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 32 results. Next

A207836 a(n) = n*A052530(n)/2.

Original entry on oeis.org

0, 3, 16, 75, 336, 1463, 6240, 26199, 108640, 445995, 1815792, 7341347, 29505840, 117982815, 469672384, 1862393775, 7359403968, 28991540051, 113892526800, 446305331451, 1744950085648, 6808253393415, 26513475730464, 103072540115975, 400058834841120
Offset: 2

Views

Author

N. J. A. Sloane, Feb 21 2012

Keywords

Crossrefs

Cf. A052530.

Programs

  • PARI
    concat(0, Vec(x^3*(x^2-8*x+3)/(x^2-4*x+1)^2 + O(x^100))) \\ Colin Barker, Jun 08 2015

Formula

a(n) = 8*a(n-1)-18*a(n-2)+8*a(n-3)-a(n-4) for n>5. - Colin Barker, Jun 08 2015
G.f.: x^3*(x^2-8*x+3) / (x^2-4*x+1)^2. - Colin Barker, Jun 08 2015

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

Views

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)

A290890 p-INVERT of the positive integers, where p(S) = 1 - S^2.

Original entry on oeis.org

0, 1, 4, 11, 28, 72, 188, 493, 1292, 3383, 8856, 23184, 60696, 158905, 416020, 1089155, 2851444, 7465176, 19544084, 51167077, 133957148, 350704367, 918155952, 2403763488, 6293134512, 16475640049, 43133785636, 112925716859, 295643364940, 774004377960
Offset: 0

Views

Author

Clark Kimberling, Aug 15 2017

Keywords

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
Note that in A290890, s = (1,2,3,4,...); i.e., A000027(n+1) for n>=0, whereas in A290990, s = (0,1,2,3,4,...); i.e., A000027(n) for n>=0.
Guide to p-INVERT sequences using s = (1,2,3,4,5,...) = A000027:
p(S) t(1,2,3,4,5,...)
1 - S A001906
1 - S^2 A290890; see A113067 for signed version
1 - S^3 A290891
1 - S^4 A290892
1 - S^5 A290893
1 - S^6 A290894
1 - S^7 A290895
1 - S^8 A290896
1 - S - S^2 A289780
1 - S - S^3 A290897
1 - S - S^4 A290898
1 - S^2 - S^4 A290899
1 - S^2 - S^3 A290900
1 - S^3 - S^4 A290901
1 - 2S A052530; (1/2)*A052530 = A001353
1 - 3S A290902; (1/3)*A290902 = A004254
1 - 4S A003319; (1/4)*A003319 = A001109
1 - 5S A290903; (1/5)*A290903 = A004187
1 - 2*S^2 A290904; (1/2)*A290904 = A290905
1 - 3*S^2 A290906; (1/3)*A290906 = A290907
1 - 4*S^2 A290908; (1/4)*A290908 = A099486
1 - 5*S^2 A290909; (1/5)*A290909 = A290910
1 - 6*S^2 A290911; (1/6)*A290911 = A290912
1 - 7*S^2 A290913; (1/7)*A290913 = A290914
1 - 8*S^2 A290915; (1/8)*A290915 = A290916
(1 - S)^2 A290917
(1 - S)^3 A290918
(1 - S)^4 A290919
(1 - S)^5 A290920
(1 - S)^6 A290921
1 - S - 2*S^2 A290922
1 - 2*S - 2*S^2 A290923; (1/2)*A290923 = A290924
1 - 3*S - 2*S^2 A290925
(1 - S^2)^2 A290926
(1 - S^2)^3 A290927
(1 - S^3)^2 A290928
(1 - S)(1 - S^2) A290929
(1 - S^2)(1 - S^4) A290930
1 - 3 S + S^2 A291025
1 - 4 S + S^2 A291026
1 - 5 S + S^2 A291027
1 - 6 S + S^2 A291028
1 - S - S^2 - S^3 A291029
1 - S - S^2 - S^3 - S^4 A201030
1 - 3 S + 2 S^3 A291031
1 - S - S^2 - S^3 + S^4 A291032
1 - 6 S A291033
1 - 7 S A291034
1 - 8 S A291181
1 - 3 S + 2 S^3 A291031
1 - 3 S + 2 S^2 A291182
1 - 4 S + 2 S^3 A291183
1 - 4 S + 3 S^3 A291184

Examples

			(See the examples at A289780.)
		

Crossrefs

Cf. A000027, A113067, A289780, A113067 (signed version of same sequence).

Programs

  • Mathematica
    z = 60; s = x/(1 - x)^2; p = 1 - s^2;
    Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
    Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290890 *)

Formula

G.f.: x/(1 - 4 x + 5 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-3) - a(n-4).

A291000 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3.

