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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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A001906 F(2n) = bisection of Fibonacci sequence: a(n) = 3*a(n-1) - a(n-2).

Original entry on oeis.org

0, 1, 3, 8, 21, 55, 144, 377, 987, 2584, 6765, 17711, 46368, 121393, 317811, 832040, 2178309, 5702887, 14930352, 39088169, 102334155, 267914296, 701408733, 1836311903, 4807526976, 12586269025, 32951280099, 86267571272, 225851433717, 591286729879, 1548008755920
Offset: 0

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Keywords

Comments

Apart from initial term, same as A088305.
Second column of array A102310 and of A028412.
Numbers k such that 5*k^2 + 4 is a square. - Gregory V. Richardson, Oct 13 2002
Apart from initial terms, also Pisot sequences E(3,8), P(3,8), T(3,8). See A008776 for definitions of Pisot sequences.
Binomial transform of A000045. - Paul Barry, Apr 11 2003
Number of walks of length 2n+1 in the path graph P_4 from one end to the other one. Example: a(2)=3 because in the path ABCD we have ABABCD, ABCBCD and ABCDCD. - Emeric Deutsch, Apr 02 2004
Simplest example of a second-order recurrence with the sixth term a square.
Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 3. - Lekraj Beedassy, Jun 11 2004
a(n) (for n > 0) is the smallest positive integer that cannot be created by summing at most n values chosen among the previous terms (with repeats allowed). - Andrew Weimholt, Jul 20 2004
All nonnegative integer solutions of Pell equation b(n)^2 - 5*a(n)^2 = +4 together with b(n) = A005248(n), n >= 0. - Wolfdieter Lang, Aug 31 2004
a(n+1) is a Chebyshev transform of 3^n (A000244), 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
a(n) is the number of distinct products of matrices A, B, C, in (A+B+C)^n where commutator [A,B] = 0 but C does not commute with A or B. - Paul D. Hanna and Max Alekseyev, Feb 01 2006
Number of binary words with exactly k-1 strictly increasing runs. Example: a(3)=F(6)=8 because we have 0|0,1|0,1|1,0|01,01|0,1|01,01|1 and 01|01. Column sums of A119900. - Emeric Deutsch, Jul 23 2006
See Table 1 on page 411 of Lukovits and Janezic paper. - Parthasarathy Nambi, Aug 22 2006
Inverse: With phi = (sqrt(5) + 1)/2, log_phi((sqrt(5) a(n) + sqrt(5 a(n)^2 + 4))/2) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007
[1,3,8,21,55,144,...] is the Hankel transform of [1,1,4,17,75,339,1558,...](see A026378). - Philippe Deléham, Apr 13 2007
The Diophantine equation a(n) = m has a solution (for m >= 1) if and only if floor(arcsinh(sqrt(5)*m/2)/log(phi)) <> floor(arccosh(sqrt(5)*m/2)/log(phi)) where phi is the golden ratio. An equivalent condition is A130259(m) = A130260(m). - Hieronymus Fischer, May 25 2007
a(n+1) = AB^(n)(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., 1=`1`, 3=`10`, 8=`100`, 21=`1000`, ..., in Wythoff code.
Equals row sums of triangles A140069, A140736 and A140737. - Gary W. Adamson, May 25 2008
a(n) is also the number of idempotent order-preserving partial transformations (of an n-element chain) of width n (width(alpha) = max(Im(alpha))). Equivalently, it is the number of idempotent order-preserving full transformations (of an n-element chain). - Abdullahi Umar, Sep 08 2008
a(n) is the number of ways that a string of 0,1 and 2 of size (n-1) can be arranged with no 12-pairs. - Udita Katugampola, Sep 24 2008
Starting with offset 1 = row sums of triangle A175011. - Gary W. Adamson, Apr 03 2010
As a fraction: 1/71 = 0.01408450... or 1/9701 = 0.0001030821.... - Mark Dols, May 18 2010
Sum of the products of the elements in the compositions of n (example for n=3: the compositions are 1+1+1, 1+2, 2+1, and 3; a(3) = 1*1*1 + 1*2 + 2*1 + 3 = 8). - Dylon Hamilton, Jun 20 2010, Geoffrey Critzer, Joerg Arndt, Dec 06 2010
a(n) relates to regular polygons with even numbers of edges such that Product_{k=1..(n-2)/2} (1 + 4*cos^2 k*Pi/n) = even-indexed Fibonacci numbers with a(n) relating to the 2*n-gons. The constants as products = roots to even-indexed rows of triangle A152063. For example: a(5) = 55 satisfies the product formula relating to the 10-gon. - Gary W. Adamson, Aug 15 2010
Alternatively, product of roots to x^4 - 12x^3 + 51x^2 - 90x + 55, (10th row of triangle A152063) = (4.618...)*(3.618...)*(2.381...)*(1.381...) = 55. - Gary W. Adamson, Aug 15 2010
a(n) is the number of generalized compositions of n when there are i different types of i, (i=1,2,...). - Milan Janjic, Aug 26 2010
Starting with "1" = row sums of triangle A180339, and eigensequence of triangle A137710. - Gary W. Adamson, Aug 28 2010
a(2) = 3 is the only prime.
Number of nonisomorphic graded posets with 0 and uniform hasse graph of rank n > 0, with exactly 2 elements of each rank level above 0. (Uniform used in the sense of Retakh, Serconek, and Wilson. Graded used in Stanley's sense that every maximal chain has the same length n.) - David Nacin, Feb 13 2012
Pisano period lengths: 1, 3, 4, 3, 10, 12, 8, 6, 12, 30, 5, 12, 14, 24, 20, 12, 18, 12, 9, 30, ... - R. J. Mathar, Aug 10 2012
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy + 1. - Michel Lagneau, Feb 01 2014
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2}. - Milan Janjic, Jan 25 2015
With a(0) = 0, for n > 1, a(n) is the smallest number not already in the sequence such that a(n)^2 - a(n-1)^2 is a Fibonacci number. - Derek Orr, Jun 08 2015
Let T be the tree generated by these rules: 0 is in T, and if p is in T, then p + 1 is in T and x*p is in T and y*p is in T. The n-th generation of T consists of A001906(n) polynomials, for n >= 0. - Clark Kimberling, Nov 24 2015
For n > 0, a(n) = exactly the maximum area of a quadrilateral with sides in order of lengths F(n), F(n), L(n), and L(n) with L(n)=A000032(n). - J. M. Bergot, Jan 20 2016
a(n) = twice the area of a triangle with vertices at (L(n+1), L(n+2)), (F(n+1), F(n+1)), and (L(n+2), L(n+1)), with L(n)=A000032(n). - J. M. Bergot, Apr 20 2016
Except for the initial 0, this is the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S - S^2; see A291000. - Clark Kimberling, Aug 24 2017
a(n+1) is the number of spanning trees of the graph T_n, where T_n is a sequence of n triangles, where adjacent triangles share an edge. - Kevin Long, May 07 2018
a(n) is the number of ways to partition [n] such that each block is a run of consecutive numbers, and each block has a fixed point, e.g., for n=3, 12|3 with 1 and 3 as fixed points is valid, but 13|2 is not valid as 1 and 3 do not form a run. Consequently, a(n) also counts the spanning trees of the graph given by taking a path with n vertices and adding another vertex adjacent to all of them. - Kevin Long, May 11 2018
From Wolfdieter Lang, May 31 2018: (Start)
The preceding comment can be paraphrased as follows. a(n) is the row sum of the array A305309 for n >= 1. The array A305309(n, k) gives the sum of the products of the block lengths of the set partition of [n] := {1, 2, ..., n} with A048996(n, k) blocks of consecutive numbers, corresponding to the compositions obtained from the k-th partition of n in Abramowitz-Stegun order. See the comments and examples at A305309.
{a(n)} also gives the infinite sequence of nonnegative numbers k for which k * ||k*phi|| < 1/sqrt(5), where the irrational number phi = A001622 (golden section), and ||x|| is the absolute value of the difference between x and the nearest integer. See, e.g., the Havil reference, pp. 171-172. (End)
a(n) is the number of tilings of two n X 1 rectangles joined orthogonally at a common end-square (so to have 2n-1 squares in a right-angle V shape) with only 1 X 1 and 2 X 1 tiles. This is a consequence of F(2n) = F(n+1)*F(n) + F(n)*F(n-1). - Nathaniel Gregg, Oct 10 2021
These are the denominators of the upper convergents to the golden ratio, tau; they are also the numerators of the lower convergents (viz. 1/1 < 3/2 < 8/5 < 21/13 < ... < tau < ... 13/8 < 5/3 < 2/1). - Clark Kimberling, Jan 02 2022
For n > 1, a(n) is the smallest Fibonacci number of unit equilateral triangle tiles needed to make an isosceles trapezoid of height F(n) triangles. - Kiran Ananthpur Bacche, Sep 01 2024

Examples

			G.f. = x + 3*x^2 + 8*x^3 + 21*x^4 + 55*x^5 + 144*x^6 + 377*x^7 + 987*x^8 + ...
a(3) = 8 because there are exactly 8 idempotent order-preserving full transformations on a 3-element chain, namely: (1,2,3)->(1,1,1),(1,2,3)->(2,2,2),(1,2,3)->(3,3,3),(1,2,3)->(1,1,3),(1,2,3)->(2,2,3),(1,2,3)->(1,2,2),(1,2,3)->(1,3,3),(1,2,3)->(1,2,3)-mappings are coordinate-wise. - _Abdullahi Umar_, Sep 08 2008
		

References

  • Mohammad K. Azarian, The Generating Function for the Fibonacci Sequence, Missouri Journal of Mathematical Sciences, Vol. 2, No. 2, Spring 1990, pp. 78-79. Zentralblatt MATH, Zbl 1097.11516.
  • Mohammad K. Azarian, A Generalization of the Climbing Stairs Problem II, Missouri Journal of Mathematical Sciences, Vol. 16, No. 1, Winter 2004, pp. 12-17.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 2,5,6,14,33,55.
  • R. J. Douglas, Tournaments that admit exactly one Hamiltonian cycle, Proc. London Math. Soc., 21 (1970), 716-730.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • A. Gerardin, Reply to Query 4389, L'Intermédiaire des Mathématiciens, 22 (1915), 23.
  • Julian Havil, The Irrationals, Princeton University Press, Princeton and Oxford, 2012, pp. 171-172.
  • Howie, J. M. Combinatorial and probabilistic results in transformation semigroups. Words, languages and combinatorics, II (Kyoto, 1992), 200--206, World Sci. Publ., River Edge, NJ, (1994).
  • Laradji, A. and Umar, A. Combinatorial results for semigroups of order-preserving full transformations. Semigroup Forum 72 (2006), 51-62.
  • I. Lukovits, A. Graovac, E. Kalman, G. Kaptay, P. Nagy, S. Nikolic, J. Sytchev and N. Trinajstich, "Nanotubes: Number of Kekulé Structures and Aromaticity", J. Chem. Inf. Comput. Sci, vol. 43 (2003), pp. 609-614. See Equation 6 on page 611.
  • T. Mansour, M. Shattuck, A statistic on n-color compositions and related sequences, Proc. Indian Acad. Sci. (Math. Sci.) Vol. 124, No. 2, May 2014, pp. 127-140.
  • H. Mathieu, Query 3932, L'Intermédiaire des Mathématiciens, 18 (1911), 222. - N. J. A. Sloane, Mar 08 2022
  • I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 101.
  • Paulo Ribenboim, Primes in Lucas sequences (Chap 4), in 'My Numbers, My Friends', Springer-Verlag 2000 NY, page 27.
  • 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. Stanley, Enumerative combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Crossrefs

Fibonacci A000045 = union of this sequence and A001519.
Inverse sequences A130259 and A130260.

