

A001835


a(n) = 4*a(n1)  a(n2), with a(0) = 1, a(1) = 1.
(Formerly M2894 N1160)


75



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
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OFFSET

0,3


COMMENTS

Number of ways of packing a 3 X 2*(n1) rectangle with dominoes.  David Singmaster.
Equivalently, number of perfect matchings of the P_3 X P_{2(n1)} 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
Number of 01avoiding 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
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..(n1)/2)} (4 + 2*cos(2*k*Pi/n).
(End)
Let M = a triangle with the evenindexed 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>infinity} M^n, the leftshifted 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*n2) X (2*n2) 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
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 nladder graph whose edges are replaced by oneohm resistors. The resistance in ohms is measured at two nodes at one end of the ladder. It approaches sqrt(3)  1 for n > infinity. 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 (n1) 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


REFERENCES

R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318321.
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. 163166, Congress. Numer., XXIIIXXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009)
L. 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), 129154.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. AddisonWesley, Reading, MA, 1990, p. 329.
Serge Lang, Introduction to Diophantine Approximations, AddisonWesley, 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.


LINKS



FORMULA

G.f.: (1  3*x)/(1  4*x + x^2).  Simon Plouffe in his 1992 dissertation
a(1n) = 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+1) = Sum_{k=0..n} (2^k * binomial(n + k, n  k), n >= 0.  Len Smiley, Dec 09 2001
a(n) = 2*A061278(n1) + 1 for n > 0.  Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
Let q(n, x) = Sum_{i=0..n} x^(ni)*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(n1, 4)  S(n2, 4) = T(2*n1, sqrt(3/2))/sqrt(3/2) = S(2*(n1), 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).  SarahMarie 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[ji+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
E.g.f.: exp^(2*x)*(3*cosh(sqrt(3)*x)  sqrt(3)*sinh(sqrt(3)*x))/3. (End)


MAPLE

f:=n>((3+sqrt(3))^(2*n1)+(3sqrt(3))^(2*n1))/6^n; [seq(simplify(expand(f(n))), n=0..20)]; # N. J. A. Sloane, Nov 10 2009


MATHEMATICA

CoefficientList[Series[(13x)/(14x+x^2), {x, 0, 24}], x] (* JeanFrançois Alcover, Jul 25 2011, after g.f. *)
LinearRecurrence[{4, 1}, {1, 1}, 30] (* Harvey P. Dale, Jun 08 2013 *)
Table[(3*ChebyshevT[n, 2]  ChebyshevU[n, 2])/2, {n, 0, 20}] (* G. C. Greubel, Dec 23 2019 *)


PROG

(PARI) {a(n) = real( (2 + quadgen(12))^n * (1  1 / quadgen(12)) )} /* Michael Somos, Sep 19 2008 */
(PARI) {a(n) = subst( (polchebyshev(n) + polchebyshev(n1)) / 3, x, 2)} /* Michael Somos, Sep 19 2008 */
(Sage) [lucas_number1(n, 4, 1)lucas_number1(n1, 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
(Haskell)
a001835 n = a001835_list !! n
a001835_list =
1 : 1 : zipWith () (map (4 *) $ tail a001835_list) a001835_list
(Magma) [n le 2 select 1 else 4*Self(n1)Self(n2): n in [1..25]]; // Vincenzo Librandi, Sep 16 2016
(GAP) a:=[1, 1];; for n in [3..20] do a[n]:=4*a[n1]a[n2]; od; a; # G. C. Greubel, Dec 23 2019


CROSSREFS



KEYWORD

nonn,easy,nice


AUTHOR



STATUS

approved