Original entry on oeis.org

1, 3, 9, 26, 74, 210, 596, 1692, 4804, 13640, 38728, 109960, 312208, 886448, 2516880, 7146144, 20289952, 57608992, 163568448, 464417728, 1318615104, 3743926400, 10630080640, 30181847168, 85694918912, 243312448256, 690833811712, 1961475291648, 5569190816256
Offset: 0

Views

Author

Clark Kimberling, Aug 22 2017

Keywords

Comments

Suppose s = (c(0), c(1), c(2),...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
In the following guide to p-INVERT sequences using s = (1,1,1,1,1,...) = A000012, in some cases t(1,1,1,1,1,...) is a shifted version of the cited sequence:
p(S) t(1,1,1,1,1,...)
1 - S A000079
1 - S^2 A000079
1 - S^3 A024495
1 - S^4 A000749
1 - S^5 A139761
1 - S^6 A290993
1 - S^7 A290994
1 - S^8 A290995
1 - S - S^2 A001906
1 - S - S^3 A116703
1 - S - S^4 A290996
1 - S^3 - S^6 A290997
1 - S^2 - S^3 A095263
1 - S^3 - S^4 A290998
1 - 2 S^2 A052542
1 - 3 S^2 A002605
1 - 4 S^2 A015518
1 - 5 S^2 A163305
1 - 6 S^2 A290999
1 - 7 S^2 A291008
1 - 8 S^2 A291001
(1 - S)^2 A045623
(1 - S)^3 A058396
(1 - S)^4 A062109
(1 - S)^5 A169792
(1 - S)^6 A169793
(1 - S^2)^2 A024007
1 - 2 S - 2 S^2 A052530
1 - 3 S - 2 S^2 A060801
(1 - S)(1 - 2 S) A053581
(1 - 2 S)(1 - 3 S) A291002
(1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S) A291003
(1 - 2 S)^2 A120926
(1 - 3 S)^2 A291004
1 + S - S^2 A000045 (Fibonacci numbers starting with -1)
1 - S - S^2 - S^3 A291000
1 - S - S^2 - S^3 - S^4 A291006
1 - S - S^2 - S^3 - S^4 - S^5 A291007
1 - S^2 - S^4 A290990
(1 - S)(1 - 3 S) A291009
(1 - S)(1 - 2 S)(1 - 3 S) A291010
(1 - S)^2 (1 - 2 S) A291011
(1 - S^2)(1 - 2 S) A291012
(1 - S^2)^3 A291013
(1 - S^3)^2 A291014
1 - S - S^2 + S^3 A045891
1 - 2 S - S^2 + S^3 A291015
1 - 3 S + S^2 A136775
1 - 4 S + S^2 A291016
1 - 5 S + S^2 A291017
1 - 6 S + S^2 A291018
1 - S - S^2 - S^3 + S^4 A291019
1 - S - S^2 - S^3 - S^4 + S^5 A291020
1 - S - S^2 - S^3 + S^4 + S^5 A291021
1 - S - 2 S^2 + 2 S^3 A175658
1 - 3 S^2 + 2 S^3 A291023
(1 - 2 S^2)^2 A291024
(1 - S^3)^3 A291143
(1 - S - S^2)^2 A209917

Crossrefs

Programs

  • Mathematica
    z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3;
    Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
    Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291000 *)

Formula

G.f.: (-1 + x - x^2)/(-1 + 4 x - 4 x^2 + 2 x^3).
a(n) = 4*a(n-1) - 4*a(n-2) + 2*a(n-3) for n >= 4.

A092184 Sequence S_6 of the S_r family.

Original entry on oeis.org

0, 1, 6, 25, 96, 361, 1350, 5041, 18816, 70225, 262086, 978121, 3650400, 13623481, 50843526, 189750625, 708158976, 2642885281, 9863382150, 36810643321, 137379191136, 512706121225, 1913445293766, 7141075053841, 26650854921600, 99462344632561, 371198523608646
Offset: 0

Views

Author

Rainer Rosenthal, Apr 03 2004

Keywords

Comments

The r-family of sequences is S_r(n) = 2*(T(n,(r-2)/2) - 1)/(r-4) provided r is not equal to 4 and S_4(n) = n^2 = A000290(n). Here T(n,x) are Chebyshev's polynomials of the first kind. See their coefficient triangle A053120. See also the R. Stephan link for the explicit formula for s_k(n) for k not equal to 4 (Stephan's s_k(n) is identical with S_r(n)).
An integer n is in this sequence iff mutually externally tangent circles with radii n, n+1, n+2 have Soddy circles (i.e., circles tangent to all three) of rational radius. - James R. Buddenhagen, Nov 16 2005
This sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. It is the case P1 = 6, P2 = 8, Q = 1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014
a(n) is the block size of the (n-1)-th design in a sequence of multi-set designs with 2 blocks, see A335649. - John P. McSorley, Jun 22 2020

Examples

			a(3)=25 because a(1)=1 and a(2)=6 and a(1)*a(3) = 1*25 = (6-1)^2 = (a(2)-1)^2.
		

Crossrefs

See A001110=S_36 for further references to S_r sequences.
Other members of this r-family are: A007877 (r=2), |A078070| (r=3), A004146 (r=5), A054493 (r=7). A098306, A100047. A001353, A001834. A001350, A052530.

