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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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Author

N. J. A. Sloane

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)
This Chebyshev sequence a(n) = S(n-1, 3) (see a formula below) is related to the bisection of Fibonacci sequences {F(a,b;n)}{n>=0} with input F(a,b;0) = a and F(a,b;1) = b, by F(a,b;2*k) = (a+b)*S(k-1, 3) - a*S(k-2, 3) and F(a,b;2*k+1) = b*S(k, 3) + (a-b)*S(k-1, 3), for k >= 0, and S(-2, 3) = -1. Proof via the o.g.f.s GFeven(a,b,t) = (a - t*(2*a-b))/(1 - 3*t + t^2) and GFodd(a,b,t) = (b + t*(a-b))/(1 - 3*t + t^2). The special case a = 0, b = 1 gives back F(2*k) = S(k-1, 3) = a(k). - _Wolfdieter Lang, Jun 07 2019
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.

Links

Crossrefs

Fibonacci A000045 = union of this sequence and A001519.
Cf. A000041, A000032, A001622, A048996, A052529, A055991, A078812, A305309, A058038 (pairwise products), A153386.
Inverse sequences A130259 and A130260.
Cf. A033888, A033890, A249450, A344212.

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

A001333 Pell-Lucas numbers: numerators of continued fraction convergents to sqrt(2).

Original entry on oeis.org

1, 1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243, 275807, 665857, 1607521, 3880899, 9369319, 22619537, 54608393, 131836323, 318281039, 768398401, 1855077841, 4478554083, 10812186007, 26102926097, 63018038201, 152139002499, 367296043199
Offset: 0

Views

Author

N. J. A. Sloane and R. K. Guy

Keywords

Comments

Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].
Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011
Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012
Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedman
a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.
In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017
Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002
a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1.
a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
Binomial transform of A077957. - Paul Barry, Feb 25 2003
For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2. - Herbert Kociemba, Jun 02 2004
For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004
Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004
Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006
Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006]
Odd-indexed prime numerators are prime RMS numbers (A140480) and also NSW primes (A088165). - Ctibor O. Zizka, Aug 13 2008
The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators here. - Clark Kimberling, Aug 26 2008
Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008
Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008
From Charlie Marion, Jan 07 2009: (Start)
In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2) and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2) and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then b(1,n)=a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
Cf. A000129, A142238-A142239, A153313-A153318.
(End)
This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel (A048211). - Sameen Ahmed Khan, Jun 28 2010
Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
Equals the INVERTi transform of A055099. - Gary W. Adamson, Aug 14 2010
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(8,2) = (0 0 1 0)
(0 1 0 1)
(1 0 2 0)
(0 2 0 1).
Then a(n) = (1/4)*Trace(U^n). (See also A084130, A006012.)
(End)
For n >= 1, row sums of triangle
m/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....2
.2..|..1.....2.....4
.3..|..1.....4.....4.....8
.4..|..1.....4....12.....8....16
.5..|..1.....6....12....32....16....32
.6..|..1.....6....24....32....80....32....64
.7..|..1.....8....24....80....80...192....64...128
which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012
The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012
Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012
This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012
a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
If p is prime, a(p) == 1 (mod p) (compare with similar comment for A000032). - Creighton Dement, Oct 11 2005, modified by Davide Colazingari, Jun 26 2016
a(n) = A000129(n) + A000129(n-1), where A000129(n) is the n-th Pell Number; e.g., a(6) = 99 = A000129(6) + A000129(5) = 70 + 29. Hence the sequence of fractions has the form 1 + A000129(n-1)/A000129(n), and the ratio A000129(n-1)/A000129(n)converges to sqrt(2) - 1. - Gregory L. Simay, Nov 30 2018
For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020
For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021
a(n) is the permanent of the n X n tridiagonal matrix defined in A332602. - Stefano Spezia, Apr 12 2022
From Greg Dresden, May 08 2023: (Start)
For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.
_
|| _
|||_|||
|_|
(End)
12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024

Examples

			Convergents are 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.
The 15 3 X 2 crossword grids, with white squares represented by an o:
  ooo ooo ooo ooo ooo ooo ooo oo. o.o .oo o.. .o. ..o oo. .oo
  ooo oo. o.o .oo o.. .o. ..o ooo ooo ooo ooo ooo ooo .oo oo.
G.f. = 1 + x + 3*x^2 + 7*x^3 + 17*x^4 + 41*x^5 + 99*x^6 + 239*x^7 + 577*x^8 + ...

References

  • M. R. Bacon and C. K. Cook, Some properties of Oresme numbers and convolutions ..., Fib. Q., 62:3 (2024), 233-240.
  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
  • J. Devillet, J.-L. Marichal, and B. Teheux, Classifications of quasitrivial semigroups, Semigroup Forum, 100 (2020), 743-764.
  • Maribel Díaz Noguera [Maribel Del Carmen Díaz Noguera], Rigoberto Flores, Jose L. Ramirez, and Martha Romero Rojas, Catalan identities for generalized Fibonacci polynomials, Fib. Q., 62:2 (2024), 100-111.
  • Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
  • R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 288.
  • A. F. Horadam, R. P. Loh, and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979.
  • Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
  • Kin Y. Li, Math Problem Book I, 2001, p. 24, Problem 159.
  • I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 102, Problem 10.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
  • 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, Volume 1 (1986), p. 203, Example 4.1.2.
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.
  • R. C. Tilley et al., The cell growth problem for filaments, Proc. Louisiana Conf. Combinatorics, ed. R. C. Mullin et al., Baton Rouge, 1970, 310-339.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 34.

