyshev(2*n, 2, x/2), polchebyshev(k, 1, I*x/2)))}

Formula

T(n,2*k) = A103997(n,k) for k > 0.

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).

A103998 Number of dimer tilings of a 4 x 2n Moebius strip.

Original entry on oeis.org

1, 11, 71, 539, 4271, 34276, 276119, 2226851, 17965151, 144948419, 1169523076, 9436433171, 76139155439, 614339685971, 4956888901511, 39995380044004, 322708555336511, 2603821045832171, 21009309912323639
Offset: 0

Views

Author

Ralf Stephan, Feb 26 2005

Keywords

Crossrefs

Cf. Second row of array A103997.

Programs

  • Mathematica
    Join[{1},LinearRecurrence[{11,-25,11,-1},{11,71,539,4271},40]] (* or *) CoefficientList[ Series[ (1-25x^2+22x^3-3x^4)/ (1-11x+ 25x^2- 11x^3+x^4),{x,0,40}],x] (* Harvey P. Dale, Jun 15 2011 *)

Formula

G.f.: (1-25x^2+22x^3-3x^4)/(1-11x+25x^2-11x^3+x^4).
a(0)=1, a(1)=11, a(2)=71, a(3)=539, a(4)=4271, a(n)=11*a(n-1)-25*a(n-2)+ 11*a(n-3)-a(n-4). - Harvey P. Dale, Jun 15 2011

A334124 a(n) = 2^n * sqrt(Resultant(U_{2*n}(x/2), T_{2*n}(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).

Original entry on oeis.org

1, 3, 71, 17753, 46069729, 1234496016491, 341133743251787719, 971684488369988888850993, 28523907708086181923163934073729, 8628515016553040037389969912341438652243, 26895841132028233579514694272575933932911355677831
Offset: 0

Views

Author

Seiichi Manyama, Apr 15 2020

Keywords

Crossrefs

Main diagonal of A103997.

Programs

  • Mathematica
    Table[2^n * Sqrt[Resultant[ChebyshevU[2*n, x/2], ChebyshevT[2*n, I*x/2], x]], {n, 0, 12}] (* Vaclav Kotesovec, Apr 16 2020 *)
  • PARI
    {a(n) = sqrtint(4^n*polresultant(polchebyshev(2*n, 2, x/2), polchebyshev(2*n, 1, I*x/2)))}
    
  • Python
    from math import isqrt
    from sympy import resultant, chebyshevt, chebyshevu, I
    from sympy.abc import x
    def A334124(n): return isqrt(resultant(chebyshevu(n<<1,x/2),chebyshevt(n<<1,I*x/2))*(1<<(n<<1))) if n else 1 # Chai Wah Wu, Nov 07 2023

Formula

a(n) = A103997(n,n).
a(n) ~ 2^(1/4) * exp(2*G*n*(2*n+1)/Pi) / (1 + sqrt(2))^n, where G is Catalan's constant A006752. - Vaclav Kotesovec, Apr 16 2020, updated Jan 03 2021

A334135 Number of dimer tilings of a 2*n x 4 Moebius strip.

Original entry on oeis.org

1, 7, 71, 769, 8449, 93127, 1027207, 11332097, 125019649, 1379271559, 15216755911, 167878292737, 1852111179521, 20433349871431, 225430197438727, 2487050546734081, 27438295728380929, 302712011022644999, 3339659377008916807, 36844671993005504257, 406487518942362537217
Offset: 0

Views

Author

Seiichi Manyama, Apr 15 2020

Keywords

Crossrefs

Column 2 of A103997.

Programs

  • Mathematica
    a[n_] := 2^n * Sqrt[Resultant[ChebyshevU[2*n, x/2], ChebyshevT[4, I*x/2], x]]; Array[a, 21, 0] (* Amiram Eldar, May 04 2021 *)
  • PARI
    a(n) = sqrtint(4^n*polresultant(polchebyshev(2*n, 2, x/2), 1+2*x^2+1/2*x^4))
    
  • PARI
    N=20; x='x+O('x^N); Vec(((1-x)*(1-6*x+x^2))/(1-14*x+34*x^2-14*x^3+x^4))

Formula

a(n)^2 = 4^n * Resultant(U(2*n,x), 1+2*x^2+1/2*x^4), where U(n,x) is a Chebyshev polynomial of the second kind and i = sqrt(-1).
G.f.: ((1-x)*(1-6*x+x^2))/(1-14*x+34*x^2-14*x^3+x^4).
a(n) = 14*a(n-1) - 34*a(n-2) + 14*a(n-3) - a(n-4) for n > 3.
Showing 1-10 of 15 results. Next