Programs

  • Magma
    [Floor(1/2*(-2+(2+Sqrt(3))^n+(2-Sqrt(3))^n)): n in [0..30]]; // Vincenzo Librandi, Oct 14 2015
  • Maple
    A092184 := proc(n)
        option remember;
        if n <= 1 then
            n;
        else
            4*procname(n-1)-procname(n-2)+2 ;
        end if ;
    end proc:
    seq(A092184(n),n=0..10) ;# Zerinvary Lajos, Mar 09 2008
  • Mathematica
    Table[Simplify[ -((2 + Sqrt[3])^n - 1)*((2 - Sqrt[3])^n - 1)]/2, {n, 0, 26}] (* Stefan Steinerberger, May 15 2007 *)
    LinearRecurrence[{5,-5,1},{0,1,6},27] (* Ray Chandler, Jan 27 2014 *)
    CoefficientList[Series[x (1 + x)/(1 - 5 x + 5 x^2 - x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 14 2015 *)
  • PARI
    Vec(x*(1+x)/(1 - 5*x + 5*x^2 - x^3) + O(x^50)) \\ Michel Marcus, Oct 14 2015
    

Formula

S_r type sequences are defined by a(0)=0, a(1)=1, a(2)=r and a(n-1)*a(n+1) = (a(n)-1)^2. This sequence emanates from r=6.
a(n) = 1/2*(-2 + (2+sqrt(3))^n + (2-sqrt(3))^n). - Ralf Stephan, Apr 14 2004
G.f.: x*(1+x)/(1 - 5*x + 5*x^2 - x^3) = x*(1+x)/((1-x)*(1 - 4*x + x^2)). [from the Ralf Stephan link]
a(n) = T(n, 2)-1 = A001075(n)-1, with Chebyshev's polynomials T(n, 2) of the first kind.
a(n) = b(n) + b(n-1), n >= 1, with b(n):=A061278(n) the partial sums of S(n, 4) = U(n, 2) = A001353(n+1) Chebyshev's polynomials of the second kind.
An integer k is in this sequence iff k is nonnegative and (k^2 + 2*k)/3 is a square. - James R. Buddenhagen, Nov 16 2005
a(0)=0, a(1)=1, a(n+1) = 3 + floor(a(n)*(2+sqrt(3))). - Anton Vrba (antonvrba(AT)yahoo.com), Jan 16 2007
a(n) = 4*a(n-1) - a(n-2) + 2. - Zerinvary Lajos, Mar 09 2008
From Peter Bala, Mar 25 2014: (Start)
a(2*n) = 6*A001353(n)^2; a(2*n+1) = A001834(n)^2.
a(n) = u(n)^2, where {u(n)} is the Lucas sequence in the quadratic integer ring Z[sqrt(6)] defined by the recurrence u(0) = 0, u(1) = 1, u(n) = sqrt(6)*u(n-1) - u(n-2) for n >= 2.
Equivalently, a(n) = U(n-1,sqrt(6)/2)^2, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = (1/2)*( ((sqrt(6) + sqrt(2))/2)^n - ((sqrt(6) - sqrt(2))/2)^n )^2.
a(n) = bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -2; 1, 3] and T(n,x) denotes the Chebyshev polynomial of the first kind. Cf. A098306.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
exp( Sum_{n >= 1} 2*a(n)*x^n/n ) = 1 + Sum_{n >= 1} A052530(n)*x^n. Cf. A001350. - Peter Bala, Mar 19 2015
E.g.f.: exp(2*x)*cosh(sqrt(3)*x) - cosh(x) - sinh(x). - Stefano Spezia, Oct 13 2019

Extensions

Extension and Chebyshev comments from Wolfdieter Lang, Sep 10 2004

A095263 a(n+3) = 3*a(n+2) - 2*a(n+1) + a(n).

Original entry on oeis.org

1, 3, 7, 16, 37, 86, 200, 465, 1081, 2513, 5842, 13581, 31572, 73396, 170625, 396655, 922111, 2143648, 4983377, 11584946, 26931732, 62608681, 145547525, 338356945, 786584466, 1828587033, 4250949112, 9882257736, 22973462017, 53406819691
Offset: 1

Views

Author

Gary W. Adamson, May 31 2004

Keywords

Comments

a(n+1) = number of n-tuples over {0,1,2} without consecutive digits. For the general case see A096261.
Diagonal sums of Riordan array (1/(1-x)^3, x/(1-x^3)), A127893. - Paul Barry, Jan 07 2008
The signed variant (-1)^(n+1)*a(n+1) is the bottom right entry of the n-th power of the matrix [[0,1,0],[0,0,1],[-1,-2,-3]]. - Roger L. Bagula, Jul 01 2007
a(n) is the number of generalized compositions of n+1 when there are i^2/2-i/2 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Dedrickson (Section 4.1) gives a bijection between colored compositions of n, where each part k has one of binomial(k,2) colors, and 0,1,2 strings of length n-2 without sequential digits (i.e., avoiding 01 and 12). Cf. A052529. - Peter Bala, Sep 17 2013
Except for the initial 0, this is the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S^2 - S^3; see A291000. - Clark Kimberling, Aug 24 2017
For n>1, a(n-1) is the number of ways to split [n] into an unspecified number of intervals and then choose 2 blocks (i.e., subintervals) from each interval. For example, for n=6, a(5)=37 since the number of ways to split [6] into intervals and then select 2 blocks from each interval is C(6,2) + C(4,2)*C(2,2) + C(3,2)*C(3,2) + C(2,2)*C(4,2) + C(2,2)*C(2,2)*C(2,2). - Enrique Navarrete, May 20 2022