Programs

  • Haskell
    a001906 n = a001906_list !! n
    a001906_list =
       0 : 1 : zipWith (-) (map (* 3) $ tail a001906_list) a001906_list
    -- Reinhard Zumkeller, Oct 03 2011
    
  • Magma
    [Fibonacci(2*n): n in [0..30]]; // Vincenzo Librandi, Sep 10 2014
  • Maple
    with(combstruct): SeqSeqSeqL := [T, {T=Sequence(S, card > 0), S=Sequence(U, card > 1), U=Sequence(Z, card >0)}, unlabeled]: seq(count(SeqSeqSeqL, size=n+1), n=0..28); # Zerinvary Lajos, Apr 04 2009
    H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -4):
    a := n -> `if`(n = 0, 0, H(2*n, 1, 1/2)):
    seq(simplify(a(n)), n=0..30); # Peter Luschny, Sep 03 2019
    A001906 := proc(n)
        combinat[fibonacci](2*n) ;
    end proc:
    seq(A001906(n),n=0..20) ; # R. J. Mathar, Jan 11 2024
  • Mathematica
    f[n_] := Fibonacci[2n]; Array[f, 28, 0] (* or *)
    LinearRecurrence[{3, -1}, {0, 1}, 28] (* Robert G. Wilson v, Jul 13 2011 *)
    Take[Fibonacci[Range[0,60]],{1,-1,2}] (* Harvey P. Dale, May 23 2012 *)
    Table[ ChebyshevU[n-1, 3/2], {n, 0, 30}] (* Jean-François Alcover, Jan 25 2013, after Michael Somos *)
    CoefficientList[Series[(x)/(1 - 3x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Sep 10 2014 *)
  • Maxima
    makelist(fib(2*n),n,0,30); /* Martin Ettl, Oct 21 2012 */
    
  • MuPAD
    numlib::fibonacci(2*n) $ n = 0..35; // Zerinvary Lajos, May 09 2008
    
  • PARI
    {a(n) = fibonacci(2*n)}; /* Michael Somos, Dec 06 2002 */
    
  • PARI
    {a(n) = subst( poltchebi(n+1)*4 - poltchebi(n)*6, x, 3/2)/5}; /* Michael Somos, Dec 06 2002 */
    
  • PARI
    {a(n) = polchebyshev( n-1, 2, 3/2)}; /* Michael Somos Jun 18 2011 */
    
  • PARI
    Vec(x/(1-3*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Oct 24 2012
    
  • Python
    def a(n, adict={0:0, 1:1}):
        if n in adict:
            return adict[n]
        adict[n]=3*a(n-1) - a(n-2)
        return adict[n] # David Nacin, Mar 04 2012
    
  • Sage
    [lucas_number1(n,3,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [fibonacci(2*n) for n in range(0, 28)] # Zerinvary Lajos, May 15 2009
    

Formula

G.f.: x / (1 - 3*x + x^2). - Simon Plouffe in his 1992 dissertation
a(n) = 3*a(n-1) - a(n-2) = A000045(2*n).
a(n) = -a(-n).
a(n) = A060921(n-1, 0), n >= 1.
a(n) = sqrt((A005248(n)^2 - 4)/5).
a(n) = A007598(n) - A007598(n-2), n > 1.
a(n) = (ap^n - am^n)/(ap-am), with ap := (3+sqrt(5))/2, am := (3-sqrt(5))/2.
Invert transform of natural numbers: a(n) = Sum_{k=1..n} k*a(n-k), a(0) = 1. - Vladeta Jovovic, Apr 27 2001
a(n) = S(n-1, 3) with S(n, x) = U(n, x/2) Chebyshev's polynomials of the 2nd kind, see A049310.
a(n) = Sum_{k=0..n} binomial(n, k)*F(k). - Benoit Cloitre, Sep 03 2002
Limit_{n->infinity} a(n)/a(n-1) = 1 + phi = (3 + sqrt(5))/2. This sequence includes all of the elements of A033888 combined with A033890.
a(0)=0, a(1)=1, a(2)=3, a(n)*a(n-2) + 1 = a(n-1)^2. - Benoit Cloitre, Dec 06 2002
a(n) = n + Sum_{k=0..n-1} Sum_{i=0..k} a(i) = n + A054452(n). - Benoit Cloitre, Jan 26 2003
a(n) = Sum_{k=1..n} binomial(n+k-1, n-k). - Vladeta Jovovic, Mar 23 2003
E.g.f.: (2/sqrt(5))*exp(3*x/2)*sinh(sqrt(5)*x/2). - Paul Barry, Apr 11 2003
Second diagonal of array defined by T(i, 1) = T(1, j) = 1, T(i, j) = Max(T(i-1, j) + T(i-1, j-1); T(i-1, j-1) + T(i, j-1)). - Benoit Cloitre, Aug 05 2003
a(n) = F(n)*L(n) = A000045(n)*A000032(n). - Lekraj Beedassy, Nov 17 2003
F(2n+2) = 1, 3, 8, ... is the binomial transform of F(n+2). - Paul Barry, Apr 24 2004
Partial sums of A001519(n). - Lekraj Beedassy, Jun 11 2004
a(n) = Sum_{i=0..n-1} binomial(2*n-1-i, i)*5^(n-i-1)*(-1)^i. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n) = Sum_{k=0..n} binomial(n+k, n-k-1) = Sum_{k=0..n} binomial(n+k, 2k+1).
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*3^(n-2*k). - Paul Barry, Oct 25 2004
a(n) = (n*L(n) - F(n))/5 = Sum_{k=0..n-1} (-1)^n*L(2*n-2*k-1).
The i-th term of the sequence is the entry (1, 2) in the i-th power of the 2 X 2 matrix M = ((1, 1), (1, 2)). - Simone Severini, Oct 15 2005
Computation suggests that this sequence is the Hankel transform of A005807. The Hankel transform of {a(n)} is Det[{{a(1), ..., a(n)}, {a(2), ..., a(n+1)}, ..., {a(n), ..., a(2n-1)}}]. - John W. Layman, Jul 21 2000
a(n+1) = (A005248(n+1) - A001519(n))/2. - Creighton Dement, Aug 15 2004
a(n+1) = Sum_{i=0..n} Sum_{j=0..n} binomial(n-i, j)*binomial(n-j, i). - N. J. A. Sloane, Feb 20 2005
a(n) = (2/sqrt(5))*sinh(2*n*psi), where psi:=log(phi) and phi=(1+sqrt(5))/2. - Hieronymus Fischer, Apr 24 2007
a(n) = ((phi+1)^n - A001519(n))/phi with phi=(1+sqrt(5))/2. - Reinhard Zumkeller, Nov 22 2007
Row sums of triangle A135871. - Gary W. Adamson, Dec 02 2007
a(n)^2 = Sum_{k=1..n} a(2*k-1). This is a property of any sequence S(n) such that S(n) = B*S(n-1) - S(n-2) with S(0) = 0 and S(1) = 1 including {0,1,2,3,...} where B = 2. - Kenneth J Ramsey, Mar 23 2008
a(n) = 1/sqrt(5)*(phi^(2*n+2) - phi^(-2*n-2)), where phi = (1+sqrt(5))/2, the golden ratio. - Udita Katugampola (SIU), Sep 24 2008
If p[i] = i 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
If p[i] = Stirling2(i,2) 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-1) = det(A). - Milan Janjic, May 08 2010
a(n) = F(2*n+10) mod F(2*n+5).
a(n) = 1 + a(n-1) + Sum_{i=1..n-1} a(i), with a(0)=0. - Gary W. Adamson, Feb 19 2011
a(n) is equal to the permanent of the (n-1) X (n-1) Hessenberg matrix with 3's along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
a(n), n > 1 is equal to the determinant of an (n-x) X (n-1) tridiagonal matrix with 3's in the main diagonal, 1's in the super and subdiagonals, and the rest 0's. - Gary W. Adamson, Jun 27 2011
a(n) = b such that Integral_{x=0..Pi/2} sin(n*x)/(3/2-cos(x)) dx = c + b*log(3). - Francesco Daddi, Aug 01 2011
a(n+1) = Sum_{k=0..n} A101950(n,k)*2^k. - Philippe Deléham, Feb 10 2012
G.f.: A(x) = x/(1-3*x+x^2) = G(0)/sqrt(5); where G(k)= 1 -(a^k)/(1 - b*x/(b*x - 2*(a^k)/G(k+1))), a = (7-3*sqrt(5))/2, b = 3+sqrt(5), if |x|<(3-sqrt(5))/2 = 0.3819660...; (continued fraction 3 kind, 3-step ). - Sergei N. Gladkovskii, Jun 25 2012
a(n) = 2^n*b(n;1/2) = -b(n;-1), where b(n;d), n=0,1,...,d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also Witula's et al. papers). - Roman Witula, Jul 12 2012
Product_{n>=1} (1 + 1/a(n)) = 1 + sqrt(5). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = (1/6)*(1 + sqrt(5)). - Peter Bala, Dec 23 2012
G.f.: x/(1-2*x) + x^2/(1-2*x)/(Q(0)-x) where Q(k) = 1 - x/(x*k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Feb 23 2013
G.f.: G(0)/2 - 1, where G(k) = 1 + 1/( 1 - x/(x + (1-x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: x*G(0)/(2-3*x), where G(k) = 1 + 1/( 1 - x*(5*k-9)/(x*(5*k-4) - 6/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013
Sum_{n>=1} 1/(a(n) + 1/a(n)) = 1. Compare with A001519, A049660 and A049670. - Peter Bala, Nov 29 2013
a(n) = U(n-1,3/2) where U(n-1,x) is Chebyshev polynomial of the second kind. - Milan Janjic, Jan 25 2015
The o.g.f. A(x) satisfies A(x) + A(-x) + 6*A(x)*A(-x) = 0. The o.g.f. for A004187 equals -A(sqrt(x))*A(-sqrt(x)). - Peter Bala, Apr 02 2015
For n > 1, a(n) = (3*F(n+1)^2 + 2*F(n-2)*F(n+1) - F(n-2)^2)/4. - J. M. Bergot, Feb 16 2016
For n > 3, a(n) = floor(MA) - 4 for n even and floor(MA) + 5 for n odd. MA is the maximum area of a quadrilateral with lengths of sides in order L(n), L(n), F(n-3), F(n+3), with L(n)=A000032(n). The ratio of the longer diagonal to the shorter approaches 5/3. - J. M. Bergot, Feb 16 2016
a(n+1) = Sum_{j=0..n} Sum_{k=0..j} binomial(n-j,k)*binomial(j,k)*2^(j-k). - Tony Foster III, Sep 18 2017
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k+i,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = Sum_{k=1..A000041(n)} A305309(n, k), n >= 1. Also row sums of triangle A078812.- Wolfdieter Lang, May 31 2018
a(n) = H(2*n, 1, 1/2) for n > 0 where H(n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -4). - Peter Luschny, Sep 03 2019
Sum_{n>=1} 1/a(n) = A153386. - Amiram Eldar, Oct 04 2020
a(n) = A249450(n) + 2. - Leo Tavares, Oct 10 2021
a(n) = -2/(sqrt(5)*tan(2*arctan(phi^(2*n)))), where phi = A001622 is the golden ratio. - Diego Rattaggi, Nov 21 2021
a(n) = sinh(2*n*arcsinh(1/2))/sqrt(5/4). - Peter Luschny, May 21 2022
From Amiram Eldar, Dec 02 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 1 + 1/sqrt(5) (A344212).
Product_{n>=2} (1 + (-1)^n/a(n)) = (5/6) * (1 + 1/sqrt(5)). (End)
a(n) = Sum_{k>=0} Fibonacci(2*n*k)/(Lucas(2*n)^(k+1)). - Diego Rattaggi, Jan 12 2025
Sum_{n>=0} a(n)/3^n = 3. - Diego Rattaggi, Jan 20 2025