Links

Crossrefs

For denominators see A000129.
See A040000 for the continued fraction expansion of sqrt(2).
See also A078057 which is the same sequence without the initial 1.
Cf. also A002203, A152113.
Row sums of unsigned Chebyshev T-triangle A053120. a(n)= A054458(n, 0) (first column of convolution triangle).
Row sums of A140750, A160756, A135837.
Equals A034182(n-1) + 2 and A084128(n)/2^n. First differences of A052937. Partial sums of A052542. Pairwise sums of A048624. Bisection of A002965.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Second row of the array in A135597.
Cf. A048211, A153588, A174283, A174284, A174285, A174286, A176499, A176500, A176501, A176502.
Cf. A055099.
Cf. A028859, A001906 / A088305, A033303, A000225, A095263, A003945, A006356, A002478, A214260, A001911 and A000217 for other restricted ternary words.
Cf. Triangle A106513 (alternating row sums).
Equals A293004 + 1.
Cf. A033539, A332602, A086395 (subseq. of primes).

Programs

  • Haskell
    a001333 n = a001333_list !! n
a001333_list = 1 : 1 : zipWith (+)
                       a001333_list (map (* 2) $ tail a001333_list)
-- Reinhard Zumkeller, Jul 08 2012
  • Magma
    [n le 2 select 1 else 2*Self(n-1)+Self(n-2): n in [1..35]]; // Vincenzo Librandi, Nov 10 2018
  • Maple
    A001333 := proc(n) option remember; if n=0 then 1 elif n=1 then 1 else 2*procname(n-1)+procname(n-2) fi end;
Digits := 50; A001333 := n-> round((1/2)*(1+sqrt(2))^n);
with(numtheory): cf := cfrac (sqrt(2),1000): [seq(nthnumer(cf,i), i=0..50)];
a:= n-> (M-> M[2, 1]+M[2, 2])(<<2|1>, <1|0>>^n):
seq(a(n), n=0..33);  # Alois P. Heinz, Aug 01 2008
A001333List := proc(m) local A, P, n; A := [1,1]; P := [1,1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-2]]);
A := [op(A), P[-1]] od; A end: A001333List(32); # Peter Luschny, Mar 26 2022
  • Mathematica
    Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[2], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 08 2006 *)
Table[((1 - Sqrt[2])^n + (1 + Sqrt[2])^n)/2, {n, 0, 29}] // Simplify (* Robert G. Wilson v, May 02 2006 *)
a[0] = 1; a[1] = 1; a[n_] := a[n] = 2a[n - 1] + a[n - 2]; Table[a@n, {n, 0, 29}] (* Robert G. Wilson v, May 02 2006 *)
Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, May 02 2006 *)
a=c=0;t={b=1}; Do[c=a+b+c; AppendTo[t,c]; a=b;b=c,{n,40}]; t (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
LinearRecurrence[{2, 1}, {1, 1}, 40] (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
Join[{1}, Numerator[Convergents[Sqrt[2], 30]]] (* Harvey P. Dale, Aug 22 2011 *)
Table[(-I)^n ChebyshevT[n, I], {n, 10}] (* Eric W. Weisstein, Apr 04 2017 *)
CoefficientList[Series[(-1 + x)/(-1 + 2 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
Table[Sqrt[(ChebyshevT[n, 3] + (-1)^n)/2], {n, 0, 20}] (* Eric W. Weisstein, Apr 17 2018 *)
  • PARI
    {a(n) = if( n<0, (-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [1, 1]}; /* Michael Somos, Sep 02 2012 */
  • PARI
    {a(n) = polchebyshev(n, 1, I) / I^n}; /* Michael Somos, Sep 02 2012 */
  • PARI
    a(n) = real((1 + quadgen(8))^n); \\ Michel Marcus, Mar 16 2021
  • PARI
    { for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[1, 1]; if (a > 10^(10^3 - 6), break); write("b001333.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 12 2009
  • Python
    from functools import cache
@cache
def a(n): return 1 if n < 2 else 2*a(n-1) + a(n-2)
print([a(n) for n in range(32)]) # Michael S. Branicky, Nov 13 2022
  • Sage
    from sage.combinat.sloane_functions import recur_gen2
it = recur_gen2(1,1,2,1)
[next(it) for i in range(30)] ## Zerinvary Lajos, Jun 24 2008
  • Sage
    [lucas_number2(n,2,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009