Examples

			a(9) = 1081 = 3*465 - 2*200 + 86.
M^9 * [1 0 0] = [a(7) a(8) a(9)] = [200 465 1081].
G.f. = x + 3*x^2 + 7*x^3 + 16*x^4 + 37*x^5 + 86*x^6 + 200*x^7 + ...
		

Crossrefs

Cf. A052921 (first differences), A137229 (partial sums).
Column k=3 of A277666.

Programs

  • Magma
    I:=[1,3,7]; [n le 3 select I[n] else 3*Self(n-1) -2*Self(n-2) +Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 12 2021
    
  • Maple
    A:= gfun:-rectoproc({a(n+3)=3*a(n+2)-2*a(n+1)+a(n),a(1)=1,a(2)=3,a(3)=7},a(n),remember):
    seq(A(n),n=1..100); # Robert Israel, Sep 15 2014
  • Mathematica
    a[1]=1; a[2]=3; a[3]=7; a[n_]:= a[n]= 3a[n-1] -2a[n-2] +a[n-3]; Table[a[n], {n, 22}] (* Or *)
    a[n_]:= (MatrixPower[{{0,1,2,3}, {1,2,3,0}, {2,3,0,1}, {3,0,1,2}}, n].{{1}, {0}, {0}, {0}})[[2, 1]]; Table[ a[n], {n, 22}] (* Robert G. Wilson v, Jun 16 2004 *)
    RecurrenceTable[{a[1]==1,a[2]==3,a[3]==7,a[n+3]==3a[n+2]-2a[n+1]+a[n]},a,{n,30}] (* Harvey P. Dale, Sep 17 2022 *)
  • Sage
    [sum( binomial(n+k+1,3*k+2) for k in (0..(n-1)//2)) for n in (1..30)] # G. C. Greubel, Apr 12 2021

Formula

Let M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 -2 3]; then M^n *[1 0 0] = [a(n-2) a(n-1) a(n)].
a(n)/a(n-1) tends to 2.3247179572..., an eigenvalue of M and a root of the characteristic polynomial. [Is that constant equal to 1 + A060006? - Michel Marcus, Oct 11 2014] [Yes, the limit is the root of the equation -1 + 2*x - 3*x^2 + x^3 = 0, after substitution x = y + 1 we have the equation for y: -1 - y + y^3 = 0, y = A060006. - Vaclav Kotesovec, Jan 27 2015]
Related to the Padovan sequence A000931 as follows : a(n)=A000931(3n+4). Also the binomial transform of A000931(n+4).
From Paul Barry, Jul 06 2004: (Start)
a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n+k, n-2*k+1).
a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n+k, 3*k-1). (End)
From Paul Barry, Jan 07 2008: (Start)
G.f.: x/(1 -3*x +2*x^2 -x^3).
a(n) = Sum_{k=0..floor(n/2)} binomial(n+k+2,3*k+2).
a(n) = Sum_{k=0..n} binomial(n,k) * Sum_{j=0..floor((k+4)/2)} binomial(j,k-2j+4). (End)
If p[i]=i(i-1)/2 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>=2, a(n-1)=det A. - Milan Janjic, May 02 2010
a(n) = A000931(3*n + 4). - Michael Somos, Sep 18 2012

Extensions

Edited by Paul Barry, Jul 06 2004
Corrected and extended by Robert G. Wilson v, Jun 05 2004

A052529 Expansion of (1-x)^3/(1 - 4*x + 3*x^2 - x^3).

Original entry on oeis.org

1, 1, 4, 13, 41, 129, 406, 1278, 4023, 12664, 39865, 125491, 395033, 1243524, 3914488, 12322413, 38789712, 122106097, 384377665, 1209982081, 3808901426, 11990037126, 37743426307, 118812495276, 374009739309, 1177344897715
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

a(n+1) is the number of distinct matrix products in (A+B+C+D)^n where commutator [A,B] = [A,C] = [B,C] = 0 but D does not commute with A, B, or C. - Paul D. Hanna and Max Alekseyev, Feb 01 2006
Starting (1, 4, 13, ...) = INVERT transform of the triangular series, (1, 3, 6, 10, ...). Example: a(5) = 129 = termwise products of (1, 1, 4, 13, 41) and (15, 10, 6, 3, 1) = (15 + 10 + 24 + 39 + 41). - Gary W. Adamson, Apr 10 2009
a(n) is the number of generalized compositions of n when there are i^2/2+i/2 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Dedrickson (Section 4.2) gives a bijection between colored compositions of n, where each part k has one of binomial(k+1,2) colors, and 0,1,2,3 strings of length n-1 avoiding 10, 20 and 21. Cf. A095263. For a refinement of this sequence counting binomial(k+1,2)-colored compositions by the number of parts see A127893. - Peter Bala, Sep 17 2013

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 80.