A002212 Number of restricted hexagonal polyominoes with n cells.

Original entry on oeis.org

1, 1, 3, 10, 36, 137, 543, 2219, 9285, 39587, 171369, 751236, 3328218, 14878455, 67030785, 304036170, 1387247580, 6363044315, 29323149825, 135700543190, 630375241380, 2938391049395, 13739779184085, 64430797069375, 302934667061301, 1427763630578197
Offset: 0

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Author

N. J. A. Sloane, Ronald C. Read

Keywords

Comments

Number of Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0) with no peaks at odd level. Example: a(2)=3 because we have UUDD, UHD and HH. - Emeric Deutsch, Dec 06 2003
Number of 3-Motzkin paths of length n-1 (i.e., lattice paths from (0,0) to (n-1,0) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and three types of steps H=(1,0)). Example: a(4)=36 because we have 27 HHH paths, 3 HUD paths, 3 UHD paths and 3 UDH paths. - Emeric Deutsch, Jan 22 2004
Number of rooted, planar trees having edges weighted by strictly positive integers (multi-trees) with weight-sum n. - Roland Bacher, Feb 28 2005
Number of skew Dyck paths of semilength n. A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of the path is defined to be the number of its steps. - Emeric Deutsch, May 10 2007
Equivalently, number of self-avoiding paths of semilength n in the first quadrant beginning at the origin, staying weakly above the diagonal, ending on the diagonal, and consisting of steps r=(+1,0) (right), U=(0,+1) (up), and D=(0,-1) (down). Self-avoidance implies that factors UD and DU and steps D reaching the diagonal before the end are forbidden. The a(3) = 10 such paths are UrUrUr, UrUUrD, UrUUrr, UUrrUr, UUrUrD, UUrUrr, UUUDrD, UUUrDD, UUUrrD, and UUUrrr. - Joerg Arndt, Jan 15 2024
Hankel transform of [1,3,10,36,137,543,...] is A000012 = [1,1,1,1,...]. - Philippe Deléham, Oct 24 2007
From Gary W. Adamson, May 17 2009: (Start)
Convolved with A026375, (1, 3, 11, 45, 195, ...) = A026378: (1, 4, 17, 75, ...)
(1, 3, 10, 36, 137, ...) convolved with A026375 = A026376: (1, 6, 30, 144, ...).
Starting (1, 3, 10, 36, ...) = INVERT transform of A007317: (1, 2, 5, 15, 51, ...). (End)
Binomial transform of A032357. - Philippe Deléham, Sep 17 2009
a(n) = number of rooted trees with n vertices in which each vertex has at most 2 children and in case a vertex has exactly one child, it is labeled left, middle or right. These are the hex trees of the Deutsch, Munarini, Rinaldi link. This interpretation yields the second MATHEMATICA recurrence below. - David Callan, Oct 14 2012
The left shift (1,3,10,36,...) of this sequence is the binomial transform of the left-shifted Catalan numbers (1,2,5,14,...). Example: 36 =1*14 + 3*5 + 3*2 + 1*1. - David Callan, Feb 01 2014
Number of Schroeder paths from (0,0) to (2n,0) with no level steps H=(2,0) at even level. Example: a(2)=3 because we have UUDD, UHD and UDUD. - José Luis Ramírez Ramírez, Apr 27 2015
This is the Riordan transform with the Riordan matrix A097805 (of the associated type) of the Catalan sequence A000108. See a Feb 17 2017 comment in A097805. - Wolfdieter Lang, Feb 17 2017
a(n) is the number of parking functions of size n avoiding the patterns 132 and 231. - Lara Pudwell, Apr 10 2023

Examples

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 36*x^4 + 137*x^5 + 543*x^6 + 2219*x^7 + 9285*x^8 + ...
		

References

  • J. Brunvoll, B. N. Cyvin, and S. J. Cyvin, Studies of some chemically relevant polygonal systems: mono-q-polyhexes, ACH Models in Chem., 133 (3) (1996), 277-298, Eq 14.
  • S. J. Cyvin, J. Brunvoll, G. Xiaofeng, and Z. Fuji, Number of perifusenes with one internal vertex, Rev. Roumaine Chem., 38(1) (1993), 65-78.
  • 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).

Crossrefs

First differences of A007317.
Row sums of triangle A104259.

Programs

  • Magma
    I:= [1,3]; [1] cat [n le 2 select I[n]  else ((6*n-3)*Self(n-1)-5*(n-2)*Self(n-2)) div (n+1): n in [1..30]]; // Vincenzo Librandi, Jun 15 2015
  • Maple
    t1 := series(1+ (1-3*x-(1-x)^(1/2)*(1-5*x)^(1/2))/(2*x), x, 50):
    A002212_list := len -> seq(coeff(t1,x,n),n=0..len): A002212_list(40);
    a[0] := 1: a[1] := 1: for n from 2 to 50 do a[n] := (3*(2*n-1)*a[n-1]-5*(n-2)*a[n-2])/(n+1) od: print(convert(a,list)); # Zerinvary Lajos, Jan 01 2007
    a := n -> `if`(n=0,1,simplify(GegenbauerC(n-1, -n, -3/2)/n)):
    seq(a(n), n=0..23); # Peter Luschny, May 09 2016
  • Mathematica
    InverseSeries[Series[(y)/(1+3*y+y^2), {y, 0, 24}], x] (* then A(x)=1+y(x) *) (* Len Smiley, Apr 14 2000 *)
    (* faster *)
    a[0]=1;a[1]=1;
    a[n_]/;n>=2 := a[n] = a[n-1] +  Sum[a[i]a[n-1-i],{i,0,n-1}];
    Table[a[n],{n,0,14}] (* See COMMENTS above, [David Callan, Oct 14 2012] *)
    (* fastest *)
    s[0]=s[1]=1;
    s[n_]/;n>=2 := s[n] = (3(2n-1)s[n-1]-5(n-2)s[n-2])/(n+1);
    Table[s[n],{n,0,14 }] (* See Deutsch, Munarini, Rinaldi link, [David Callan, Oct 14 2012] *)
    (* 2nd fastest *)
    a[n_] := Hypergeometric2F1[3/2, 1-n, 3, -4]; a[0]=1; Table[a[n], {n, 0, 14}]  (* Jean-François Alcover, May 16 2013 *)
    CoefficientList[Series[(1 - x - Sqrt[1 - 6x + 5x^2])/(2x), {x, 0, 20}], x] (* Nikolaos Pantelidis, Jan 30 2023 *)
  • Maxima
    makelist(sum(binomial(n,k)*binomial(n-k,k)*3^(n-2*k)/(k+1),k,0,n/2),n,0,24); /* for a(n+1) */ /* Emanuele Munarini, May 18 2011 */
    
  • PARI
    {a(n) = polcoeff( (1 - x - sqrt(1 - 6*x + 5*x^2 + x^2 * O(x^n))) / 2, n+1)};
    
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( serreverse( x / (1 + 3*x + x^2) + x * O(x^n)), n))}; /* Michael Somos */
    
  • PARI
    my(N=66,x='x+O('x^N)); Vec((1 - x - sqrt(1-6*x+5*x^2))/(2*x)) \\ Joerg Arndt, Jan 13 2024
    
  • Sage
    def A002212():
        x, y, n = 1, 1, 1
        while True:
            yield x
            n += 1
            x, y = y, ((6*n - 3)*y - (5*n - 10)*x) / (n + 1)
    a = A002212()
    [next(a) for i in range(24)]  # Peter Luschny, Oct 12 2013
    

