A049670 -id:A049670 - cp's OEIS frontend

A097843 First differences of Chebyshev polynomials S(n,123) = A049670(n+1) with Diophantine property.

1, 122, 15005, 1845493, 226980634, 27916772489, 3433536035513, 422297015595610, 51939099382224517, 6388086926998019981, 785682752921374233146, 96632590522402032656977, 11885022951502528642575025, 1461761190444288621004071098, 179784741401695997854858170029

Offset: 0

Wolfdieter Lang, Sep 10 2004

(11*b(n))^2 - 5*(5*a(n))^2 = -4 with b(n)=A097842(n) give all positive solutions of this Pell equation.

			All positive solutions of Pell equation x^2 - 125*y^2 = -4 are (11 = 11*1,1), (1364 = 11*124,122), (167761 = 11*15251,15005), (20633239 = 11*1875749,1845493), ...

Colin Barker, Table of n, a(n) for n = 0..478
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (123,-1).

a:=[1,122];; for n in [3..20] do a[n]:=123*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2019

m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-123*x+x^2) )); // G. C. Greubel, Jan 14 2019

LinearRecurrence[{123,-1}, {1,122}, 20] (* G. C. Greubel, Jan 14 2019 *)

Vec((1-x)/(1-123*x+x^2) + O(x^30)) \\ Colin Barker, Jun 15 2015

((1-x)/(1-123*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 14 2019

a(n) = ((-1)^n)*S(2*n, 11*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.
G.f.: (1-x)/(1-123*x+x^2).
a(n) = S(n, 123) - S(n-1, 123) = T(2*n+1, 5*sqrt(5)/2)/(5*sqrt(5)/2), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x) and T(n, x) Chebyshev's polynomials of the first kind, A053120.
a(n) = 123*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=122. - Philippe Deléham, Nov 18 2008
a(n) = (F(10*(n+1)) - F(10*n))/F(10), with F=A000045 (Fibonacci). F(10*n)/F(10) = A049670. - Wolfdieter Lang, Oct 11 2012
a(n) = (1/5)*F(10*n + 5). Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/11^2. Compare with A001519 and A007805. - Peter Bala, Nov 29 2013
From Peter Bala, Mar 23 2015: (Start)
a(n) = A049666(2*n + 1).
a(n) = ( Fibonacci(10*n + 10 - 2*k) - Fibonacci(10*n + 2*k) )/( Fibonacci(10 - 2*k) - Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(10*n + 10 - 2*k - 1) + Fibonacci(10*n + 2*k + 1) )/( Fibonacci(10 - 2*k - 1) + Fibonacci(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n))n>=1 = [1, 0, 122, 0, 15005, 0, 1845493, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -125, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)

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

N. J. A. Sloane

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

			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

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Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042
Index entries for "core" sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-1).
Index entries for two-way infinite sequences

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.

a001906 n = a001906_list !! n
a001906_list =
   0 : 1 : zipWith (-) (map (* 3) $ tail a001906_list) a001906_list
-- Reinhard Zumkeller, Oct 03 2011

[Fibonacci(2*n): n in [0..30]]; // Vincenzo Librandi, Sep 10 2014

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

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 *)

makelist(fib(2*n),n,0,30); /* Martin Ettl, Oct 21 2012 */

numlib::fibonacci(2*n) $ n = 0..35; // Zerinvary Lajos, May 09 2008

{a(n) = fibonacci(2*n)}; /* Michael Somos, Dec 06 2002 */

{a(n) = subst( poltchebi(n+1)*4 - poltchebi(n)*6, x, 3/2)/5}; /* Michael Somos, Dec 06 2002 */