Crossrefs

Trisection of A000930.
First differences of A052544.
Row sums of triangle A127893.

Programs

  • Magma
    I:=[1, 1, 4, 13, 41, 129]; [n le 6 select I[n] else 4*Self(n-1) -3*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 22 2012
    
  • Maple
    spec := [S,{S=Sequence(Prod(Z,Sequence(Z),Sequence(Z),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
    f:= gfun:-rectoproc({a(n+4)-4*a(n+3)+3*a(n+2)-a(n+1), a(0) = 1, a(1) = 1, a(2) = 4, a(3) = 13},a(n),`remember`):
    seq(f(n),n=0..40); # Robert Israel, Dec 19 2014
  • Mathematica
    CoefficientList[Series[(-1+x)^3/(-1+4*x-3*x^2+x^3),{x,0,40}],x] (* Vincenzo Librandi, Jun 22 2012 *)
    LinearRecurrence[{4,-3,1},{1,1,4,13},30] (* Harvey P. Dale, Oct 04 2015 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-x)^3/(1-4*x+3*x^2-x^3)) \\ G. C. Greubel, May 12 2019
    
  • Sage
    ((1-x)^3/(1-4*x+3*x^2-x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 12 2019

Formula

a(n) = Sum_{a=0..n} (Sum_{b=0..n} (Sum_{c=0..n} C(n-b-c,a)*C(n-a-c,b)*C(n-a-b,c))).
G.f.: (1 - x)^3/(1 - 4*x + 3*x^2 - x^3).
a(n) = 4*a(n-1) - 3*a(n-2) + a(n-3) for n>=4.
a(n) = Sum_{alpha = RootOf(-1+4*x-3*x^2+x^3)} (1/31)*(6 - 5*alpha - 3*alpha^2) * alpha^(-1-n).
For n>0, a(n) = Sum_{k=0..n-1} Sum_{i=0..k} Sum_{j=0..i} a(j). - Benoit Cloitre, Jan 26 2003
a(n) = Sum_{k=0..n} binomial(n+2*k-1, n-k). - Vladeta Jovovic, Mar 23 2003
If p[i]=i(i+1)/2 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)=det A. - Milan Janjic, May 02 2010
Recurrence equation: a(n) = Sum_{k = 1..n} 1/2*k*(k+1)*a(n-k) with a(0) = 1. - Peter Bala, Sep 19 2013
a(n) = Sum_{i=0..n} (n-i)*A052544(i) = A052544(n) - A052544(n-1) for n>=1. - Areebah Mahdia, Jul 07 2020

A079935 a(n) = 4*a(n-1) - a(n-2) with a(1) = 1, a(2) = 3.

Original entry on oeis.org

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
Offset: 1

Views

Author

Benoit Cloitre and Paul D. Hanna, Jan 20 2003

Keywords

Comments

See A001835 for another version.
Greedy frac multiples of sqrt(3): a(1)=1, Sum_{n>0} frac(a(n)*x) < 1 at x=sqrt(3).
The n-th greedy frac multiple of x is the smallest integer that does not cause Sum_{k=1..n} frac(a(k)*x) to exceed unity; an infinite number of terms appear as the denominators of the convergents to the continued fraction of x.
Binomial transform of A002605. - Paul Barry, Sep 17 2003
In general, Sum_{k=0..n} binomial(2n-k,k)*j^(n-k) = (-1)^n* U(2n, i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
The Hankel transform of this sequence is [1,2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
From Richard Choulet, May 09 2010: (Start)
This sequence is a particular case of the following situation:
a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3) = (a(n+2)*a(n+1)+q)/a(n)
where q is given in Z to have Q = (a*b^2 + q*b + a + q)/(a*b) itself in Z.
The g.f is f: f(z) = (1 + a*z + (b-Q)*z^2 + (a*b + q - a*Q)*z^3)/(1 - Q*z^2 + z^4);
so we have the linear recurrence: a(n+4) = Q*a(n+2) - a(n).
The general form of a(n) is given by:
a(2*m) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (b-Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p) and
a(2*m+1) = a*Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (a*b+q-a*Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p).
(End)
x-values in the solution to 3*x^2 - 2 = y^2. - Sture Sjöstedt, Nov 25 2011
From Wolfdieter Lang, Oct 12 2020: (Start)
[X(n) = S(n, 4) - S(n-1, 4), Y(n) = X(n-1)] gives all positive solutions of X^2 + Y^2 - 4*X*Y = -2, for n = -oo..+oo, where the Chebyshev S-polynomials are given in 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 = +12. There is only this family representing -2 properly with X and Y positive, and there are no improper solutions.
See also the preceding comment by Sture Sjöstedt.
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)
a(n) is also the output of Tesler's formula for the number of perfect matchings of an m x n Mobius band where m and n are both even, specialized to m=2. (The twist is on the length-n side.) - Sarah-Marie Belcastro, Feb 15 2022

Examples

			a(4) = 41 since frac(1*x) + frac(3*x) + frac(11*x) + frac(41*x) < 1, while frac(1*x) + frac(3*x) + frac(11*x) + frac(k*x) > 1 for all k > 11 and k < 41.
		