Formula

a(0)=1, for n > 0: a(n) = Sum_{j=0..n-1} Sum_{i=0..j} a(i)*a(j-i). G.f.: A(x) = 1 + x*A(x)^2/(1-x). - Mario Catalani (mario.catalani(AT)unito.it), Jun 19 2003
a(n) = Sum_{i=ceiling((n-1)/2)..n-1} (3^(2i+1-n)*binomial(n, i)*binomial(i, n-i-1))/n. - Emeric Deutsch, Jul 23 2002
a(n) = Sum_{k=1..n} binomial(2k, k)*binomial(n-1, k-1)/(k+1), i.e., binomial transform of the Catalan numbers 1, 2, 5, 14, 42, ... (A000108). a(n) = Sum_{k=0..floor((n-1)/2)} 3^(n-1-2*k)*binomial(2k, k)*binomial(n-1, 2k)/(k+1). - Emeric Deutsch, Aug 05 2002
D-finite with recurrence: a(1)=1, a(n) = (3(2n-1)*a(n-1)-5(n-2)*a(n-2))/(n+1) for n > 1. - Emeric Deutsch, Dec 18 2002
a(n) is asymptotic to c*5^n/n^(3/2) with c=0.63.... - Benoit Cloitre, Jun 23 2003
In closed form, c = (1/2)*sqrt(5/Pi) = 0.63078313050504... - Vaclav Kotesovec, Oct 04 2012
Reversion of Sum_{n>0} a(n)x^n = -Sum_{n>0} A001906(n)(-x)^n.
G.f. A(x) satisfies xA(x)^2 + (1-x)(1-A(x)) = 0.
G.f.: (1 - x - sqrt(1 - 6x + 5x^2))/(2x). For n > 1, a(n) = 3*a(n-1) + Sum_{k=1..n-2} a(k)*a(n-k-1). - John W. Layman, Feb 22 2001
The Hankel transform of this sequence gives A001519 = 1, 2, 5, 13, 34, 89, ... E.g., Det([1, 1, 3, 10, 36; 1, 3, 10, 36, 137; 3, 10, 36, 137, 543; 10, 36, 137, 543, 2219; 36, 137, 543, 2219, 9285 ])= 34. - Philippe Deléham, Jan 25 2004
a(m+n+1) = Sum_{k>=0} A091965(m, k)*A091965(n, k) = A091965(m+n, 0). - Philippe Deléham, Sep 14 2005
a(n+1) = Sum_{k=0..n} 2^(n-k)*M(k)*binomial(n,k), where M(k) = A001006(k) is the k-th Motzkin number (from here it follows that a(n+1) and M(n) have the same parity). - Emeric Deutsch, May 10 2007
a(n+1) = Sum_{k=0..n} A097610(n,k)*3^k. - Philippe Deléham, Oct 02 2007
G.f.: 1/(1-x/(1-x-x/(1-x/(1-x-x/(1-x/(1-x-x/(1-... (continued fraction). - Paul Barry, May 16 2009
G.f.: (1-x)/(1-2x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-.... (continued fraction). - Paul Barry, Oct 17 2009
G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1-x) (continued fraction); more generally g.f. C(x/(1-x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) = -5^(1/2)/(10*(n+1)) * (5*hypergeom([1/2, n], [1], 4/5) -3*hypergeom([1/2, n+1], [1], 4/5)) (for n>0). - Mark van Hoeij, Nov 12 2009
For n >= 1, a(n) = (1/(2*Pi))*Integral_{x=1..5} x^(n-1)*sqrt((x-1)*(5-x)) dx. - Groux Roland, Mar 16 2011
a(n+1) = [x^n](1-x^2)(1+3*x+x^2)^n. - Emanuele Munarini, May 18 2011
From Gary W. Adamson, Jul 21 2011: (Start)
a(n) = upper left term in M^(n-1), M = an infinite square production matrix as follows (with 3,2,2,2,... as the main diagonal):
3, 1, 0, 0, 0, 0, ...
1, 2, 1, 0, 0, 0, ...
1, 1, 2, 1, 0, 0, ...
1, 1, 1, 2, 1, 0, ...
1, 1, 1, 1, 2, 0, ...
...
Alternatively, let M = the previous matrix but change the 3 to a 2. Then a(n) = sum of top row terms of M^(n-1). (End)
a(n) = hypergeometric([1-n,3/2],[3],-4), for n>0. - Peter Luschny, Aug 15 2012
a(n) = GegenbauerC(n-1, -n, -3/2)/n for n >= 1. - Peter Luschny, May 09 2016
E.g.f.: 1 + Integral (exp(3*x) * BesselI(1,2*x) / x) dx. - Ilya Gutkovskiy, Jun 01 2020
G.f.: 1 + x/G(0) with G(k) = (1 - 3*x - x^2/G(k+1)) (continued fraction). - Nikolaos Pantelidis, Dec 12 2022
From Peter Bala, Feb 03 2024: (Start)
G.f.: 1 + x/(1 - x) * c(x/(1 - x))^2 = 1 + x/(1 - 5*x) * c(-x/(1 - 5*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
a(n+1) = Sum_{k = 0..n} binomial(n, k)*Catalan(k+1).
a(n+1) = hypergeom([-n, 3/2], [3], -4).
a(n+1) = 5^n * Sum_{k = 0..n} (-5)^(-k)*binomial(n, k)*Catalan(k+1).
a(n+1) = 5^n * hypergeom([-n, 3/2], [3], 4/5). (End)

A007317 Binomial transform of Catalan numbers.

Original entry on oeis.org

1, 2, 5, 15, 51, 188, 731, 2950, 12235, 51822, 223191, 974427, 4302645, 19181100, 86211885, 390248055, 1777495635, 8140539950, 37463689775, 173164232965, 803539474345, 3741930523740, 17481709707825, 81912506777200, 384847173838501, 1812610804416698
Offset: 1

Views

Author

Keywords

Comments

Partial sums of A002212 (the restricted hexagonal polyominoes with n cells). Number of Schroeder paths (i.e., consisting of steps U=(1,1),D=(1,-1),H=(2,0) and never going below the x-axis) from (0,0) to (2n-2,0), with no peaks at even level. Example: a(3)=5 because among the six Schroeder paths from (0,0) to (4,0) only UUDD has a peak at an even level. - Emeric Deutsch, Dec 06 2003
Number of binary trees of weight n where leaves have positive integer weights. Non-commutative Non-associative version of partitions of n. - Michael Somos, May 23 2005
Appears also as the number of Euler trees with total weight n (associated with even switching class of matrices of order 2n). - David Garber, Sep 19 2005
Number of symmetric hex trees with 2n-1 edges; also number of symmetric hex trees with 2n-2 edges. A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a median child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read reference). A hex tree is symmetric if it is identical with its reflection in a bisector through the root. - Emeric Deutsch, Dec 19 2006
The Hankel transform of [1, 2, 5, 15, 51, 188, ...] is [1, 1, 1, 1, 1, ...], see A000012 ; the Hankel transform of [2, 5, 15, 51, 188, 731, ...] is [2, 5, 13, 34, 89, ...], see A001519. - Philippe Deléham, Dec 19 2006
a(n) = number of 321-avoiding partitions of [n]. A partition is 321-avoiding if the permutation obtained from its canonical form (entries in each block listed in increasing order and blocks listed in increasing order of their first entries) is 321-avoiding. For example, the only partition of [5] that fails to be 321-avoiding is 15/24/3 because the entries 5,4,3 in the permutation 15243 form a 321 pattern. - David Callan, Jul 22 2008
The sequence 1,1,2,5,15,51,188,... has Hankel transform A001519. - Paul Barry, Jan 13 2009
From Gary W. Adamson, May 17 2009: (Start)
Equals INVERT transform of A033321: (1, 1, 2, 6, 21, 79, 311, ...).
Equals INVERTi transform of A002212: (1, 3, 10, 36, 137, ...).
Convolved with A026378, (1, 4, 17, 75, 339, ...) = A026376: (1, 6, 30, 144, ...)
(End)
a(n) is the number of vertices of the composihedron CK(n). The composihedra are a sequence of convex polytopes used to define maps of certain homotopy H-spaces. They are cellular quotients of the multiplihedra and cellular covers of the cubes. - Stefan Forcey (sforcey(AT)gmail.com), Dec 17 2009
a(n) is the number of Motzkin paths of length n-1 in which the (1,0)-steps at level 0 come in 2 colors and those at a higher level come in 3 colors. Example: a(4)=15 because we have 2^3 = 8 paths of shape UHD, 2 paths of shape HUD, 2 paths of shape UDH, and 3 paths of shape UHD; here U=(1,1), H=(1,0), and D=(1,-1). - Emeric Deutsch, May 02 2011
REVERT transform of (1, 2, -3, 5, -8, 13, -21, 34, ... ) where the entries are Fibonacci numbers, A000045. Equivalently, coefficients in the series reversion of x(1-x)/(1+x-x^2). This means that the substitution of the gf (1-x-(1-6x+5x^2)^(1/2))/(2(1-x)) for x in x(1-x)/(1+x-x^2) will simplify to x. - David Callan, Nov 11 2012
The number of plane trees with nodes that have positive integer weights and whose total weight is n. - Brad R. Jones, Jun 12 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)] = (x-x^2) / (1 + x - x^2).
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1].
(End)
Starting with offset 0, a(n) is also the number of Schröder paths of semilength n avoiding UH (an up step directly followed by a long horizontal step). Example: a(2)=5 because among the six possible Schröder paths of semilength 2 only UHD contains UH. - Valerie Roitner, Jul 23 2020

Examples

			a(3)=5 since {3, (1+2), (1+(1+1)), (2+1), ((1+1)+1)} are the five weighted binary trees of weight 3.
G.f. = x + 2*x^2 + 5*x^3 + 15*x^4 + 51*x^5 + 188*x^6 + 731*x^7 + 2950*x^8 + 12235*x^9 + ... _Michael Somos_, Jan 17 2018
		

References

  • J. Brunvoll et al., Studies of some chemically relevant polygonal systems: mono-q-polyhexes, ACH Models in Chem., 133 (3) (1996), 277-298, Eq. 15.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A181768 for another version. - N. J. A. Sloane, Nov 12 2010
First column of triangle A104259. Row sums of absolute values of A091699.
Number of vertices of multiplihedron A121988.
m-th binomial transform of the Catalan numbers: A126930 (m = -2), A005043 (m = -1), A000108 (m = 0), A064613 (m = 2), A104455 (m = 3), A104498 (m = 4) and A154623 (m = 5).

Programs

  • Maple
    G := (1-sqrt(1-4*z/(1-z)))*1/2: Gser := series(G, z = 0, 30): seq(coeff(Gser, z, n), n = 1 .. 26); # Emeric Deutsch, Aug 12 2007
    seq(round(evalf(JacobiP(n-1,1,-n-1/2,9)/n,99)),n=1..25); # Peter Luschny, Sep 23 2014
  • Mathematica
    Rest@ CoefficientList[ InverseSeries[ Series[(y - y^2)/(1 + y - y^2), {y, 0, 26}], x], x] (* then A(x)=y(x); note that InverseSeries[Series[y-y^2, {y, 0, 24}], x] produces A000108(x) *) (* Len Smiley, Apr 10 2000 *)
    Range[0, 25]! CoefficientList[ Series[ Exp[ 3x] (BesselI[0, 2x] - BesselI[1, 2x]), {x, 0, 25}], x] (* Robert G. Wilson v, Apr 15 2011 *)
    a[n_] := Sum[ Binomial[n, k]*CatalanNumber[k], {k, 0, n}]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Aug 07 2012 *)
    Rest[CoefficientList[Series[3/2 - (1/2) Sqrt[(1 - 5 x)/(1 - x)], {x, 0, 40}], x]] (* Vincenzo Librandi, Nov 03 2014 *)
    Table[Hypergeometric2F1[1/2, -n+1, 2, -4], {n, 1, 30}] (* Vaclav Kotesovec, May 12 2022 *)
  • PARI
    {a(n) = my(A); if( n<2, n>0, A=vector(n); for(j=1,n, A[j] = 1 + sum(k=1,j-1, A[k]*A[j-k])); A[n])}; /* Michael Somos, May 23 2005 */
    
  • PARI
    {a(n) = if( n<1, 0, polcoeff( serreverse( (x - x^2) / (1 + x - x^2) + x * O(x^n)), n))}; /* Michael Somos, May 23 2005 */
    
  • PARI
    /* Offset = 0: */ {a(n)=local(A=1+x);for(i=1,n, A=sum(m=0,n, x^m*sum(k=0,m,A^k)+x*O(x^n))); polcoeff(A,n)} \\ Paul D. Hanna