Formula

a(n) = A055642(A125058(n)). - Reinhard Zumkeller, Feb 02 2007
a(n) = 2a(n-1) + a(n-2);
a(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2.
a(n)+a(n+1) = 2 A000129(n+1). 2*a(n) = A002203(n).
G.f.: (1 - x) / (1 - 2*x - x^2) = 1 / (1 - x / (1 - 2*x / (1 + x))). - Simon Plouffe in his 1992 dissertation.
A000129(2n) = 2*A000129(n)*a(n). - John McNamara, Oct 30 2002
a(n) = (-i)^n * T(n, i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
a(n) = a(n-1) + A052542(n-1), n>1. a(n)/A052542(n) converges to sqrt(1/2). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
E.g.f.: exp(x)cosh(x*sqrt(2)). - Paul Barry, May 08 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)2^k. - Paul Barry, May 13 2003
For n > 0, a(n)^2 - (1 + (-1)^(n))/2 = Sum_{k=0..n-1} ((2k+1)*A001653(n-1-k)); e.g., 17^2 - 1 = 288 = 1*169 + 3*29 + 5*5 + 7*1; 7^2 = 49 = 1*29 + 3*5 + 5*1. - Charlie Marion, Jul 18 2003
a(n+2) = A078343(n+1) + A048654(n). - Creighton Dement, Jan 19 2005
a(n) = A000129(n) + A000129(n-1) = A001109(n)/A000129(n) = sqrt(A001110(n)/A000129(n)^2) = ceiling(sqrt(A001108(n))). - Henry Bottomley, Apr 18 2000
Also the first differences of A000129 (the Pell numbers) because A052937(n) = A000129(n+1) + 1. - Graeme McRae, Aug 03 2006
a(n) = Sum_{k=0..n} A122542(n,k). - Philippe Deléham, Oct 08 2006
For another recurrence see A000129.
a(n) = Sum_{k=0..n} A098158(n,k)*2^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1,1; 2,1]^n. - Gary W. Adamson, Mar 12 2008
If p[1]=1, and 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)=det A. - Milan Janjic, Apr 29 2010
For n>=2, a(n)=F_n(2)+F_(n+1)(2), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(-n) = (-1)^n * a(n). - Michael Somos, Sep 02 2012
Dirichlet g.f.: (PolyLog(s,1-sqrt(2)) + PolyLog(s,1+sqrt(2)))/2. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A000129(n) - A000129(n-1), where A000129(n) is the n-th Pell Number. Hence the continued fraction is of the form 1-(A000129(n-1)/A000129(n)). - Gregory L. Simay, Nov 09 2018
a(n) = (A000129(n+3) + A000129(n-3))/10, n>=3. - Paul Curtz, Jun 16 2021
a(n) = (A000129(n+6) - A000129(n-6))/140, n>=6. - Paul Curtz, Jun 20 2021
a(n) = round((1/2)*sqrt(Product_{k=1..n} 4*(1 + sin(k*Pi/n)^2))), for n>=1. - Greg Dresden, Dec 28 2021
a(n)^2 + a(n+1)^2 = A075870(n+1) = 2*(b(n)^2 + b(n+1)^2) for all n in Z where b(n) := A000129(n). - Michael Somos, Apr 02 2022
a(n) = 2*A048739(n-2)+1. - R. J. Mathar, Feb 01 2024
Sum_{n>=1} 1/a(n) = 1.5766479516393275911191017828913332473... - R. J. Mathar, Feb 05 2024
From Peter Bala, Jul 06 2025: (Start)
G.f.: Sum_{n >= 1} (-1)^(n+1) * x^(n-1) * Product_{k = 1..n} (1 - k*x)/(1 - 3*x + k*x^2).
The following series telescope:
Sum_{n >= 1} (-1)^(n+1)/(a(2*n) + 1/a(2*n)) = 1/4, since 1/(a(2*n) + 1/a(2*n)) = 1/A077445(n) + 1/A077445(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) - 1/a(2*n+1)) = 1/8, since. 1/(a(2*n+1) - 1/a(2*n+1)) = 1/(4*Pell(2*n)) + 1/(4*Pell(2*n+2)), where Pell(n) = A000129(n).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) + 9/a(2*n+1)) = 1/10, since 1/(a(2*n+1) + 9/a(2*n+1)) = b(n) + b(n+1), where b(n) = A001109(n)/(2*Pell(2*n-1)*Pell(2*n+1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 1 - sqrt(2)/2 = A268682, since (-1)^(n+1)/(a(n)*a(n+1)) = Pell(n)/a(n) - Pell(n+1)/a(n+1). (End)

Extensions

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A030267 Compose the natural numbers with themselves, A(x) = B(B(x)) where B(x) = x/(1-x)^2 is the generating function for natural numbers.

Original entry on oeis.org

1, 4, 14, 46, 145, 444, 1331, 3926, 11434, 32960, 94211, 267384, 754309, 2116936, 5914310, 16458034, 45638101, 126159156, 347769719, 956238170, 2623278946, 7181512964, 19622668679, 53522804976, 145753273225, 396323283724, 1076167858046, 2918447861686
Offset: 1

Views

Author

Christian G. Bower

Keywords

Comments

Sum of pyramid weights of all nondecreasing Dyck paths of semilength n. (A pyramid in a Dyck word (path) is a factor of the form U^h D^h, where U=(1,1), D=(1,-1) and h is the height of the pyramid. A pyramid in a Dyck word w is maximal if, as a factor in w, it is not immediately preceded by a u and immediately followed by a d. The pyramid weight of a Dyck path (word) is the sum of the heights of its maximal pyramids.) Example: a(4) = 46. Indeed, there are 14 Dyck paths of semilength 4. One of them, namely UUDUDDUD is not nondecreasing because the valleys are at heights 1 and 0. The other 13, with the maximal pyramids shown between parentheses, are: (UD)(UD)(UD)(UD), (UD)(UD)(UUDD), (UD)(UUDD)(UD), (UD)U(UD)(UD)D, (UD)(UUUDDD), (UUDD)(UD)(UD), (UUDD)(UUDD), (UUUDDD)(UD), U(UD)(UD)(UD)D, U(UD)(UUDD)D, U(UUDD)(UD)D, UU(UD)(UD)DD and (UUUUDDDD). The pyramid weights of these paths are 4, 4, 4, 3, 4, 4, 4, 4, 3, 3, 3, 2, and 4, respectively. Their sum is 46. a(n) = Sum_{k = 1..n} k*A121462(n, k). - Emeric Deutsch, Jul 31 2006
Number of 1s in all compositions of n, where compositions are understood with two different kinds of 1s, say 1 and 1' (n >= 1). Example: a(2) = 4 because the compositions of 2 are 11, 11', 1'1, 1'1', 2, having a total of 2 + 1 + 1 + 0 + 0 = 4 1s. Also number of k's in all compositions of n + k (k = 2, 3, ...). - Emeric Deutsch, Jul 21 2008
From Petros Hadjicostas, Jun 24 2019: (Start)
If c = (c(m): m >= 1) is the input sequence and b_k = (b_k(n): n >= 1) is the output sequence under the AIK[k] = INVERT[k] transform (see Bower's web link below), then the bivariate g.f. of the list of sequences (b_k: k >= 1) = ((b_k(n): n >= 1): k >= 1) is Sum_{n, k >= 1} b_k(n)*x^n*y^k = y*C(x)/(1 - y*C(x)), where C(x) = Sum_{m >= 1} c(m)*x^m is the g.f. of the input sequence.
Here, b_k(n) is the number of all (linear) compositions of n with k parts where a part of size m is colored with one of c(m) colors. Thus, Sum_{k = 1..n} k*b_k(n) is the total number of parts in all compositions of n.
If we differentiate the bivariate g.f. function above, i.e., Sum_{n, k >= 1} b_k(n)*x^n*y^k, with respect to y and set y = 1, we get the g.f. of the sequence (Sum_{k = 1..n} k*b_k(n): n >= 1). It is C(x)/(1 - C(x))^2.
When c(m) = m for all m >= 1, we have m-color compositions of n that were first studied by Agarwal (2000). The cyclic version of these m-color compositions were studied by Gibson (2017) and Gibson et al. (2018).
When c(m) = m for each m >= 1, we have C(x) = x/(1 - x)^2, and so C(x)/(1 - C(x))^2 = x * (1 - x)^2/(1 - 3*x + x^2)^2, which is the g.f. of the current sequence.
Hence, a(n) is the total number of parts in all m-color compositions of n (in the sense of Agarwal (2000)).
(End)
Series reversal gives A153294 starting from index 1, with alternating signs: 1, -4, 18, -86, 427, -2180, ... - Vladimir Reshetnikov, Aug 03 2019

