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A001353
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a(n) = 4*a(n-1)-a(n-2) with a(0) = 0, a(1) = 1.
(Formerly M3499 N1420)
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113
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0, 1, 4, 15, 56, 209, 780, 2911, 10864, 40545, 151316, 564719, 2107560, 7865521, 29354524, 109552575, 408855776, 1525870529, 5694626340, 21252634831, 79315912984, 296011017105, 1104728155436, 4122901604639, 15386878263120
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OFFSET
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0,3
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COMMENTS
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3*a(n)^2 + 1 is a perfect square. Moreover, 3*a(n)^2 + 1 = (2*a(n) - a(n-1))^2.
Consecutive terms give nonnegative solutions to x^2 - 4*x*y + y^2 = 1. - Max Alekseyev, Dec 12 2012
Values y solving the Pellian x^2 - 3*y^2 = 1; Corresponding x given by A001075(n). Moreover, we have a(n) = 2*a(n-1) + A001075(n-1). - Lekraj Beedassy, Jul 13 2006
Number of spanning trees in 2 X n grid: by examining what happens at the right-hand end we see that a(n) = 3*a(n-1) + 2*a(n-2) + 2*a(n-3) + ... + 2*a(1) + 1, where the final 1 corresponds to the tree ==...=| !. Solving this we get a(n) = 4a(n-1) - a(n-2).
Complexity of 2 X n grid.
A016064 also describes triangles whose sides are consecutive integers and in which an inscribed circle has an integer radius. A001353 is exactly and precisely mapped to the integer radii of such inscribed circles, i.e. for each term of A016064, the corresponding term of A001353 gives the radius of the inscribed circle. - Harvey P. Dale, Dec 28 2000
If M is any term of the sequence, the next one is 2M + sqrt(3M^2 + 1). - Lekraj Beedassy, Feb 18 2002
n such that 3*n^2=floor(sqrt(3)*n*ceil(sqrt(3)*n)). - Benoit Cloitre, May 10 2003
For n>0, ratios a(n+1)/a(n) may be obtained as convergents of the continued fraction expansion of 2+sqrt(3): either as successive convergents of [4;-4] or as odd convergents of [3;1, 2]. - Lekraj Beedassy, Sep 19 2003
Ways of packing a 3 X (2n-1) rectangle with dominoes, after attaching an extra square to the end of one of the sides of length 3. With reference to A001835, therefore: a(n) = a(n-1) + A001835(n-1) and A001835(n) = 3*A011835(n-1) + 2*a(n-1). - Joshua Zucker and the Castilleja School Math Club, Oct 28 2003
a(n+1) is a Chebyshev transform of 4^n, where the sequence with g.f. G(x) is sent to the sequence with g.f. (1/(1+x^2))G(x/(1+x^2)). - Paul Barry, Oct 25 2004
This sequence generates many brilliant (A078972) numbers for a(p) with prime p: a(2) = 4 = 2 * 2 a(3) = 15 = 3 * 5 a(5) = 209 = 11 * 19 a(7) = 2911 = 41 * 71 a(19) = 21252634831 = 110771 * 191861 a(37) = 419245718107612602961 = 15558008491 * 26947261171. Is this a prime-free sequence? If not, what is its first prime? - Jonathan Vos Post, Feb 08 2005
Numbers such that there is an m with t(n+m)=3t(m), where t(n) are the triangular numbers A000217. For instance t(35)=3t(20)=630, so 35-20=15 is in the sequence. - comment by Floor van Lamoen (fvlamoen(AT)hotmail.com), Oct 13 2005
a(n) = number of unique matrix products in (A+B+C+D)^n where commutator [A,B]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Max Alekseyev, Feb 01 2006
For n>1, middle side (or long leg) of primitive Pythagorean triangles having an angle nearing pi/3 with larger values of sides. [Complete triple (X, Y, Z), X<Y<Z, is given by X=A120892(n), Y=a(n), Z=A120893(n), with recurrence relations X(i+1)=2*{X(i) - (-1)^i} + a(i) ; Z(i+1)=2*{Z(i) + a(i)} - (-1)^i] - Lekraj Beedassy, Jul 13 2006
Number of 2 X n simple rectangular mazes. A simple rectangular m X n maze is a graph G with vertex set {0,1,...,m} X {0,1,...,n} that satisfies the following two properties: (i) G consists of two orthogonal trees; (ii) one tree has a path that sequentially connects (0,0),(0,1),...,(0,n),(1,n),...,(m-1, n) and the other tree has a path that sequentially connects (1,0),(2,0),...,(m,0),(m,1),...,(m,n). For example, a(2)=4 because there are four 2X2 simple rectangular mazes:
.__.............__ ...........__.............__
|..|..|........|__...|.......|.....|........|...__|
|...__|........|...__|.......|..|__|........|...__|. - Dennis P. Walsh, Oct 04 2006
[1,4,15,56,209,...] is the Hankel transform of [1,1,5,26,139,758,...](see A005573). - Philippe DELEHAM, Apr 14 2007
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
Contribution from Gary W. Adamson, Jun 21 2009: (Start)
A001353 and A001835 = bisection of continued fraction [1,2,1,2,1,2,...] i.e. of [1, 3, 4, 11, 15, 41,...].