Crossrefs

Cf. A002530 (denominators of convergents to sqrt(3)), A079934, A079936, A001353.
Cf. A001835 (same except for the first term).
Row 4 of array A094954.
Cf. similar sequences listed in A238379.

Programs

  • Haskell
    a079935 n = a079935_list !! (n-1)
    a079935_list =
       1 : 3 : zipWith (-) (map (4 *) $ tail a079935_list) a079935_list
    -- Reinhard Zumkeller, Aug 14 2011
    
  • Magma
    I:=[1,3]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jun 06 2015
    
  • Maple
    f:= gfun:-rectoproc({a(n) = 4*a(n-1) - a(n-2),a(1)=1,a(2)=3}, a(n), remember):
    seq(f(n),n=1..30); # Robert Israel, Jun 05 2015
  • Mathematica
    a[n_] := (MatrixPower[{{1, 2}, {1, 3}}, n].{{1}, {1}})[[1, 1]]; Table[ a[n], {n, 0, 23}] (* Robert G. Wilson v, Jan 13 2005 *)
    LinearRecurrence[{4,-1},{1,3},30] (* or *) CoefficientList[Series[ (1-x)/(1-4x+x^2),{x,0,30}],x]  (* Harvey P. Dale, Apr 26 2011 *)
    a[n_] := Sqrt[2/3] Cosh[(-1 - 2 n) ArcCsch[Sqrt[2]]];
    Table[Simplify[a[n-1]], {n, 1, 12}] (* Peter Luschny, Oct 13 2020 *)
  • PARI
    a(n)=([0,1; -1,4]^(n-1)*[1;3])[1,1] \\ Charles R Greathouse IV, Mar 18 2017
    
  • PARI
    my(x='x+O('x^30)); Vec((1-x)/(1-4*x+x^2)) \\ G. C. Greubel, Feb 25 2019
  • Sage
    [lucas_number1(n,4,1)-lucas_number1(n-1,4,1) for n in range(1, 25)] # Zerinvary Lajos, Apr 29 2009
    

Formula

For n > 0, a(n) = ceiling( (2+sqrt(3))^n/(3+sqrt(3)) ).
From Paul Barry, Sep 17 2003: (Start)
G.f.: (1-x)/(1-4*x+x^2).
E.g.f.: exp(2*x)*(sinh(sqrt(3)*x)/sqrt(3) + cosh(sqrt(3)*x)).
a(n) = ( (3+sqrt(3))*(2+sqrt(3))^n + (3-sqrt(3))*(2-sqrt(3))^n )/6 (offset 0). (End)
a(n) = Sum_{k=0..n} binomial(2*n-k, k)*2^(n-k). - Paul Barry, Jan 22 2005 [offset 0]
a(n) = (-1)^n*U(2*n, i*sqrt(2)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005 [offset 0]
a(n) = Jacobi_P(n,-1/2,1/2,2)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006 [offset 0]
a(n) = sqrt(2+(2-sqrt(3))^(2*n-1) + (2+sqrt(3))^(2*n-1))/sqrt(6). - Gerry Martens, Jun 05 2015
a(n) = (1/2 + sqrt(3)/6)*(2-sqrt(3))^n + (1/2 - sqrt(3)/6)*(2+sqrt(3))^n. - Robert Israel, Jun 05 2015
a(n) = S(n-1,4) - S(n-2,4) = (-1)^(n-1)*S(2*(n-1), i*sqrt(2)), with Chebyshev S-polynomials (A049310), the imaginary unit i, S(-1, x) = 0, for n >= 1. See also the formula above by Paul Barry (with offset 0). - Wolfdieter Lang, Oct 12 2020
a(n) = sqrt(2/3)*cosh((-1 - 2*n) arccsch(sqrt(2))), where arccsch is the inverse hyperbolic cosecant function (with offset 0). - Peter Luschny, Oct 13 2020
From Peter Bala, May 04 2025: (Start)
a(n) = (1/sqrt(3)) * sqrt(1 - T(2*n-1, -2)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(n) divides a(3*n-1); a(n) divides a(5*n-2); in general, for k >= 0, a(n) divides a((2*k+1)*n - k).
The aerated sequence [b(n)]n>=1 = [1, 0, 3, 0, 11, 0, 41, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -6, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 2} 1/(a(n) - 1/a(n)) = 1/2 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A052530(n-1) - 1/A052530(n).) (End)

A003699 Number of Hamiltonian cycles in C_4 X P_n.