{a(n) = polchebyshev( n-1, 2, 3/2)}; /* Michael Somos Jun 18 2011 */

Vec(x/(1-3*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Oct 24 2012

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

[lucas_number1(n,3,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

[fibonacci(2*n) for n in range(0, 28)] # Zerinvary Lajos, May 15 2009

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

A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 14, 14, 6, 1, 42, 48, 27, 8, 1, 132, 165, 110, 44, 10, 1, 429, 572, 429, 208, 65, 12, 1, 1430, 2002, 1638, 910, 350, 90, 14, 1, 4862, 7072, 6188, 3808, 1700, 544, 119, 16, 1, 16796, 25194, 23256, 15504, 7752, 2907, 798, 152, 18, 1

Offset: 0

N. J. A. Sloane

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges. - Emeric Deutsch, Jan 15 2005
Riordan array ((1-2x-sqrt(1-4x))/(2x^2),(1-2x-sqrt(1-4x))/(2x)). Inverse array is A053122. - Paul Barry, Mar 17 2005
T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper half-plane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW. - Emeric Deutsch, Apr 15 2005
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) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 30 2007
Number of (2n+1)-step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,-1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu. - Philippe Deléham, Apr 16 2007, Apr 18 2007
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 two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths. - Philippe Deléham, Sep 25 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
With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)-1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008
T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3. - Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008
The signed triangle S(n,k) := (-1)^(n-k)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1 - 5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n- and k-summation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1 - c(-x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1-L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n) - 5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(-x) = 1 + 2*x - (x*c(-x))^2. - Wolfdieter Lang, Aug 27 2012
O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:
  ((1+x) - C(z))/(x - (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)-1)/x,C(x)-1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section. - Wolfdieter Lang, Nov 13 2012
The A-sequence for this Riordan triangle is [1,2,1] and the Z-sequence is [2,1]. See a W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 13 2012
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)-th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even-indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310): rho(N)^(2*n+1) = Sum_{k=0..n} T(n, k)*R(N, 2*(k+1)), n >= 0, identical in N >= 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears. For the even powers of rho(n) see A039599. (End)
The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122). -  Tom Copeland, Dec 11 2015 (revised Dec 28 2015)
T(n,k) is the number of pairs of non-intersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976. - Peter Bala, Apr 12 2017
Also the convolution triangle of the Catalan numbers A000108. - Peter Luschny, Oct 07 2022