Formula

(n+2)*a(n+2) = (6n+4)*a(n+1) - 5n*a(n).
G.f.: 3/2-(1/2)*sqrt((1-5*x)/(1-x)) [Gessel-Kim]. - N. J. A. Sloane, Jul 05 2014
G.f. for sequence doubled: (1/(2*x))*(1+x-(1-x)^(-1)*(1-x^2)^(1/2)*(1-5*x^2)^(1/2)).
a(n) = hypergeom([1/2, -n], [2], -4), n=0, 1, 2...; Integral representation as n-th moment of a positive function on a finite interval of the positive half-axis: a(n)=int(x^n*sqrt((5-x)/(x-1))/(2*Pi), x=1..5), n=0, 1, 2... This representation is unique. - Karol A. Penson, Sep 24 2001
a(1)=1, a(n)=1+sum(i=1, n-1, a(i)*a(n-i)). - Benoit Cloitre, Mar 16 2004
a(n) = Sum_{k=0..n} (-1)^k*3^(n-k)*binomial(n, k)*binomial(k, floor(k/2)) [offset 0]. - Paul Barry, Jan 27 2005
G.f. A(x) satisfies 0=f(x, A(x)) where f(x, y)=x-(1-x)(y-y^2). - Michael Somos, May 23 2005
G.f. A(x) satisfies 0=f(x, A(x), A(A(x))) where f(x, y, z)=x(z-z^2)+(x-1)y^2 . - Michael Somos, May 23 2005
G.f. (for offset 0): (-1+x+(1-6*x+5*x^2)^(1/2))/(2*(-x+x^2)).
G.f. =z*c(z/(1-z))/(1-z) = 1/2 - (1/2)sqrt(1-4z/(1-z)), where c(z)=(1-sqrt(1-4z))/(2z) is the Catalan function (follows from Michael Somos' first comment). - Emeric Deutsch, Aug 12 2007
G.f.: 1/(1-2x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-3x-x^2/(1-.... (continued fraction). - Paul Barry, Apr 19 2009
a(n) = Sum_{k, 0<=k<=n} A091965(n,k)*(-1)^k. - Philippe Deléham, Nov 28 2009
E.g.f.: exp(3x)*(I_0(2x)-I_1(2x)), where I_k(x) is a modified Bessel function of the first kind. - Emanuele Munarini, Apr 15 2011
If we prefix sequence with an additional term a(0)=1, g.f. is (3-3*x-sqrt(1-6*x+5*x^2))/(2*(1-x)). [See Kim, 2011] - N. J. A. Sloane, May 13 2011
From Gary W. Adamson, Jul 21 2011: (Start)
a(n) = upper left term in M^(n-1), M = an infinite square production matrix as follows:
2, 1, 0, 0, 0, 0, ...
1, 2, 1, 0, 0, 0, ...
1, 1, 2, 1, 0, 0, ...
1, 1, 1, 2, 1, 0, ...
1, 1, 1, 1, 2, 1, ...
1, 1, 1, 1, 1, 2, ...
... (End)
G.f. satisfies: A(x) = Sum_{n>=0} x^n * (1 - A(x)^(n+1))/(1 - A(x)); offset=0. - Paul D. Hanna, Nov 07 2011
G.f.: 1/x - 1/x/Q(0), where Q(k)= 1 + (4*k+1)*x/((1-x)*(k+1) - x*(1-x)*(2*k+2)*(4*k+3)/(x*(8*k+6)+(2*k+3)*(1-x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
G.f.: (1-x - (1-5*x)*G(0))/(2*x*(1-x)), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2)*(1-x) - 2*x*(1-x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1-x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 25 2013
Asymptotics (for offset 0): a(n) ~ 5^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 28 2013
G.f.: G(0)/(1-x), where G(k) = 1 + (4*k+1)*x/((k+1)*(1-x) - 2*x*(1-x)*(k+1)*(4*k+3)/(2*x*(4*k+3) + (2*k+3)*(1-x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 29 2014
a(n) = JacobiP(n-1,1,-n-1/2,9)/n. - Peter Luschny, Sep 23 2014
0 = +a(n)*(+25*a(n+1) -50*a(n+2) +15*a(n+3)) +a(n+1)*(-10*a(n+1) +31*a(n+2) -14*a(n+3)) +a(n+2)*(+2*a(n+2) +a(n+3)) for all n in Z. - Michael Somos, Jan 17 2018
a(n+1) = (2/Pi) * Integral_{x = -1..1} (m + 4*x^2)^n*sqrt(1 - x^2) dx at m = 1. In general, the integral, qua sequence in n, gives the m-th binomial transform of the Catalan numbers. - Peter Bala, Jan 26 2020

A035263 Trajectory of 1 under the morphism 0 -> 11, 1 -> 10; parity of 2-adic valuation of 2n: a(n) = A000035(A001511(n)).

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1
Offset: 1

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Author

Keywords

Comments

First Feigenbaum symbolic (or period-doubling) sequence, corresponding to the accumulation point of the 2^{k} cycles through successive bifurcations.
To construct the sequence: start with 1 and concatenate: 1,1, then change the last term (1->0; 0->1) gives: 1,0. Concatenate those 2 terms: 1,0,1,0, change the last term: 1,0,1,1. Concatenate those 4 terms: 1,0,1,1,1,0,1,1 change the last term: 1,0,1,1,1,0,1,0, etc. - Benoit Cloitre, Dec 17 2002
Let T denote the present sequence. Here is another way to construct T. Start with the sequence S = 1,0,1,,1,0,1,,1,0,1,,1,0,1,,... and fill in the successive holes with the successive terms of the sequence T (from paper by Allouche et al.). - Emeric Deutsch, Jan 08 2003 [Note that if we fill in the holes with the terms of S itself, we get A141260. - N. J. A. Sloane, Jan 14 2009]
From N. J. A. Sloane, Feb 27 2009: (Start)
In more detail: define S to be 1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1,0,1___...
If we fill the holes with S we get A141260:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........0.......1.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.0.1.... = A141260.
But instead, if we define T recursively by filling the holes in S with the terms of T itself, we get A035263:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........1.......0.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.1.1.0.1.0.1..0..1.1.1..0..1.0.1.. = A035263. (End)
Characteristic function of A003159, i.e., A035263(n)=1 if n is in A003159 and A035263(n)=0 otherwise (from paper by Allouche et al.). - Emeric Deutsch, Jan 15 2003
This is the sequence of R (=1), L (=0) moves in the Towers of Hanoi puzzle: R, L, R, R, R, L, R, L, R, L, R, R, R, ... - Gary W. Adamson, Sep 21 2003
Manfred Schroeder, p. 279 states, "... the kneading sequences for unimodal maps in the binary notation, 0, 1, 0, 1, 1, 1, 0, 1..., are obtained from the Morse-Thue sequence by taking sums mod 2 of adjacent elements." On p. 278, in the chapter "Self-Similarity in the Logistic Parabola", he writes, "Is there a closer connection between the Morse-Thue sequence and the symbolic dynamics of the superstable orbits? There is indeed. To see this, let us replace R by 1 and C and L by 0." - Gary W. Adamson, Sep 21 2003
Partial sums modulo 2 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... . - Philippe Deléham, Jan 02 2004
Parity of A007913, A065882 and A065883. - Philippe Deléham, Mar 28 2004
The length of n-th run of 1's in this sequence is A080426(n). - Philippe Deléham, Apr 19 2004
Also parity of A005043, A005773, A026378, A104455, A117641. - Philippe Deléham, Apr 28 2007
Equals parity of the Towers of Hanoi, or ruler sequence (A001511), where the Towers of Hanoi sequence (1, 2, 1, 3, 1, 2, 1, 4, ...) denotes the disc moved, labeled (1, 2, 3, ...) starting from the top; and the parity of (1, 2, 1, 3, ...) denotes the direction of the move, CW or CCW. The frequency of CW moves converges to 2/3. - Gary W. Adamson, May 11 2007
A conjectured identity relating to the partition sequence, A000041: p(x) = A(x) * A(x^2) when A(x) = the Euler transform of A035263 = polcoeff A174065: (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...). - Gary W. Adamson, Mar 21 2010
a(n) is 1 if the number of trailing zeros in the binary representation of n is even. - Ralf Stephan, Aug 22 2013
From Gary W. Adamson, Mar 25 2015: (Start)
A conjectured identity relating to the partition sequence, A000041 as polcoeff p(x); A003159, and its characteristic function A035263: (1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); and A036554 indicating n-th terms with zeros in A035263: (2, 6, 8, 10, 14, 18, 22, ...).
The conjecture states that p(x) = A(x) = A(x^2) when A(x) = polcoeffA174065 = the Euler transform of A035263 = 1/(1-x)*(1-x^3)*(1-x^4)*(1-x^5)*... = (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...) and the aerated variant = the Euler transform of the complement of A035263: 1/(1-x^2)*(1-x^6)*(1-x^8)*... = (1 + x^2 + x^4 + 2x^6 + 3x^8 + 4x^10 + ...).
(End)
The conjecture above was proved by Jean-Paul Allouche on Dec 21 2013.
Regarded as a column vector, this sequence is the product of A047999 (Sierpinski's gasket) regarded as an infinite lower triangular matrix and A036497 (the Fredholm-Rueppel sequence) where the 1's have alternating signs, 1, -1, 0, 1, 0, 0, 0, -1, .... - Gary W. Adamson, Jun 02 2021
The numbers of 1's through n (A050292) can be determined by starting with the binary (say for 19 = 1 0 0 1 1) and writing: next term is twice current term if 0, otherwise twice plus 1. The result is 1, 2, 4, 9, 19. Take the difference row, = 1, 1, 2, 5, 10; and add the odd-indexed terms from the right: 5, 4, 3, 2, 1 = 10 + 2 + 1 = 13. The algorithm is the basis for determining the disc configurations in the tower of Hanoi game, as shown in the Jul 24 2021 comment of A060572. - Gary W. Adamson, Jul 28 2021

References

  • Karamanos, Kostas. "From symbolic dynamics to a digital approach." International Journal of Bifurcation and Chaos 11.06 (2001): 1683-1694. (Full version. See p. 1685)
  • Karamanos, K. (2000). From symbolic dynamics to a digital approach: chaos and transcendence. In Michel Planat (Ed.), Noise, Oscillators and Algebraic Randomness (Lecture Notes in Physics, pp. 357-371). Springer, Berlin, Heidelberg. (Short version. See p. 359)
  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W. H. Freeman, 1991
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 892, column 2, Note on p. 84, part (a).

Crossrefs

Parity of A001511. Anti-parity of A007814.
Absolute values of first differences of A010060. Apart from signs, same as A029883. Essentially the same as A056832.
Swapping 0 and 1 gives A096268.
Cf. A033485, A050292 (partial sums), A089608, A088172, A019300, A039982, A073675, A121701, A141260, A000041, A174065, A220466, A154269 (Mobius transform).
Limit of A317957(n) for large n.