Examples

			From _Petros Hadjicostas_, Jun 24 2019: (Start)
Recall that with m-color compositions, a part of size m may be colored with one of m colors.
We have a(1) = 1 because we only have one colored composition, namely 1_1, that has only 1 part.
We have a(2) = 4 because we have the following colored compositions of n = 2: 2_1, 2_2, 1_1 + 1_1; hence, a(2) = 1 + 1 + 2 = 4.
We have a(3) = 14 because we have the following colored compositions of n = 3: 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 2_1 + 1_1, 2_2 + 1_1, 1_1 + 1_1 + 1_1; hence, a(3) = 1 + 1 + 1 + 2 + 2 + 2 + 2 + 3 = 14.
We have a(14) = 46 because we have the following colored compositions of n = 4:
(i) 4_1, 4_2, 4_3, 4_4; with a total of 4 parts.
(ii) 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 3_1 + 1_1, 3_2 + 1_1, 3_3 + 1_1, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_1, 2_2 + 2_2; with a total of 2 x 10 = 20 parts.
(iii) 1_1 + 1_1 + 2_1, 1_1 + 1_1 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 2_1 + 1_1 + 1_1, 2_2 + 1_1 + 1_1; with a total of 3 x 6 = 18 parts.
(iv) 1_1 + 1_1 + 1_1 + 1_1; with a total of 4 parts.
Hence, a(4) = 4 + 20 + 18 + 4 = 46.
(End)

References

  • R. P. Grimaldi, Compositions and the alternate Fibonacci numbers, Congressus Numerantium, 186, 2007, 81-96.

Links

Crossrefs

Partial sums of A038731. First differences of A001870.
Cf. A001629 (right-shifted inverse Binomial Transform), A023610 (inverse Binomial Transform of left-shifted sequence), A030279, A045623, A088305, A121462, A153294, A279282, A307415, A308723.

Programs

  • Maple
    with(combinat): L[0]:=2: L[1]:=1: for n from 2 to 60 do L[n]:=L[n-1] +L[n-2] end do: seq(2*fibonacci(2*n)*1/5+(1/5)*n*L[2*n],n=1..30); # Emeric Deutsch, Jul 21 2008
  • Mathematica
    Table[Sum[k Binomial[n+k-1,2k-1],{k,n}],{n,30}] (* or *) LinearRecurrence[ {6,-11,6,-1},{1,4,14,46},30] (* Harvey P. Dale, Aug 01 2011 *)
  • PARI
    a(n)=(2*n*fibonacci(2*n+1)+(2-n)*fibonacci(2*n))/5

Formula

a(n) = -a(-n) = (2n * F(2n+1) + (2 - n) * F(2n))/5 with F(n) = A000045(n) (Fibonacci numbers).
G.f.: x * (1 - x)^2/(1 - 3*x + x^2)^2.
a(n) = Sum_{k = 1..n} k*C(n + k - 1, 2*k - 1).
a(n) = (2/5)*F(2*n) + (1/5)*n*L(2*n), where F(k) are the Fibonacci numbers (F(0)=0, F(1)=1) and L(k) are the Lucas numbers (L(0) = 2, L(1) = 1). - Emeric Deutsch, Jul 21 2008
a(0) = 1, a(1) = 4, a(2) = 14, a(3) = 46, a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). - Harvey P. Dale, Aug 01 2011
a(n) = ((3 - sqrt(5))^n*(5*n - 2*sqrt(5)) + (3 + sqrt(5))^n*(5*n + 2*sqrt(5)))/ (25*2^n). - Peter Luschny, Mar 07 2022
E.g.f.: exp(3*x/2)*(15*x*cosh(sqrt(5)*x/2) + sqrt(5)*(4 + 5*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025

Extensions

Name clarified using a comment of the author by Peter Luschny, Aug 03 2019

A032198 "CIK" (necklace, indistinct, unlabeled) transform of 1,2,3,4,...

Original entry on oeis.org

1, 3, 6, 13, 25, 58, 121, 283, 646, 1527, 3601, 8678, 20881, 50823, 124054, 304573, 750121, 1855098, 4600201, 11442085, 28527446, 71292603, 178526881, 447919418, 1125750145, 2833906683, 7144450566, 18036423973
Offset: 1