Original entry on oeis.org

1, 6, 22, 82, 306, 1142, 4262, 15906, 59362, 221542, 826806, 3085682, 11515922, 42978006, 160396102, 598606402, 2234029506, 8337511622, 31116016982, 116126556306, 433390208242, 1617434276662, 6036346898406, 22527953316962, 84075466369442, 313773912160806
Offset: 1

Views

Author

Keywords

Comments

a(n) is the number of generalized compositions of n when there are i^2+i-1 different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
Is this the same as the sequence visible in Table 5 of Pettersson, 2014? - N. J. A. Sloane, Jun 05 2015

References

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

Crossrefs

Column k=4 of A359855.
First differences of A052530 and A071954.

Programs

  • GAP
    a:=[6,22];; for n in [3..20] do a[n]:=4a[n-1]-a[n-2]; od; Concatenation([1], a); # G. C. Greubel, Dec 23 2019
  • Magma
    I:=[1,6,22]; [n le 3 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 13 2018
    
  • Maple
    seq( simplify( `if`(n=1, 1, 2*(ChebyshevU(n-1,2) - ChebyshevU(n-2,2))) ), n=1..30); # G. C. Greubel, Dec 23 2019
  • Mathematica
    Join[{1},LinearRecurrence[{4,-1},{6,22},30]] (* Harvey P. Dale, Jul 19 2011 *)
    Table[If[n<2, n, 2*(ChebyshevU[n-1, 2] - ChebyshevU[n-2, 2])], {n,30}] (* G. C. Greubel, Dec 23 2019 *)
  • Maxima
    (a[1] : 1, a[2] : 6, a[3] : 22, a[n] := 4*a[n - 1] - a[n - 2], makelist(a[n], n, 1, 50)); /* Franck Maminirina Ramaharo, Nov 12 2018 */
    
  • PARI
    vector(30, n, if(n==1, 1, 2*(polchebyshev(n-1, 2, 2) - polchebyshev(n-2, 2, 2))) ) \\ G. C. Greubel, Dec 23 2019
    
  • Sage
    [1]+[2*(chebyshev_U(n-1,2) - chebyshev_U(n-2,2)) for n in (2..30)] # G. C. Greubel, Dec 23 2019
    

Formula

a(n) = 2 * A001835(n), n > 1.
From Benoit Cloitre, Mar 28 2003: (Start)
a(n) = ceiling((1 - sqrt(1/3))*(2 + sqrt(3))^n) for n > 1.
a(1) = 1, a(2) = 6, a(3) = 22 and for n > 3, a(n) = 4*a(n-1) - a(n-2). (End)
O.g.f.: x*(1 + 2*x - x^2)/(1-4*x+x^2) = -2 - x + 2*(1 - 3*x)/(1-4*x+x^2). - R. J. Mathar, Nov 23 2007
From Franck Maminirina Ramaharo, Nov 12 2018: (Start)
a(n) = ((1 + sqrt(3))*(2 - sqrt(3))^n - (1 - sqrt(3))*(2 + sqrt(3))^n)/sqrt(3), n > 1.
E.g.f.: ((1 + sqrt(3))*exp((2 - sqrt(3))*x) - (1 - sqrt(3))*exp((2 + sqrt(3))*x) - (2 + x)*sqrt(3))/sqrt(3). (End)
a(n) = 2*(ChebyshevU(n-1, 2) - ChebyshevU(n-2, 2)) for n >1, with a(1)=1. - G. C. Greubel, Dec 23 2019

A048788 a(2n+1) = a(2n) + a(2n-1), a(2n) = 2*a(2n-1) + a(2n-2); a(n) = n for n = 0, 1.

Original entry on oeis.org

0, 1, 2, 3, 8, 11, 30, 41, 112, 153, 418, 571, 1560, 2131, 5822, 7953, 21728, 29681, 81090, 110771, 302632, 413403, 1129438, 1542841, 4215120, 5757961, 15731042, 21489003, 58709048, 80198051, 219105150, 299303201, 817711552, 1117014753
Offset: 0

Views

Author

Robin Trew (trew(AT)hcs.harvard.edu), Dec 11 1999

Keywords

Comments

Numerators of continued fraction convergents to sqrt(3) - 1 (A160390). See A002530 for denominators. - N. J. A. Sloane, Dec 17 2007
Convergents are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, ... - Clark Kimberling, Sep 21 2013
A strong divisibility sequence, that is gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. - Peter Bala, Jun 06 2014
From Sarah-Marie Belcastro, Feb 15 2022: (Start)
a(n) is also the number of perfect matchings of an edge-labeled 2 X (n-1) Mobius band grid graph, or equivalently the number of domino tilings of a 2 X (n-1) Mobius band grid. (The twist is on the length-n side.)
a(n) is also the output of Lu and Wu's formula for the number of perfect matchings of an m X n Mobius band grid, specialized to m = 2 with the twist on the length-n side.
2*a(n) is the number of perfect matchings of an edge-labeled 2 X (n-1) projective planar grid graph, or equivalently the number of domino tilings of a 2 X (n-1) projective planar grid. (End)

References

  • Russell Lyons, A bird's-eye view of uniform spanning trees and forests, in Microsurveys in Discrete Probability, AMS, 1998.