Programs

  • Haskell
    import Data.Bits (xor)
    a035263 n = a035263_list !! (n-1)
    a035263_list = zipWith xor a010060_list $ tail a010060_list
    -- Reinhard Zumkeller, Mar 01 2012
    
  • Maple
    nmax:=105: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (p+1) mod 2 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 07 2013
    A035263 := n -> 1 - padic[ordp](n, 2) mod 2:
    seq(A035263(n), n=1..105); # Peter Luschny, Oct 02 2018
  • Mathematica
    a[n_] := a[n] = If[ EvenQ[n], 1 - a[n/2], 1]; Table[ a[n], {n, 1, 105}] (* Or *)
    Rest[ CoefficientList[ Series[ Sum[ x^(2^k)/(1 + (-1)^k*x^(2^k)), {k, 0, 20}], {x, 0, 105}], x]]
    f[1] := True; f[x_] := Xor[f[x - 1], f[Floor[x/2]]]; a[x_] := Boole[f[x]] (* Ben Branman, Oct 04 2010 *)
    a[n_] := If[n == 0, 0, 1 - Mod[ IntegerExponent[n, 2], 2]]; (* Jean-François Alcover, Jul 19 2013, after Michael Somos *)
    Nest[ Flatten[# /. {0 -> {1, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v, Jul 23 2014 *)
    SubstitutionSystem[{0->{1,1},1->{1,0}},1,{7}][[1]] (* Harvey P. Dale, Jun 06 2022 *)
  • PARI
    {a(n) = if( n==0, 0, 1 - valuation(n, 2)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); subst( Pol(binary(n)) - Pol(binary(n-1)), x, 1)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); direuler(p=2, n, 1 / (1 - X^((p<3) + 1)))[n])}; /* Michael Somos, Sep 04 2006 */
    
  • Python
    def A035263(n): return (n&-n).bit_length()&1 # Chai Wah Wu, Jan 09 2023
  • Scheme
    (define (A035263 n) (let loop ((n n) (i 1)) (cond ((odd? n) (modulo i 2)) (else (loop (/ n 2) (+ 1 i)))))) ;; (Use mod instead of modulo in R6RS) Antti Karttunen, Sep 11 2017
    

Formula

Absolute values of first differences (A029883) of Thue-Morse sequence (A001285 or A010060). Self-similar under 10->1 and 11->0.
Series expansion: (1/x) * Sum_{i>=0} (-1)^(i+1)*x^(2^i)/(x^(2^i)-1). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 17 2003
a(n) = Sum_{k>=0} (-1)^k*(floor((n+1)/2^k)-floor(n/2^k)). - Benoit Cloitre, Jun 03 2003
Another g.f.: Sum_{k>=0} x^(2^k)/(1+(-1)^k*x^(2^k)). - Ralf Stephan, Jun 13 2003
a(2*n) = 1-a(n), a(2*n+1) = 1. - Ralf Stephan, Jun 13 2003
a(n) = parity of A033485(n). - Philippe Deléham, Aug 13 2003
Equals A088172 mod 2, where A088172 = 1, 2, 3, 7, 13, 26, 53, 106, 211, 422, 845, ... (first differences of A019300). - Gary W. Adamson, Sep 21 2003
a(n) = a(n-1) - (-1)^n*a(floor(n/2)). - Benoit Cloitre, Dec 02 2003
a(1) = 1 and a(n) = abs(a(n-1) - a(floor(n/2))). - Benoit Cloitre, Dec 02 2003
a(n) = 1 - A096268(n+1); A050292 gives partial sums. - Reinhard Zumkeller, Aug 16 2006
Multiplicative with a(2^k) = 1 - (k mod 2), a(p^k) = 1, p > 2. Dirichlet g.f.: Product_{n = 4 or an odd prime} (1/(1-1/n^s)). - Christian G. Bower, May 18 2005
a(-n) = a(n). a(0)=0. - Michael Somos, Sep 04 2006
Dirichlet g.f.: zeta(s)*2^s/(2^s+1). - Ralf Stephan, Jun 17 2007
a(n+1) = a(n) XOR a(ceiling(n/2)), a(1) = 1. - Reinhard Zumkeller, Jun 11 2009
Let D(x) be the generating function, then D(x) + D(x^2) == x/(1-x). - Joerg Arndt, May 11 2010
a(n) = A010060(n) XOR A010060(n+1); a(A079523(n)) = 0; a(A121539(n)) = 1. - Reinhard Zumkeller, Mar 01 2012
a((2*n-1)*2^p) = (p+1) mod 2, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 07 2013
a(n) = A000035(A001511(n)). - Omar E. Pol, Oct 29 2013
a(n) = 2-A056832(n) = (5-A089608(n))/4. - Antti Karttunen, Sep 11 2017, after Benoit Cloitre
For n >= 0, a(n+1) = M(2n) mod 2 where M(n) is the Motzkin number A001006 (see Deutsch and Sagan 2006 link). - David Callan, Oct 02 2018
a(n) = A038712(n) mod 3. - Kevin Ryde, Jul 11 2019
Given any n in the form (k * 2^m, k odd), extract k and m. Categorize the results into two outcomes of (k, m, even or odd). If (k, m) is (odd, even) substitute 1. If (odd, odd), denote the result 0. Example: 5 = (5 * 2^0), (odd, even, = 1). (6 = 3 * 2^1), (odd, odd, = 0). - Gary W. Adamson, Jun 23 2021

Extensions

Alternative description added to the name by Antti Karttunen, Sep 11 2017

A091965 Triangle read by rows: T(n,k) = number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and three types of steps H=(1,0) (left factors of 3-Motzkin steps).

Original entry on oeis.org

1, 3, 1, 10, 6, 1, 36, 29, 9, 1, 137, 132, 57, 12, 1, 543, 590, 315, 94, 15, 1, 2219, 2628, 1629, 612, 140, 18, 1, 9285, 11732, 8127, 3605, 1050, 195, 21, 1, 39587, 52608, 39718, 19992, 6950, 1656, 259, 24, 1, 171369, 237129, 191754, 106644, 42498, 12177, 2457
Offset: 0

Views

Author

Emeric Deutsch, Mar 13 2004

Keywords

Comments

T(n,0) = A002212(n+1), T(n,1) = A045445(n+1); row sums give A026378.
The inverse is A207815. - Gary W. Adamson, Dec 17 2006 [corrected by Philippe Deléham, Feb 22 2012]
Reversal of A084536. - Philippe Deléham, Mar 23 2007
Triangle T(n,k), 0 <= k <= n, read by rows given by T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 3*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 27 2007
This triangle belongs to the family of triangles defined by T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k >= 1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
5^n = (n-th row terms) dot (first n+1 terms in (1,2,3,...)). Example for row 4: 5^4 = 625 = (137, 132, 57, 12, 1) dot (1, 2, 3, 4, 5) = (137 + 264 + 171 + 48 + 5) = 625. - Gary W. Adamson, Jun 15 2011
Riordan array ((1-3*x-sqrt(1-6*x+5*x^2))/(2*x^2), (1-3*x-sqrt(1-6*x+5*x^2))/(2*x)). - Philippe Deléham, Feb 19 2012

Examples

			Triangle begins:
     1;
     3,    1;
    10,    6,    1;
    36,   29,    9,    1;
   137,  132,   57,   12,    1;
   543,  590,  315,   94,   15,    1;
  2219, 2628, 1629,  612,  140,   18,    1;
T(3,1)=29 because we have UDU, UUD, 9 HHU paths, 9 HUH paths and 9 UHH paths.
Production matrix begins
  3, 1;
  1, 3, 1;
  0, 1, 3, 1;
  0, 0, 1, 3, 1;
  0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 1;
- _Philippe Deléham_, Nov 07 2011
		

References

  • A. Nkwanta, Lattice paths and RNA secondary structures, DIMACS Series in Discrete Math. and Theoretical Computer Science, 34, 1997, 137-147.

Crossrefs

Programs

  • Mathematica
    nmax = 9; t[n_, k_] := ((k+1)*n!*Hypergeometric2F1[k+3/2, k-n, 2k+3, -4]) / ((k+1)!*(n-k)!); Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 14 2011, after Vladimir Kruchinin *)
    T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0,
    T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]];
    Table[T[n, k, 3, 3], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, May 22 2017 *)
  • Maxima
    T(n,k):=(k+1)*sum((binomial(2*(m+1),m-k)*binomial(n,m))/(m+1),m,k,n); /* Vladimir Kruchinin, Oct 08 2011 */
    
  • Sage
    @CachedFunction
    def A091965(n,k):
        if n==0 and k==0: return 1
        if k<0 or k>n: return 0
        if k==0: return 3*A091965(n-1,0)+A091965(n-1,1)
        return A091965(n-1,k-1)+3*A091965(n-1,k)+A091965(n-1,k+1)
    for n in (0..7):
        [A091965(n,k) for k in (0..n)] # Peter Luschny, Nov 05 2012

Formula

G.f.: G = 2/(1 - 3*z - 2*t*z + sqrt(1-6*z+5*z^2)). Alternatively, G = M/(1 - t*z*M), where M = 1 + 3*z*M + z^2*M^2.
Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0) = A002212(m+n+1). - Philippe Deléham, Sep 14 2005
The triangle may also be generated from M^n * [1,0,0,0,...], where M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [3,3,3,...] in the main diagonal. - Gary W. Adamson, Dec 17 2006
Sum_{k=0..n} T(n,k)*(k+1) = 5^n. - Philippe Deléham, Mar 27 2007
Sum_{k=0..n} T(n,k)*x^k = A117641(n), A033321(n), A007317(n), A002212(n+1), A026378(n+1) for x = -3, -2, -1, 0, 1 respectively. - Philippe Deléham, Nov 28 2009
T(n,k) = (k+1)*Sum_{m=k..n} binomial(2*(m+1),m-k)*binomial(n,m)/(m+1). - Vladimir Kruchinin, Oct 08 2011
The n-th row polynomial R(n,x) equals the n-th degree Taylor polynomial of the function (1 - x^2)*(1 + 3*x + x^2)^n expanded about the point x = 0. - Peter Bala, Sep 06 2022

A091867 Triangle read by rows: T(n,k) = number of Dyck paths of semilength n having k peaks at odd height.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 3, 4, 6, 0, 1, 6, 15, 10, 10, 0, 1, 15, 36, 45, 20, 15, 0, 1, 36, 105, 126, 105, 35, 21, 0, 1, 91, 288, 420, 336, 210, 56, 28, 0, 1, 232, 819, 1296, 1260, 756, 378, 84, 36, 0, 1, 603, 2320, 4095, 4320, 3150, 1512, 630, 120, 45, 0, 1, 1585, 6633, 12760, 15015, 11880, 6930, 2772, 990, 165, 55, 0, 1
Offset: 0