For n>0, a(n) equals the determinant of an (n-1) X (n-1) tridiagonal matrix with one's in the super and subdiagonals and (4,4,4,...) as the main diagonal. [Corrected: Johannes Boot, Sep 4 2011]
A001835 and A001353 = right and next to right borders of triangle A125077 (End)
Number of units of a(n) belongs to a periodic sequence: 0, 1, 4, 5, 6, 9. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 04 2009]
a(n) is equal to the permanent of the (n-1) X (n-1) Hessenberg matrix with 4's along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. [From John M. Campbell, June 9 2011]
2a(n) is the number of n-color compositions of 2n consisting of only even parts; see Guo in references. [From Brian Hopkins, 19 July 2011]
Pisano period lengths: 1, 2, 6, 4, 3, 6, 8, 4, 18, 6, 10, 12, 12, 8, 6, 8, 18, 18, 5, 12,... - R. J. Mathar, Aug 10 2012
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REFERENCES
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Bastida, Julio R. 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) - From N. J. A. Sloane, May 30 2012
M. N. Deshpande, One Interesting Family of Diophantine Triplets, International Journal of Mathematical Education In Science and Technology, Vol. 33 (No. 2, Mar-Apr), 2002.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; p. 163.
E. I. Emerson, Recurrent sequences in the equation DQ^2 = R^2 + N, Fib. Quart., 7 (1969), 231-242.
F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.
D. Fortin, B-SPLINE TOEPLITZ INVERSE UNDER CORNER PERTURBATIONS, International Journal of Pure and Applied Mathematics, Volume 77, No. 1, 2012, 107-118; http://ijpam.eu/contents/2012-77-1/11/11.pdf. - From N. J. A. Sloane, Oct 22 2012
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
T. N. E. Greville, Table for third-degree spline interpolations with equally spaced arguments, Math. Comp., 24 (1970), 179-183.
Y.-H. Guo, n-Colour even self-inverse compositions, Proc. Indian Acad. Sci. (Math. Sci.), 120 (2010), 27-33.
B. Hopkins, Spotted tilings and n-color compositions, INTEGERS 12B (2012/2013), #A6.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=4, q=-1.
W. D. Hoskins, Table for third-degree spline interpolation using equi-spaced knots, Math. Comp., 25 (1971), 797-801.
Clark Kimberling, "Best lower and upper approximates to irrational numbers," Elemente der Mathematik, 52 (1997) 122-126.
J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.
G. Kreweras, Complexite et circuits Euleriens dans les sommes tensorielles de graphes, J. Combin. Theory, B 24 (1978), 202-212.
Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq. (44) lhs, m=6.
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).
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..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 Hamilton cycles in product graphs
F. Faase, Results from the counting program
F. Faase, Counting Hamilton cycles in product graphs
Tanya Khovanova, Recursive Sequences
Hojoo Lee, Problems in elementary number theory Problem I 18.
_Simon Plouffe_, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
_Simon Plouffe_, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
P. Raff, Analysis of the Number of Spanning Trees of K_2 x P_n. Contains sequence, recurrence, generating function, and more. [From Paul Raff, Mar 06 2009]
D. P. Walsh, Counting n x 2 Simple Rectangular Mazes
Eric Weisstein's World of Mathematics, Ladder Graph
Eric Weisstein's World of Mathematics, Spanning Tree
Index entries for sequences related to linear recurrences with constant coefficients, signature (4,-1)
Index entries for sequences related to Chebyshev polynomials.
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FORMULA
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G.f.: x/(1-4*x+x^2).
a(n)=((2+sqrt(3))^n-(2-sqrt(3))^n)/(2*sqrt(3)).
a(-n) = -a(n). - Michael Somos, Sep 19 2008
Limit as n-> infinity of a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 06 2002
Binomial transform of A002605.
E.g.f.: exp(2*x)*sinh(sqrt(3)*x)/sqrt(3).
a(n) = S(n-1, 4) = U(n-1, 2), S(-1, x) := 0, Chebyshev's polynomials of the second kind A049310.