			Triangle T(n,k) starts:
n\k     0      1      2      3      4     5    6    7   8  9 10
0:      1
1:      2      1
2:      5      4      1
3:     14     14      6      1
4:     42     48     27      8      1
5:    132    165    110     44     10     1
6:    429    572    429    208     65    12    1
7:   1430   2002   1638    910    350    90   14    1
8:   4862   7072   6188   3808   1700   544  119   16   1
9:  16796  25194  23256  15504   7752  2907  798  152  18  1
10: 58786  90440  87210  62016  33915 14364 4655 1120 189 20  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012.
Production matrix begins:
2, 1
1, 2, 1
0, 1, 2, 1
0, 0, 1, 2, 1
0, 0, 0, 1, 2, 1
0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 0, 1, 2, 1
- _Philippe Deléham_, Nov 07 2011
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: T(5,1) = 165 = 1*42 + 2*48 +1*27. The Riordan A-sequence is [1,2,1].
Recurrence from Riordan Z-sequence [2,1]: T(5,0) = 132 = 2*42 + 1*48. (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
  Example for rho(N) = 2*cos(Pi/N) powers:
  n=2: rho(N)^5 = 5*R(N, 2) + 4*R(N, 4) + 1*R(N, 6) = 5*S(1, rho(N)) + 4*S(3, rho(N)) + 1*S(5, rho(N)), identical in N >= 1. For N=5 (the pentagon with only one distinct diagonal) the degree delta(5) = 2, hence R(5, 4) and R(5, 6) can be reduced, namely to R(5, 1) = 1 and R(5, 6) = -R(5,1) = -1, respectively. Thus rho(5)^5 = 5*R(N, 2) + 4*1  + 1*(-1) = 3 + 5*R(N, 2) = 3 + 5*rho(5), with the golden section rho(5). (End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
José Agapito, Ângela Mestre, Maria M. Torres, and Pasquale Petrullo, On One-Parameter Catalan Arrays, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.1.
M. Aigner, Enumeration via ballot numbers, Discrete Math., 308 (2008), 2544-2563.
Quang T. Bach and Jeffrey B. Remmel, Generating functions for descents over permutations which avoid sets of consecutive patterns, arXiv:1510.04319 [math.CO], 2015 (see p. 25).
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Jean-Luc Baril, José L. Ramírez, and Lina M. Simbaqueba, Counting prefixes of skew Dyck paths, J. Int. Seq., Vol. 24 (2021), Article 21.8.2.
Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.
Paul Barry, Generalized Catalan Numbers Associated with a Family of Pascal-like Triangles, J. Int. Seq., Vol. 22 (2019), Article 19.5.8.
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Paul Barry, Chebyshev moments and Riordan involutions, arXiv:1912.11845 [math.CO], 2019.
Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
B. A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 29.
Eduardo H. M. Brietzke, Generalization of an identity of Andrews, Fibonacci Quart. 44 (2006), no. 2, 166-171.
Eduardo H. M. Brietzke, Recurrence Relations for Sums of Binomial Coefficients and Some Generalizations, Journal of Integer Sequences, Vol. 27 (2024), Article 24.3.4.
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Naiomi T. Cameron and Asamoah Nkwanta, On Some (Pseudo) Involutions in the Riordan Group, Journal of Integer Sequences, Vol. 8 (2005), Article 05.3.7.
Marc Chamberland, Factored matrices can generate combinatorial identities, Linear Algebra and its Applications, Volume 438, Issue 4, 15 Feb. 2013, pp. 1667-1677.
Xi Chen, H. Liang, and Y. Wang, Total positivity of recursive matrices, arXiv:1601.05645 [math.CO], 2016.
Xi Chen, H. Liang, and Y. Wang, Total positivity of recursive matrices, Linear Algebra and its Applications, Volume 471, Apr 15 2015, Pages 383-393.
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 7.
S. J. Cyvin, J. Brunvoll, E. Brendsdal, B. N. Cyvin, and E. K. Lloyd, Enumeration of polyene hydrocarbons: a complete mathematical solution, J. Chem. Inf. Comput. Sci., 35 (1995) 743-751. [Annotated scanned copy]
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Rigoberto Flórez, Leandro Junes, Luisa M. Montoya, and José L. Ramírez, Counting Subwords in Non-Decreasing Dyck Paths, J. Int. Seq. (2025) Vol. 28, Art. No. 25.1.6. See p. 19.
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Mirror image of A050166. Row sums are A001700.

Cf. A000108, A008313, A039599, A183134, A094527, A033184, A035324, A053122, A172417.

/* As triangle: */ [[Binomial(2*n,n-k) - Binomial(2*n,n-k-2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2015

T:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); # N. J. A. Sloane, Aug 26 2013
# Uses function PMatrix from A357368. Adds row and column above and to the left.
PMatrix(10, n -> binomial(2*n, n) / (n + 1)); # Peter Luschny, Oct 07 2022

Flatten[Table[Binomial[2n, n-k] - Binomial[2n, n-k-2], {n,0,9}, {k,0,n}]] (* Jean-François Alcover, May 03 2011 *)

T(n,k)=binomial(2*n,n-k) - binomial(2*n,n-k-2) \\ Charles R Greathouse IV, Nov 07 2016

# Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle.
def A039598_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True; h = 1
    for i in range(2*n) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        b = not b
        if b : print([D[z] for z in (1..h-1) ])
A039598_triangle(10)  # Peter Luschny, May 01 2012