Views

Author

Emeric Deutsch, Mar 10 2004

Keywords

Comments

Number of ordered trees with n edges having k leaves at odd height. Row sums are the Catalan numbers (A000108). T(n,0)=A005043(n). Sum_{k=0..n} k*T(n,k) = binomial(2n-2,n-1).
T(n,k)=number of Dyck paths of semilength n and having k ascents of length 1 (an ascent is a maximal string of consecutive up steps). Example: T(4,2)=6 because we have UdUduud, UduuddUd, uuddUdUd, uudUdUdd, UduudUdd and uudUddUd (the ascents of length 1 are indicated by U instead of u).
T(n,k) is the number of Łukasiewicz paths of length n having k level steps (i.e., (1,0)). A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1) (see R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps). Example: T(4,1)=4 because we have HU(2)DD, U(2)HDD, U(2)DHD and U(2)DDH, where H=(1,0), U(1,1), U(2)=(1,2) and D=(1,-1). - Emeric Deutsch, Jan 06 2005
T(n,k) = number of noncrossing partitions of [n] containing k singleton blocks. Also, T(n,k) = number of noncrossing partitions of [n] containing k adjacencies. An adjacency is an occurrence of 2 consecutive integers in the same block (here 1 and n are considered consecutive). In fact, the statistics # singletons and # adjacencies have a symmetric joint distribution.
Exponential Riordan array [e^x*(Bessel_I(0,2x)-Bessel_I(1,2x)),x]. - Paul Barry, Mar 03 2011
T(n,k) is the number of ordered trees having n edges and exactly k nodes with one child. - Geoffrey Critzer, Feb 25 2013
From Tom Copeland, Nov 04 2014: (Start)
Summing the coeff. of the partitions in A134264 for a Lagrange inversion formula (see also A249548) containing (h_1)^k = (1')^k gives this triangle, so this array's o.g.f. H(x,t) = x + t * x^2 + (1 + t^2) * x^3 ... is the inverse of the o.g.f. of A104597 with a sign change, i.e., H^(-1)(x,t) = (x-x^2) / [1 + (t-1)(x-x^2)] = Cinv(x)/[1 + (t-1)Cinv(x)] = P[Cinv(x),t-1] where Cinv(x)= x * (1-x) is the inverse of C(x) = [1-sqrt(1-4*x)]/2, an o.g.f. for the Catalan numbers A000108, and P(x,t) = x/(1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x/(1-t*x). Therefore,
O.g.f.: H(x,t) = C[Pinv(x,t-1)] = C[P(x,1-t)] = C[x/(1-(t-1)x)] = {1-sqrt[1-4*x/(1-(t-1)x)]}/2 (for A091867). Reprising,
Inverse O.g.f.: H^(-1)(x,t) = x*(1-x) / [1 + (t-1)x(1-x)] = P[Cinv(x),t-1].
From general arguments in A134264, the row polynomials are an Appell sequence with lowering operator d/dt, having the umbral property (p(.,t)+a)^n=p(n,t+a) with e.g.f. = e^(x*t)/w(x), where 1/w(x)= e.g.f. of first column for the Motzkin numbers in A005043. (Mislabeled argument corrected on Jan 31 2016.)
Cf. A124644 (t-shifted polynomials), A026378 (t=-4), A001700 (t=-3), A005773 (t=-2), A126930 (t=-1) and A210736 (t=-1, a(0)=0, unsigned), A005043 (t=0), A000108 (t=1), A007317 (t=2), A064613 (t=3), A104455 (t=4), A030528 (for inverses).
(End)
The sequence of binomial transforms A126930, A005043, A000108, ... in the above comment appears in A126930 and the link therein to a paper by F. Fite et al. on page 42. - Tom Copeland, Jul 23 2016

Examples

			T(4,2)=6 because we have (ud)uu(ud)dd, uu(ud)dd(ud), uu(ud)(ud)dd, (ud)(ud)uudd, (ud)uudd(ud) and uudd(ud)(ud) (here u=(1,1), d=(1,-1) and the peaks at odd height are shown between parentheses).
Triangle begins:
   1;
   0,   1;
   1,   0,   1;
   1,   3,   0,   1;
   3,   4,   6,   0,  1;
   6,  15,  10,  10,  0,  1;
  15,  36,  45,  20, 15,  0, 1;
  36, 105, 126, 105, 35, 21, 0, 1;
  ...
		

References

  • R. Sedgewick and P. Flajolet, Analysis of Algorithms, Addison and Wesley, 1996, page 254 (first edition)

Crossrefs

Programs

  • Maple
    T := proc(n,k) if k>n then 0 elif k=n then 1 else (binomial(n+1,k)/(n+1))*sum(binomial(n+1-k,j)*binomial(n-k-j-1,j-1),j=1..floor((n-k)/2)) fi end: seq(seq(T(n,k),k=0..n),n=0..12);
    T := (n,k) -> (-1)^(n+k)*binomial(n,k)*hypergeom([-n+k,1/2],[2],4): seq(seq(simplify(T(n, k)), k=0..n), n=0..10); # Peter Luschny, Jul 27 2016
    # alternative Maple program:
    b:= proc(x, y, t) option remember; expand(`if`(x=0, 1,
          `if`(y>0, b(x-1, y-1, 0)*z^irem(t*y, 2), 0)+
          `if`(y (p-> seq(coeff(p, z, i), i=0..n))(b(2*n, 0$2)):
    seq(T(n), n=0..16);  # Alois P. Heinz, May 12 2017
  • Mathematica
    nn=10;cy = ( 1 + x - x y - ( -4x(1+x-x y) + (-1 -x + x y)^2)^(1/2))/(2(1+x-x y)); Drop[CoefficientList[Series[cy,{x,0,nn}],{x,y}],1]//Grid  (* Geoffrey Critzer, Feb 25 2013 *)
    Table[Which[k == n, 1, k > n, 0, True, (Binomial[n + 1, k]/(n + 1)) Sum[Binomial[n + 1 - k, j] Binomial[n - k - j - 1, j - 1], {j, Floor[(n - k)/2]}]], {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 25 2016 *)

Formula

T(n, k) = [binomial(n+1, k)/(n+1)]*Sum_{j=1..floor((n-k)/2)} binomial(n+1-k, j)*binomial(n-k-j-1, j-1) for kn. G.f.=G=G(t, z) satisfies z(1+z-tz)G^2-(1+z-tz)G+1=0. T(n, k)=r(n-k)*binomial(n, k), where r(n)=A005043(n) are the Riordan numbers.
G.f.: 1/(1-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-... (continued fraction). - Paul Barry, Aug 03 2009
Sum_{k=0..n} T(n,k)*x^k = A126930(n), A005043(n), A000108(n), A007317(n), A064613(n), A104455(n) for x = -1,0,1,2,3,4 respectively. - Philippe Deléham, Dec 03 2009
Sum_{k=0..n} (-1)^(n-k)*T(n,k)*x^k = A168491(n), A099323(n+1), A001405(n), A005773(n+1), A001700(n), A026378(n+1), A005573(n), A122898(n) for x = -1, 0, 1, 2, 3, 4, 5, 6 respectively. - Philippe Deléham, Dec 03 2009
E.g.f.: e^(x+xy)*(Bessel_I(0,2x)-Bessel_I(1,2x)). - Paul Barry, Mar 10 2010
From Tom Copeland, Nov 06 2014: (Start)
O.g.f.: H(x,t) = {1-sqrt[1-4x/(1-(t-1)x)]}/2 (shifted index, as given in Copeland's comment, see comp. inverse there).
H(x,t)= x / [1-(C.+(t-1))x] = Sum_{n>=1} (C.+ (t-1))^(n-1)*x^n umbrally, e.g., (a.+b.)^2 = a_0*b_2 + 2 a_1*b1_+ a_0*b_2, where (C.)^n = C_n are the Catalan numbers (1,1,2,5,14,..) of A000108.
This shows directly that the lowering operator for the polynomials is D=d/dt, i.e., D p(n,t)= D(C. + (t-1))^n = n * (C. + (t-1))^(n-1) = n*p(n-1,t), so that the polynomials form an Appell sequence, and that p(n,0) gives a Motzkin sum, or Riordan, number A005043.
(End)
T(n,k) = (-1)^(n+k)*binomial(n,k)*hypergeom([k-n,1/2],[2],4). - Peter Luschny, Jul 27 2016

A126791 Binomial matrix applied to A111418.

Original entry on oeis.org

1, 4, 1, 17, 7, 1, 75, 39, 10, 1, 339, 202, 70, 13, 1, 1558, 1015, 425, 110, 16, 1, 7247, 5028, 2400, 771, 159, 19, 1, 34016, 24731, 12999, 4872, 1267, 217, 22, 1, 160795, 121208, 68600, 28882, 8890, 1940, 284, 25, 1, 764388, 593019, 355890, 164136
Offset: 0

Views

Author

Philippe Deléham, Mar 14 2007

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 4*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + T(n-1,k+1) for k >= 1.
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k >= 1. Other triangles arise from choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
From R. J. Mathar, Mar 12 2013: (Start)
The matrix inverse starts
1;
-4, 1;
11, -7, 1;
-29, 31, -10, 1;
76, -115, 60, -13, 1;
-199, 390, -285, 98, -16, 1;
521, -1254, 1185, -566, 145, -19, 1;
-1364, 3893, -4524, 2785, -985, 201, -22, 1; ... (End)

Examples

			Triangle begins:
      1;
      4,     1;
     17,     7,     1;
     75,    39,    10,    1;
    339,   202,    70,   13,    1;
   1558,  1015,   425,  110,   16,   1;
   7247,  5028,  2400,  771,  159,  19,  1;
  34016, 24731, 12999, 4872, 1267, 217, 22, 1; ...
From _Philippe Deléham_, Nov 07 2011: (Start)
Production matrix begins:
  4, 1
  1, 3, 1
  0, 1, 3, 1
  0, 0, 1, 3, 1
  0, 0, 0, 1, 3, 1
  0, 0, 0, 0, 1, 3, 1
  0, 0, 0, 0, 0, 1, 3, 1
  0, 0, 0, 0, 0, 0, 1, 3, 1
  0, 0, 0, 0, 0, 0, 0, 1, 3, 1 (End)
		

Programs

  • Maple
    A126791 := proc(n,k)
        if n=0 and k = 0 then
            1 ;
        elif k <0 or k>n then
            0;
        elif k= 0 then
            4*procname(n-1,0)+procname(n-1,1) ;
        else
            procname(n-1,k-1)+3*procname(n-1,k)+procname(n-1,k+1) ;
        end if;
    end proc: # R. J. Mathar, Mar 12 2013
    T := (n,k) -> (-1)^(n-k)*simplify(GegenbauerC(n-k,-n+1,3/2) - GegenbauerC(n-k-1, -n+1, 3/2)): seq(seq(T(n,k),k=1..n),n=1..10); # Peter Luschny, May 13 2016
  • Mathematica
    T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0,
    T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]];
    Table[T[n, k, 4, 3], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, May 22 2017 *)

Formula

Sum_{k>=0} T(m,k)*T(n,k) = T(m+n,0) = A026378(m+n+1).
Sum_{k=0..n} T(n,k) = 5^n = A000351(n).
T(n,k) = (-1)^(n-k)*(GegenbauerC(n-k,-n+1,3/2) - GegenbauerC(n-k-1,-n+1,3/2)). - Peter Luschny, May 13 2016
The n-th row polynomial R(n,x) equals the n-th degree Taylor polynomial of the function (1 + x )*(1 + 3*x + x^2)^n expanded about the point x = 0. - Peter Bala, Sep 06 2022

A005573 Number of walks on cubic lattice (starting from origin and not going below xy plane).