Views

Author

Christian G. Bower

Keywords

Examples

			From _Petros Hadjicostas_, Jan 07 2018: (Start)
We give some examples to illustrate the theory of C. G. Bower about transforms given in the weblink above. We assume we have boxes of different sizes and colors that we place on a circle to form a necklace. Two boxes of the same size and same color are considered identical (indistinct and unlabeled). We do, however, change the roles of the sequences (a(n): n>=1) and (b(n): n>=1) that appear in the weblink above. We assume (a(n): n>=1) = CIK((b(n): n>=1)).
Since b(1) = 1, b(2) = 2, b(3) = 3, etc., a box that can hold 1 ball only can be of 1 color only, a box that can hold 2 balls only can be one of 2 colors only, a box that can hold 3 balls can be one of 3 colors, and so on.
To prove that a(3) = 6, we consider three cases. In the first case, we have a single box that can hold 3 balls, and thus we have 3 possibilities for the 3 colors the box can be. In the second case, we have a box that can hold 2 balls and a box that can hold 1 ball. Here, we have 2 x 1 = 2 possibilities. In the third case, we have 3 identical boxes, each of which can hold 1 ball. This gives rise to 1 possibility. Hence, a(3) = 3 + 2 + 1 = 6.
To prove that a(4) = 13, we consider 5 cases: a box with 4 balls (4 possibilities), one box with 3 balls and one box with 1 ball (3 possibilities), two identical boxes each with 2 balls (3 possibilities), one box with 2 balls and two identical boxes each with 1 ball (2 possibilities), and four identical boxes each with 1 ball (1 possibility). Thus, a(4) = 4 + 3 + 3 + 2 + 1 = 13.
(End)

Links

Crossrefs

Cf. A000032, A001906, A004146, A005248, A032170, A032287, A088305, A308723, A308747.
Equals A005594(n) - 1.

Programs

  • Mathematica
    nmax = 30;
f[x_] = Sum[n*x^n, {n, 1, nmax}];
gf = Sum[(EulerPhi[n]/n)*Log[1/(1 - f[x^n])] + O[x]^nmax, {n, 1, nmax}]/x;
CoefficientList[gf, x] (* Jean-François Alcover, Jul 29 2018, after Joerg Arndt *)
  • PARI
    N = 66;  x = 'x + O('x^N);
f(x)=sum(n=1, N, n*x^n );
gf = sum(n=1, N, eulerphi(n)/n*log(1/(1-f(x^n)))  );
v = Vec(gf)
/* Joerg Arndt, Jan 21 2013 */

Formula

a(n) = 1/n*Sum_{d divides n} phi(n/d)*A004146(d). - Vladeta Jovovic, Feb 15 2003
From Petros Hadjicostas, Jan 07 2018: (Start)
a(n) = -2 + (1/n)*Sum_{d|n} phi(n/d)*A005248(d) = -2 + (1/n)*Sum_{d|n} phi(n/d)*L(2*d), where L(n) = A000032(n) is the usual Lucas sequence.
G.f.: -Sum_{n >= 1} (phi(n)/n)*log(1 - B(x^n)), where B(x) = x + 2*x^2 + 3*x^3 + 4*x^4 + ... = x/(1-x)^2.
G.f.: -2*x/(1 - x) - Sum_{n>=1} (phi(n)/n)*log(1 - 3*x^n + x^(2*n)).
(End)
From Petros Hadjicostas, Jun 19 2019: (Start)
According to Gibson et al. (2018), a(n) is the number of m-color cyclic compositions of n where each part of size m has m possible colors. This is nothing else than the CIK transform of the sequence 1, 2, 3, 4, ...
Using the theory of Flajolet and Soria (1991), Gibson et al. (2018, Eq. (1.1)) proved that the g.f. of a(n) is Sum_{s >= 1} (phi(s)/s) * log((1 - x^s)^2/(1 - 3*x^s + x^(2*s))), which is exactly the same g.f. as the ones above.
Gibson et al. (2018, p. 3210) also proved that a(n) ~ (2/(3-sqrt(5)))^n/n for large n. See also Chapter 3 in Gibson (2017).
(End)

A226205 a(n) = F(n)^2 - F(n-1)^2 or F(n+1) * F(n-2) where F(n) = A000045(n), the Fibonacci numbers.

Original entry on oeis.org

1, 0, 3, 5, 16, 39, 105, 272, 715, 1869, 4896, 12815, 33553, 87840, 229971, 602069, 1576240, 4126647, 10803705, 28284464, 74049691, 193864605, 507544128, 1328767775, 3478759201, 9107509824, 23843770275, 62423800997, 163427632720, 427859097159, 1120149658761
Offset: 1

Views

Author

Michael Somos, Jun 06 2013

Keywords

Comments

A001519(n)^2 = A079472(n)^2 + a(n)^2 and (A001519(n), A079472(n), a(n)) is a Pythagorean triple.
INVERT transform is A052156. PSUM transform is A007598. SUMADJ transform is A088305. BINOMIAL transform is A039717. BINOMIAL transform with 0 prepended is A112091 with 0 prepended. BINOMIAL transform inverse is A084179(n+1).
In general, the difference between squares of two consecutive terms of a second order linear recurrence having a signature of (c,d) will be a third order recurrence with signature (c^2+d,(c^2+d)*d,-d^3). - Gary Detlefs, Mar 13 2025

Examples

			G.f. = x + 3*x^3 + 5*x^4 + 16*x^5 + 39*x^6 + 105*x^7 + 272*x^8 + 715*x^9 + ...

Links

Crossrefs

Cf. A001519, A001622, A001654, A007598, A033999, A039717, A052156, A079472, A084179, A088305, A112091, A121646, A290565.
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.
Cf. A260259: numbers of the form F(n)*F(n+1)-(-1)^n. - Bruno Berselli, Nov 02 2015

Programs

  • Magma
    [Fibonacci(n)^2-Fibonacci(n-1)^2: n in [1..40]]; // Vincenzo Librandi, Jun 18 2014
  • Maple
    a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<1,0,3>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 28 2016
  • Mathematica
    a[ n_] := Fibonacci[n + 1] Fibonacci[n - 2]; (* Michael Somos, Jun 17 2014 *)
CoefficientList[Series[(1 - x)^2/((1 + x) (1 - 3 x + x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 17 2014 *)
  • PARI
    {a(n) = fibonacci( n + 1) * fibonacci( n - 2)};
  • PARI
    a(n) = round(2^(-1-n)*(-(-1)^n*2^(3+n)-(3-sqrt(5))^n*(1+sqrt(5))+(-1+sqrt(5))*(3+sqrt(5))^n)/5) \\ Colin Barker, Sep 28 2016
  • PARI
    lista(nn) = {my(p = (3*x-1)/(x^3-2*x^2-2*x+1)); for (n=1, nn, p = deriv(p, x); print1(subst(p, x, 0)/n!, ", "); ); } \\ Michel Marcus, May 22 2018