a(n+1)=sum{k=0..floor(n/2), binomial(n-k, k)(-1)^k*4^(n-2k)} - Paul Barry, Oct 25 2004
a(n)=sum{k=0..n-1, binomial(n+k, 2k+1)2^k} - Paul Barry, Nov 30 2004
a(n)=3*a(n-1)+3*a(n-2)-a(n-3); a(0)=0, a(1)=1, a(2)=4. - Lekraj Beedassy, Jul 13 2006
a(n) = 2*a(n-1)+sqrt(3*a(n-1)^2+1). a(n) = -A106707(n). - R. J. Mathar, Jul 07 2006
a(n) = 3*(a(n-1)+a(n-2))-a(n-3), a(n) = 5*(a(n-1)-a(n-2))+a(n-3). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 20 2006
M^n * [1,0] = [A001075(n), A001353(n)], where M = the 2 X 2 matrix [2,3; 1,2]; e.g., a(4) = 56 since M^4 * [1,0] = [97, 56] = [A001075(4), A001353(4)]. - Gary W. Adamson, Dec 27 2006
Sequence satisfies 1 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
Rational recurrence: a(n)=(17*a(n-1)*a(n-2)-4*(a(n-1)^2+a(n-2)^2))/a(n-3) for n>3 [From Jaume Oliver Lafont, Dec 05 2009]
If p[i]=fibonacci(2i) and if A is the 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)= det A. [From Milan Janjic, May 08 2010]
a(n) = C_{n-1}^{(1)}(2), where C_n^{(m)}(x) is the Gegenbauer polynomial [From Eric Weisstein, Jul 16 2011]
a(n) = -i*sin(n*arccos(2))/sqrt(3) [From Eric Weisstein, Jul 16 2011]
a(n) = sinh(n arccosh(2))/sqrt(3) [From Eric Weisstein, Jul 16 2011]
a(n) = b such that Integral_{x=0..Pi/2} (sin(n*x))/(2-cos(x)) dx = c+b*ln(2). [From Francesco Daddi, Aug 02 2011]
a(n) = sqrt(A098301(n))=sqrt([A055793 / 3]), base 3 analogon of A031150. - M. F. Hasler, Jan 16 2012
a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*3^k. - Philippe Deléham, Feb 10 2012
1, 4, 15, 56, 209,... = INVERT(INVERT(1,2,3,4,5,...)). [David Callan, Oct 13 2012]
Product {n >= 1} (1 + 1/a(n)) = 1 + sqrt(3). - Peter Bala, Dec 23 2012
Product {n >= 2} (1 - 1/a(n)) = 1/4*(1 + sqrt(3)). - Peter Bala, Dec 23 2012
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EXAMPLE
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For example, when n=3:
****
.***
.***
can be packed with dominoes in 4 different ways: 3 in which the top row is tiled with two horizontal dominoes and 1 in which the top row has two vertical and one horizontal domino, as shown below, so a(2) = 4.
---- ---- ---- ||--
.||| .--| .|-- .|||
.||| .--| .|-- .|||
x + 4*x^2 + 15*x^3 + 56*x^4 + 209*x^5 + 780*x^6 + 2911*x^7 + 10864*x^8 + ...
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MAPLE
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A001353 := proc(n) option remember; if n <= 1 then 1+3*n else 4*A001353(n-1)-A001353(n-2); fi; end;
A001353:=z/(1-4*z+z**2); [Simon Plouffe in his 1992 dissertation.]
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MATHEMATICA
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a[n_] := (MatrixPower[{{1, 2}, {1, 3}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, 0, 23}] (* from Robert G. Wilson v, Jan 13 2005 *)
Table[GegenbauerC[n-1, 1, 2]], {n, 0, 23}] (* From Zerinvary Lajos, Jul 14 2009 *)
Table[-((I Sin[n ArcCos[2]])/Sqrt[3]), {n, 0, 23}] // FunctionExpand (* from Eric Weisstein, Jul 16 2011 *)
Table[Sinh[n ArcCosh[2]]/Sqrt[3], {n, 0, 23}] // FunctionExpand (* from Eric Weisstein, Jul 16 2011 *)
Table[ChebyshevU[-1 + n, 2], {n, 0, 23}] (* from Eric Weisstein, Jul 16 2011 *)
a[0] := 0; a[1] := 1; a[n_] := a[n] = 4a[n - 1] - a[n - 2]; Table[a[n], {n, 0, 23}] (* From Alonso del Arte, Jul 19 2011 *)
LinearRecurrence[{4, -1}, {0, 1}, 50] (* Sture Sjöstedt, Dec 06 2011 *)
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PROG
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(PARI) M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=0, 30, print1(([1, 0, 0]*M^i)[2], ", ")) - from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005
(PARI) {a(n) = real( (2 + quadgen(12))^n / quadgen(12) )} /* Michael Somos, Sep 19 2008 */
(PARI) {a(n) = subst( polchebyshev(n-1, 2), x, 2)} /* Michael Somos, Sep 19 2008 */
(Sage) [lucas_number1(n, 4, 1) for n in xrange(0, 25)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 22 2009]
(Haskell)
a001353 n = a001353_list !! n
a001353_list =
0 : 1 : zipWith (-) (map (4 *) $ tail a001353_list) a001353_list
-- Reinhard Zumkeller, Aug 14 2011
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CROSSREFS
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a(n) = sqrt((A001075(n)^2-1)/3).
Cf. A003500, A001835, A001075, A001571, A001834, A002531, A005246, A016064, A082840, A079935, A078972.
A bisection of A002530.
Cf. A125077 [From Gary W. Adamson, Jun 21 2009]
Cf. A001542, A001835.
A row of A116469.
Sequence in context: A191606 A221859 A010905 * A106707 A125905 A195503
Adjacent sequences: A001350 A001351 A001352 * A001354 A001355 A001356
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KEYWORD
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nonn,easy,nice,changed
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AUTHOR
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N. J. A. Sloane.
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STATUS
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approved
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