Original entry on oeis.org

1, 5, 26, 139, 758, 4194, 23460, 132339, 751526, 4290838, 24607628, 141648830, 817952188, 4736107172, 27487711752, 159864676803, 931448227590, 5435879858958, 31769632683132, 185918669183370, 1089302293140564
Offset: 0

Views

Author

Keywords

Comments

Binomial transform of A026378, second binomial transform of A001700. - Philippe Deléham, Jan 28 2007
The Hankel transform of [1,1,5,26,139,758,...] is [1,4,15,56,209,...](see A001353). - Philippe Deléham, Apr 13 2007

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (Sqrt((1-2*x)/(1-6*x)) -1)/(2*x) )); // G. C. Greubel, May 02 2019
    
  • Mathematica
    CoefficientList[Series[(Sqrt[(1-2x)/(1-6x)]-1)/(2x),{x,0,20}],x] (* Harvey P. Dale, Jun 24 2011 *)
    a[n_] := 6^n Hypergeometric2F1[1/2, -n, 2, 2/3]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Apr 11 2017 *)
  • PARI
    my(x='x+O('x^30)); Vec((sqrt((1-2*x)/(1-6*x)) -1)/(2*x)) \\ G. C. Greubel, May 02 2019
    
  • Sage
    ((sqrt((1-2*x)/(1-6*x)) -1)/(2*x)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 02 2019

Formula

From Emeric Deutsch, Jan 09 2003; corrected by Roland Bacher: (Start)
a(n) = Sum_{i=0..n} (-1)^i*6^(n-i)*binomial(n, i)*binomial(2*i, i)/(i+1);
g.f. A(x) satisfies: x(1-6x)A^2 + (1-6x)A - 1 = 0. (End)
From Henry Bottomley, Aug 23 2001: (Start)
a(n) = 6*a(n-1) - A005572(n-1).
a(n) = Sum_{j=0..n} 4^(n-j)*binomial(n, floor(n/2))*binomial(n, j). (End)
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(2*k+1, k)*2^(n-k).
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*Catalan(k)*6^(n-k).
D-finite with recurrence (n+1)*a(n) = (8*n+2)*a(n-1)-(12*n-12)*a(n-2). - Vladeta Jovovic, Jul 16 2004
a(n) = Sum_{k=0..n} A052179(n,k). - Philippe Deléham, Jan 28 2007
Conjecture: a(n)= 6^n * hypergeom([1/2,-n],[2], 2/3). - Benjamin Phillabaum, Feb 20 2011
From Paul Barry, Apr 21 2009: (Start)
G.f.: (sqrt((1-2*x)/(1-6*x)) - 1)/(2*x).
G.f.: 1/(1-5*x-x^2/(1-4*x-x^2/(1-4*x-x^2/(1-4*x-x^2/(1-... (continued fraction). (End)
G.f.: 1/(1 - 4*x - x*(1 - 2*x)/(1 - 2*x - x*(1 - 2*x)/(1 - 2*x - x*(1 - 2*x)/(1 - 2*x - x*(1 - 2*x)/(1...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Jul 02 2010
a(n) ~ 6^(n+1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 05 2012
G.f.: G(0)/(2*x) - 1/(2*x), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2)*(1-2*x) - 2*x*(1-2*x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1-2*x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 24 2013
a(n) = 2^n*hypergeom([-n, 3/2], [2], -2). - Peter Luschny, Apr 26 2016
E.g.f.: exp(4*x)*(BesselI(0,2*x) + BesselI(1,2*x)). - Ilya Gutkovskiy, Sep 20 2017

Extensions

More terms from Henry Bottomley, Aug 23 2001

A026376 a(n) is the number of integer strings s(0),...,s(n) counted by array T in A026374 that have s(n)=2; also a(n) = T(2n,n-1).

Original entry on oeis.org

1, 6, 30, 144, 685, 3258, 15533, 74280, 356283, 1713690, 8263596, 39938616, 193419915, 938430990, 4560542550, 22195961280, 108171753355, 527816696850, 2578310320610, 12607504827600, 61706212037295, 302275142049870, 1481908332595625, 7270432009471224
Offset: 1

Views

Author

Keywords

Comments

Number of Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1) and H=(2,0) and never going below the x-axis) from (0,0) to (2n+2,0), with exactly one peak at an even level. E.g., a(2)=6 because we have UUDDH, HUUDD, UDUUDD, UUDDUD, UUDHD and UHUDD. - Emeric Deutsch, Dec 28 2003
Number of left steps in all skew Dyck paths of semilength n+1. A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of the path is defined to be the number of its steps. Example: a(2)=6 because in the 10 (=A002212(3)) skew Dyck paths of semilength 3 ( namely UDUUDL, UUUDLD, UUDUDL, UUUDDL, UUUDLL and five Dyck paths that have no left steps) we have altogether 6 left steps. - Emeric Deutsch, Aug 05 2007
From Gary W. Adamson, May 17 2009: (Start)
Equals A026378 (1, 4, 17, 75, ...) convolved with A007317 (1, 2, 5, 15, 51, ...).
Equals A081671 (1, 3, 11, 45, ...) convolved with A002212 (1, 3, 10, 36, 137, ...).
(End)

Crossrefs

Programs

  • Maple
    a := n -> simplify(GegenbauerC(n-1, -n, -3/2)):
    seq(a(n), n=1..24); # Peter Luschny, May 09 2016
  • Mathematica
    Rest[CoefficientList[Series[(1-3*x-Sqrt[1-6*x+5*x^2])/(2*x*Sqrt[1-6*x+5*x^2]), {x, 0, 20}], x]] (* Vaclav Kotesovec, Feb 13 2014 *)
  • PARI
    a(n)=if(n<0,0,polcoeff((1+3*x+x^2)^n,n-1))
    
  • Sage
    A026376 = lambda n : n*hypergeometric([1, 3/2, 1-n], [1, 3], -4)
    [round(A026376(n).n(100)) for n in (1..24)] # Peter Luschny, Sep 16 2014
    
  • Sage
    # Recurrence:
    def A026376():
        x, y, n = 1, 1, 1
        while True:
            x, y = y, ((6*n + 3)*y - (5*n - 5)*x) / (n + 2)
            yield n*x
            n += 1
    a = A026376()
    [next(a) for i in (1..24)] # Peter Luschny, Sep 16 2014

Formula

E.g.f.: exp(3x)*I_1(2x), where I_1 is Bessel function. - Michael Somos, Sep 09 2002
G.f.: (1 - 3*z - t)/(2*z*t) where t = sqrt(1-6*z+5*z^2). - Emeric Deutsch, May 25 2003
a(n) = [t^(n+1)](1+3t+t^2)^n. a := n -> Sum_{j=ceiling((n+1)/2)..n} 3^(2j-n-1)*binomial(n, j)*binomial(j, n+1-j). - Emeric Deutsch, Jan 30 2004
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(2k, k+1). - Paul Barry, Sep 20 2004
a(n) = n*A002212(n). - Emeric Deutsch, Aug 05 2007
D-finite with recurrence (n+1)*a(n) - 9*n*a(n-1) + (23*n-27)*a(n-2) + 15*(-n+2)*a(n-3) = 0. - R. J. Mathar, Dec 02 2012
a(n) ~ 5^(n+1/2) / (2*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 13 2014
a(n) = n*hypergeometric([1, 3/2, 1-n],[1, 3],-4). - Peter Luschny, Sep 16 2014
a(n) = GegenbauerC(n-1, -n, -3/2). - Peter Luschny, May 09 2016

A039717 Row sums of convolution triangle A030523.

Original entry on oeis.org

1, 4, 15, 55, 200, 725, 2625, 9500, 34375, 124375, 450000, 1628125, 5890625, 21312500, 77109375, 278984375, 1009375000, 3651953125, 13212890625, 47804687500, 172958984375, 625771484375, 2264062500000, 8191455078125
Offset: 1

Views

Author

Keywords

Comments

Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 3, s(2n) = 5.
With offset 0 = INVERT transform of A001792: (1, 3, 8, 20, 48, 112, ...). - Gary W. Adamson, Oct 26 2010
From Tom Copeland, Nov 09 2014: (Start)
The array belongs to a family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the o.g.f. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x*(1-x)/(1 + (t-1)*x*(1-x)). See A091867 for more info on this family. Here t = -4 (mod signs in the results).
Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).
O.g.f.: G(x) = x*(1-x)/(1 - 5x*(1-x)) = P(Cinv(x),-5).
Inverse O.g.f.: Ginv(x) = (1 - sqrt(1 - 4*x/(1+5x)))/2 = C(P(x,5)) (signed A026378). Cf. A030528. (End)
p-INVERT of (2^n), where p(s) = 1 - s - s^2; see A289780. - Clark Kimberling, Aug 10 2017

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 - x) / (1 - 5 x + 5 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)
  • PARI
    Vec(x*(1-x)/(1-5*x+5*x^2) + O(x^40)) \\ Altug Alkan, Nov 20 2015

Formula

G.f.: x*(1-x)/(1-5*x+5*x^2) = g1(3, x)/(1-g1(3, x)), g1(3, x) := x*(1-x)/(1-2*x)^2 (g.f. first column of A030523).
From Paul Barry, Apr 16 2004: (Start)
Binomial transform of Fibonacci(2n+2).
a(n) = (sqrt(5)/2 + 5/2)^n*(3*sqrt(5)/10 + 1/2) - (5/2 - sqrt(5)/2)^n*(3*sqrt(5)/10 - 1/2). (End)
a(n) = (1/5)*Sum_{r=1..9} sin(3*r*Pi/10)*sin(r*Pi/2)*(2*cos(r*Pi/10))^(2*n).
a(n) = 5*a(n-1) - 5*a(n-2).
a(n) = Sum_{k=0..n} Sum_{i=0..n} binomial(n, i)*binomial(k+i+1, 2k+1). - Paul Barry, Jun 22 2004
From Johannes W. Meijer, Jul 01 2010: (Start)
Limit_{k->oo} a(n+k)/a(k) = (A020876(n) + A093131(n)*sqrt(5))/2.
Limit_{n->oo} A020876(n)/A093131(n) = sqrt(5).
(End)
From Benito van der Zander, Nov 19 2015: (Start)
Limit_{k->oo} a(k+1)/a(k) = 1 + phi^2 = (5 + sqrt(5)) / 2.
a(n) = a(n-1) * 3 + A081567(n-2) (not proved).
(End)
E.g.f.: exp(x*5/2) * (cosh(x*sqrt(5)/2) + (3/sqrt(5))*sinh(x*sqrt(5)/2)). - Fabian Pereyra, Oct 29 2024
Showing 1-10 of 27 results. Next