Formula

G.f.: x * (1 - x)^2 / ((1 + x) * (1 -3*x + x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = - A121646(n).
a(n) = -a(1-n) for all n in Z.
a(n) = A121801(n+1) / 2. - Michael Somos, Jun 17 2014
a(n) = a(n-1) + A000045(n-1)^2 - 2*(-1)^n, for n>1. - Alexander Samokrutov, Sep 07 2015
a(n) = F(n-1)*F(n) - (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = 2^(-1-n)*(-(-1)^n*2^(3+n)-(3-sqrt(5))^n*(1+sqrt(5))+(-1+sqrt(5))*(3+sqrt(5))^n)/5. - Colin Barker, Sep 28 2016
From Amiram Eldar, Oct 06 2020: (Start)
Sum_{n>=3} 1/a(n) = (1/2) * A290565 - 1/4.
Sum_{n>=3} (-1)^(n+1)/a(n) = (3/2) * (1/phi - 1/2), where phi is the golden ratio (A001622). (End)

A128908 Riordan array (1, x/(1-x)^2).

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 3, 4, 1, 0, 4, 10, 6, 1, 0, 5, 20, 21, 8, 1, 0, 6, 35, 56, 36, 10, 1, 0, 7, 56, 126, 120, 55, 12, 1, 0, 8, 84, 252, 330, 220, 78, 14, 1, 0, 9, 120, 462, 792, 715, 364, 105, 16, 1, 0, 10, 165, 792, 1716, 2002, 1365, 560, 136, 18, 1
Offset: 0

Views

Author

Philippe Deléham, Apr 22 2007

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,2,-1/2,1/2,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.
Row sums give A088305. - Philippe Deléham, Nov 21 2007
Column k is C(n,2k-1) for k > 0. - Philippe Deléham, Jan 20 2012
From R. Bagula's comment in A053122 (cf. Damianou link p. 10), this array gives the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014
T is the convolution triangle of the positive integers (see A357368). - Peter Luschny, Oct 19 2022

Examples

			The triangle T(n,k) begins:
   n\k  0    1    2    3    4    5    6    7    8    9   10
   0:   1
   1:   0    1
   2:   0    2    1
   3:   0    3    4    1
   4:   0    4   10    6    1
   5:   0    5   20   21    8    1
   6:   0    6   35   56   36   10    1
   7:   0    7   56  126  120   55   12    1
   8:   0    8   84  252  330  220   78   14    1
   9:   0    9  120  462  792  715  364  105   16    1
  10:   0   10  165  792 1716 2002 1365  560  136   18    1
  ... reformatted by _Wolfdieter Lang_, Jul 31 2017
From _Peter Luschny_, Mar 06 2022: (Start)
The sequence can also be seen as a square array read by upwards antidiagonals.
   1, 1,   1,    1,    1,     1,     1,      1,      1, ...  A000012
   0, 2,   4,    6,    8,    10,    12,     14,     16, ...  A005843
   0, 3,  10,   21,   36,    55,    78,    105,    136, ...  A014105
   0, 4,  20,   56,  120,   220,   364,    560,    816, ...  A002492
   0, 5,  35,  126,  330,   715,  1365,   2380,   3876, ... (A053126)
   0, 6,  56,  252,  792,  2002,  4368,   8568,  15504, ... (A053127)
   0, 7,  84,  462, 1716,  5005, 12376,  27132,  54264, ... (A053128)
   0, 8, 120,  792, 3432, 11440, 31824,  77520, 170544, ... (A053129)
   0, 9, 165, 1287, 6435, 24310, 75582, 203490, 490314, ... (A053130)
    A27,A292, A389, A580,  A582, A1288, A10966, A10968, A165817       (End)

Links

Crossrefs

Cf. A002450, A007318, A034008, A053122, A078812, A084938, A088305.
Cf. Columns : A000007, A000027, A000292, A000389, A000580, A000582, A001288, A010966 ..(+2).. A011000, A017713 ..(+2).. A017763.
Cf. A000007, A001352, A008574, A054888, A084099, A084103, A163810, A357368.
Cf. A165817 (the main diagonal of the array).

Programs

  • Maple
    # Computing the rows of the array representation:
S := proc(n,k) option remember;
if n = k then 1 elif k < 0 or k > n then 0 else
S(n-1, k-1) + 2*S(n-1, k) - S(n-2, k) fi end:
Arow := (n, len) -> seq(S(n+k-1, k-1), k = 0..len-1):
for n from 0 to 8 do Arow(n, 9) od; # Peter Luschny, Mar 06 2022
# Uses function PMatrix from A357368.
PMatrix(10, n -> n); # Peter Luschny, Oct 19 2022
  • Mathematica
    With[{nmax = 10}, CoefficientList[CoefficientList[Series[(1 - x)^2/(1 - (2 + y)*x + x^2), {x, 0, nmax}, {y, 0, nmax}], x], y]] // Flatten (* G. C. Greubel, Nov 22 2017 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(if(n==0 && k==0, 1, if(k==0, 0, binomial(n+k-1,2*k-1))), ", "))) \\ G. C. Greubel, Nov 22 2017
  • Python
    from functools import cache
@cache
def A128908(n, k):
    if n == k: return 1
    if (k <= 0 or k > n): return 0
    return A128908(n-1, k-1) + 2*A128908(n-1, k) - A128908(n-2, k)
for n in range(10):
    print([A128908(n, k) for k in range(n+1)]) # Peter Luschny, Mar 07 2022
  • Sage
    @cached_function
def T(k,n):
    if k==n: return 1
    if k==0: return 0
    return sum(i*T(k-1,n-i) for i in (1..n-k+1))
A128908 = lambda n,k: T(k,n)
for n in (0..10): print([A128908(n,k) for k in (0..n)]) # Peter Luschny, Mar 12 2016