Crossrefs

Bisections are A001835 and A052530.

Programs

  • GAP
    a:=[0,1,2,3];; for n in [5..40] do a[n]:=4a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Magma
    I:=[0,1,2,3]; [n le 4 select I[n] else 4*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Dec 10 2013
    
  • Maple
    seq( simplify( `if`(`mod`(n,2)=0, 2*ChebyshevU((n-2)/2, 2), ChebyshevU((n-1)/2, 2) - ChebyshevU((n-3)/2, 2)) ), n=0..40); # G. C. Greubel, Dec 23 2019
  • Mathematica
    Numerator[NestList[(2/(2 + #))&, 0, 40]] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)
    CoefficientList[Series[x(1+2x-x^2)/(1-4x^2+x^4), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 10 2013 *)
    a0[n_]:= ((3+Sqrt[3])*(2-Sqrt[3])^n-((-3+Sqrt[3])*(2+Sqrt[3])^n))/6 // Simplify
    a1[n_]:= 2*Sum[a0[i], {i, 1, n}]
    Flatten[MapIndexed[{a1[#-1],a0[#]}&,Range[20]]] (* Gerry Martens, Jul 10 2015 *)
    Round@Table[With[{r=1+Sqrt[2], s=1+Sqrt[3]}, ((r + (-1)^n/r) s^n/2^(n/2) - (1/r + (-1)^n r) 2^(n/2)/s^n) Sqrt[6]/12], {n, 0, 20}] (* or *) LinearRecurrence[ {0,4,0,-1}, {0,1,2,3}, 40] (* Vladimir Reshetnikov, May 11 2016 *)
    Table[If[EvenQ[n], 2*ChebyshevU[(n-2)/2, 2], ChebyshevU[(n-1)/2, 2] - ChebyshevU[(n-3)/2, 2]], {n, 0, 40}] (* G. C. Greubel, Dec 23 2019 *)
  • PARI
    main(size)=v=vector(size); v[1]=0;v[2]=1;v[3]=2;v[4]=3;for(i=5, size, v[i]=4*v[i-2] - v[i-4]); v; \\ Anders Hellström, Jul 11 2015
    
  • PARI
    a=vector(50); a[1]=1; a[2]=2; for(n=3, #a, if(n%2==1, a[n]=a[n-1]+a[n-2], a[n]=2*a[n-1]+a[n-2])); concat(0, a) \\ Colin Barker, Jan 30 2016
    
  • PARI
    a(n)=([0,1,0,0;0,0,1,0;0,0,0,1;-1,0,4,0]^n*[0;1;2;3])[1,1] \\ Charles R Greathouse IV, Mar 16 2017
    
  • PARI
    apply( {A048788(n)=imag((2+quadgen(12))^(n\/2)*if(bittest(n, 0), quadgen(12)-1, 2))}, [0..30]) \\ M. F. Hasler, Nov 04 2019
    
  • PARI
    {a(n) = my(s=1,m=n); if(n<0,s=-(-1)^n; m=-n); polcoeff(x*(1+2*x-x^2)/(1-4*x^2+x^4) + x*O(x^m), m)*s}; /* Michael Somos, Sep 17 2020 */
    
  • Sage
    @CachedFunction
    def a(n):
        if (mod(n,2)==0): return 2*chebyshev_U((n-2)/2, 2)
        else: return chebyshev_U((n-1)/2, 2) - chebyshev_U((n-3)/2, 2)
    [a(n) for n in (0..40)] # G. C. Greubel, Dec 23 2019
    

Formula

G.f.: x*(1+2*x-x^2)/(1-4*x^2+x^4). - Paul Barry, Sep 18 2009
a(n) = 4*a(n-2) - a(n-4). - Vincenzo Librandi, Dec 10 2013
a(2*n-1) = A001835(n); a(2*n) = 2*A001353(n). - Peter Bala, Jun 06 2014
From Gerry Martens, Jul 11 2015: (Start)
Interspersion of 2 sequences [a1(n-1),a0(n)] for n>0:
a0(n) = ((3+sqrt(3))*(2-sqrt(3))^n-((-3+sqrt(3))*(2+sqrt(3))^n))/6.
a1(n) = 2*Sum_{i=1..n} a0(i). (End)
a(n) = ((r + (-1)^n/r)*s^n/2^(n/2) - (1/r + (-1)^n*r)*2^(n/2)/s^n)*sqrt(6)/12, where r = 1 + sqrt(2), s = 1 + sqrt(3). - Vladimir Reshetnikov, May 11 2016
a(n) = 2*ChebyshevU(n-1, 2) if n is even and ChebyshevU(n, 2) - ChebyshevU(n-1, 2) if n in odd. - G. C. Greubel, Dec 23 2019
a(n) = -(-1)^n*a(-n) for all n in Z. - Michael Somos, Sep 17 2020

Extensions

Denominator of g.f. corrected by Paul Barry, Sep 18 2009
Incorrect g.f. deleted by Colin Barker, Aug 10 2012
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