Formula

T(n,0) = 0^n, T(n,k) = binomial(n+k-1, 2k-1) for k >= 1.
Sum_{k=0..n} T(n,k)*2^(n-k) = A002450(n) = (4^n-1)/3 for n>=1. - Philippe Deléham, Oct 19 2008
G.f.: (1-x)^2/(1-(2+y)*x+x^2). - Philippe Deléham, Jan 20 2012
Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A001352(n), (-1)^(n+1)*A054888(n+1), (-1)^n*A008574(n), (-1)^n*A084103(n), (-1)^n*A084099(n), A163810(n), A000007(n), A088305(n) for x = -6, -5, -4, -3, -2, -1, 0, 1 respectively. - Philippe Deléham, Jan 20 2012
Riordan array (1, x/(1-x)^2). - Philippe Deléham, Jan 20 2012

A369809 Expansion of 1/(1 - x^6/(1-x)^7).

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 1, 7, 28, 84, 210, 462, 925, 1730, 3108, 5565, 10388, 20944, 45697, 104673, 242481, 553455, 1229305, 2650221, 5565127, 11465758, 23397041, 47757235, 98317135, 205108561, 433747259, 926655972, 1989584722, 4271185538, 9133958765, 19421679515
Offset: 0

Views

Author

Seiichi Manyama, Feb 01 2024

Keywords

Comments

Number of compositions of 7*n-6 into parts 6 and 7.

Links

Crossrefs

Cf. A099253, A369805, A369806, A369807, A369808.
Cf. A088305, A095263, A290998, A368475, A369794.
Cf. A000579, A017847.

Programs

  • PARI
    my(N=40, x='x+O('x^N)); Vec(1/(1-x^6/(1-x)^7))
  • PARI
    a(n) = sum(k=0, n\6, binomial(n-1+k, n-6*k));

Formula

G.f. (1-x)^7/((1-x)^7-x^6).
a(n) = A017847(7*n-6) = Sum_{k=0..floor((7*n-6)/6)} binomial(k,7*n-6-6*k) for n > 0.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 6*a(n-6) + a(n-7) for n > 7.
a(n) = Sum_{k=0..floor(n/6)} binomial(n-1+k,n-6*k).
a(n) = A373912(n)-A373912(n-1). - R. J. Mathar, Jun 24 2024

A323339 Numerator of the sum of inverse products of parts in all compositions of n.

Original entry on oeis.org

1, 1, 3, 7, 11, 347, 3289, 1011, 38371, 136553, 4320019, 12528587, 40771123, 29346499543, 129990006917, 1927874590951, 903657004321, 437445829053473, 12456509813711881, 187206004658210129, 1974369484466728177, 1967745662306280217, 21401375717067880189
Offset: 0

Views

Author

Alois P. Heinz, Jan 11 2019

Keywords

Comments

Numerators of the INVERT transform of reciprocal integers.

Examples

			1/1, 1/1, 3/2, 7/3, 11/3, 347/60, 3289/360, 1011/70, 38371/1680, 136553/3780, 4320019/75600, 12528587/138600, 40771123/285120, ... = A323339/A323340

Links

Crossrefs

Denominators: A323340.
Cf. A000142, A007840, A011782, A088305, A177208, A177209, A322364, A322365, A322380, A322381, A323290, A323291.

Programs

  • Maple
    b:= proc(n) option remember;
     `if`(n=0, 1, add(b(n-j)/j, j=1..n))
    end:
a:= n-> numer(b(n)):
seq(a(n), n=0..25);
  • Mathematica
    nmax = 20; Numerator[CoefficientList[Series[1/(1 + Log[1-x]), {x, 0, nmax}], x]] (* Vaclav Kotesovec, Feb 12 2024 *)

Formula

G.f. for fractions: 1 / (1 + log(1 - x)). - Ilya Gutkovskiy, Nov 12 2019
a(n) = numerator( A007840(n)/n! ). - Alois P. Heinz, Jan 04 2024
A323339(n)/A323340(n) ~ exp(n) / (exp(1) - 1)^(n+1). - Vaclav Kotesovec, Feb 12 2024

A147600 Expansion of 1/(1 - 3*x^2 + x^4).

Original entry on oeis.org

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

Views

Author

Roger L. Bagula, Nov 08 2008

Keywords

Comments

S(n,sqrt(5)), with the Chebyshev polynomials A049310, is an integer sequence in the real quadratic number field Q(sqrt(5)) with basis numbers <1,phi>, phi:=(1+sqrt(5))/2. S(n,sqrt(5)) = A(n) + 2*B(n)*phi, with A(n) = A005013(n+1)*(-1)^n and B(n) = a(n-1), n>=0, with a(-1)=0. - Wolfdieter Lang, Nov 24 2010
The sequence (s(n)) given by s(0) = 0 and s(n) = a(n-1) for n > 0 is the p-INVERT of (0, 1, 0, 1, 0, 1, ...) using p(S) = 1 - S^2; see A291219. - Clark Kimberling, Aug 30 2017
From Jean-François Alcover, Sep 24 2017: (Start)
Consider this array of successive differences:
0, 0, 0, 1, 0, 3, 0, 8, 0, 21, ...
0, 0, 1, -1, 3, -3, 8, -8, 21, -21, ...
0, 1, -2, 4, -6, 11, -16, 29, -42, 76, ...
1, -3, 6, -10, 17, -27, 45, -71, 118, -186, ...
-4, 9, -16, 27, -44, 72, -116, 189, -304, 495, ...
13, -25, 43, -71, 116, -188, 305, -493, 799, -1291, ...
-38, 68, -114, 187, -304, 493, -798, 1292, -2090, 3383, ...
...
First row = even-index Fibonacci numbers with interleaved zeros = this sequence right-shifted 3 positions.
Main diagonal = 0,0,-2,-10,-44,-188,-798,... = -A099919 right-shifted.
First upper subdiagonal = 0,1,4,17,72,305,1292,... = A001076 right-shifted.
Second upper subdiagonal = 0,-1,-6,-27,-116,-493,-2090,... = -A049651.
Third upper subdiagonal = 1,3,11,45,189,799,3383,... = A292278.
(End) (Comment based on an e-mail from Paul Curtz)

Examples

			G.f. = 1 + 3*x^2 + 8*x^4 + 21*x^6 + 55*x^8 + 144*x^10 + 377*x^12 + 987*x^14 + ...

Links

Crossrefs

Cf. A000045, A001076, A001906, A005013, A049310.
Cf. A049651, A088305, A099919, A291219, A292278.

Programs

  • Magma
    [(1+(-1)^n)*Fibonacci(n+2)/2: n in [0..60]]; // G. C. Greubel, Oct 25 2022
  • Mathematica
    f[x_]= -1 -x +x^2; CoefficientList[Series[-1/(x^2*f[x]*f[1/x]), {x,0,60}], x]
(* or *)
M={{0,1,0,0}, {0,0,1,0}, {0,0,0,1}, {-1,0,3,0}}; v[0]= {1,0,3,0}; v[n_]:= v[n]= M.v[n-1]; Table[v[n][[1]], {n,0,60}]
LinearRecurrence[{0,3,0,-1}, {1,0,3,0}, 60] (* Jean-François Alcover, Sep 23 2017 *)
  • PARI
    Vec(1/(1 - 3*x^2 + x^4)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012
  • SageMath
    [((n+1)%2)*fibonacci(n+2) for n in range(60)] # G. C. Greubel, Oct 25 2022

Formula

O.g.f.: 1/(1 - 3*x^2 + x^4).
a(2*k) = F(2*(k+1)), a(2*k+1) = 0, k>=0, with F(n)=A000045(n). - Richard Choulet, Nov 13 2008
a(n) + a(n-1) + a(n-2) = A005013(n + 1). - Michael Somos, Apr 13 2012
a(n) = (2^(-2-n)*((1 + (-1)^n)*((-3+sqrt(5))*(-1+sqrt(5))^n + (1+sqrt(5))^n*(3+sqrt(5)))))/sqrt(5). - Colin Barker, Mar 28 2016

A368475 Expansion of o.g.f. (1-x)^5/((1-x)^5 - x^4).

Original entry on oeis.org

1, 0, 0, 0, 1, 5, 15, 35, 71, 136, 265, 550, 1211, 2732, 6126, 13485, 29191, 62648, 134408, 289656, 627401, 1363124, 2963186, 6434484, 13951852, 30221185, 65442625, 141745045, 307137901, 665732417, 1443184210, 3128438335, 6780867186, 14696002913, 31848721632
Offset: 0

Views

Author

Enrique Navarrete, Dec 26 2023

Keywords

Comments

For n > 0, a(n) is the number of ways to split [n] into an unspecified number of intervals and then choose 4 blocks (i.e., subintervals) from each interval. For example, for n=12, a(12)=1211 since the number of ways to split [12] into intervals and then select 4 blocks from each interval is C(12,4) + C(8,4)*C(4,4) + C(7,4)*C(5,4) + C(6,4)*C(6,4) + C(5,4)*C(7,4) + C(4,4)*C(8,4) + C(4,4)*C(4,4)*C(4,4) for a total of 1211 ways.
For n > 0, a(n) is also the number of compositions of n using parts of size at least 4 where there are binomial(i,4) types of i, i >= 4 (see example).
Number of compositions of 5*n-4 into parts 4 and 5. - Seiichi Manyama, Feb 01 2024

Examples

			Since there are C(4,4) = 1 type of 4, C(5,4) = 5 types of 5, C(6,4) = 15 types of 6, C(7,4) = 35 types of 7, C(8,4) = 70 types of 8, and (12,4) = 495 types of 12, we can write 12 in the following ways:
  12: 495 ways;
  8+4: 70 ways;
  7+5: 175 ways;
  6+6: 225 ways;
  5+7: 175 ways;
  4+8: 70 ways;
  4+4+4: 1 way, for a total of 1211 ways.

Links

Crossrefs

Cf. A000332, A017827, A099131, A290998.
Cf. A079675, A369803, A369804.
Cf. A088305, A095263, A290998, A369794, A369809.

Programs

  • Mathematica
    CoefficientList[Series[(1 - x)^5/((1 - x)^5 - x^4), {x, 0, 50}], x] (* Wesley Ivan Hurt, Dec 26 2023 *)
  • PARI
    Vec((1-x)^5/((1-x)^5 - x^4) + O(x^40)) \\ Michel Marcus, Dec 27 2023

Formula

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 4*a(n-4) + a(n-5), n>=6; a(0)=1, a(1)=a(2)=a(3)=0, a(4)=1, a(5)=5.
G.f.: 1/(1-Sum_{k>=4} binomial(k,4)*x^k).
G.f.: 1/p(S), where p(S) = 1 - S^4 - S^5 and S = x/(1-x).
First differences of A099131. - R. J. Mathar, Jan 29 2024
a(n) = A017827(5*n-4) = Sum_{k=0..floor((5*n-4)/4)} binomial(k,5*n-4-4*k) for n > 0. - Seiichi Manyama, Feb 01 2024
a(n) = Sum_{k=0..floor(n/4)} binomial(n-1+k,n-4*k). - Seiichi Manyama, Feb 02 2024
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