Row n: C(2n, n-k) - C(2n, n-k-2).
a(n, k) = C(2n+1, n-k)*2*(k+1)/(n+k+2) = A050166(n, n-k) = a(n-1, k-1) + 2*a(n-1, k)+ a (n-1, k+1) [with a(0, 0) = 1 and a(n, k) = 0 if n<0 or nHenry Bottomley, Sep 24 2001
From Philippe Deléham, Feb 14 2004: (Start)
T(n, 0) = A000108(n+1), T(n, k) = 0 if n0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+2) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
Sum_{k>=0} T(m, k)*T(n, k) = A000108(m+n+1). (End)
T(n, k) = A009766(n+k+1, n-k) = A033184(n+k+2, 2k+2). - Philippe Deléham, Feb 14 2004
Sum_{j>=0} T(k, j)*A039599(n-k, j) = A028364(n, k). - Philippe Deléham, Mar 04 2004
Antidiagonal Sum_{k=0..n} T(n-k, k) = A000957(n+3). - Gerald McGarvey, Jun 05 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 [2,2,2,...] in the main diagonal. - Gary W. Adamson, Dec 17 2006
G.f.: G(t,x) = C^2/(1-txC^2), where C = (1-sqrt(1-4x))/(2x) is the Catalan function. From here G(-1,x)=C, i.e., the alternating row sums are the Catalan numbers (A000108). - Emeric Deutsch, Jan 20 2007
Sum_{k=0..n} T(n,k)*x^k = A000957(n+1), A000108(n), A000108(n+1), A001700(n), A049027(n+1), A076025(n+1), A076026(n+1) for x=-2,-1,0,1,2,3,4 respectively (see square array in A067345). - Philippe Deléham, Mar 21 2007, Nov 04 2011
Sum_{k=0..n} T(n,k)*(k+1) = 4^n. - Philippe Deléham, Mar 30 2007
Sum_{j>=0} T(n,j)*binomial(j,k) = A035324(n,k), A035324 with offset 0 (0 <= k <= n). - Philippe Deléham, Mar 30 2007
T(n,k) = A053121(2*n+1,2*k+1). - Philippe Deléham, Apr 16 2007, Apr 18 2007
T(n,k) = A039599(n,k) + A039599(n,k+1). - Philippe Deléham, Sep 11 2007
Sum_{k=0..n+1} T(n+1,k)*k^2 = A029760(n). - Philippe Deléham, Dec 16 2007
Sum_{k=0..n} T(n,k)*A059841(k) = A000984(n). - Philippe Deléham, Nov 12 2008
G.f.: 1/(1-xy-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-.... (continued fraction).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A001700(n), A194723(n+1), A194724(n+1), A194725(n+1), A194726(n+1), A194727(n+1), A194728(n+1), A194729(n+1), A194730(n+1) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Nov 03 2011
From Peter Bala, Dec 21 2014: (Start)
This triangle factorizes in the Riordan group as ( C(x), x*C(x) ) * ( 1/(1 - x), x/(1 - x) ) = A033184 * A007318, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
Let U denote the lower unit triangular array with 1's on or below the main diagonal and zeros elsewhere. For k = 0,1,2,... define U(k) to be the lower unit triangular block array
/I_k 0\
\ 0  U/ having the k X k identity matrix I_k as the upper left block; in particular, U(0) = U. Then this array equals the bi-infinite product (...*U(2)*U(1)*U(0))*(U(0)*U(1)*U(2)*...). (End)
From Peter Bala, Jul 21 2015: (Start)
O.g.f. G(x,t) = (1/x) * series reversion of ( x/f(x,t) ), where f(x,t) = ( 1 + (1 + t)*x )^2/( 1 + t*x ).
1 + x*d/dx(G(x,t))/G(x,t) = 1 + (2 + t)*x + (6 + 4*t + t^2)*x^2 + ... is the o.g.f for A094527. (End)
Conjecture: Sum_{k=0..n} T(n,k)/(k+1)^2 = H(n+1)*A000108(n)*(2*n+1)/(n+1), where H(n+1) = Sum_{k=0..n} 1/(k+1). - Werner Schulte, Jul 23 2015
From Werner Schulte, Jul 25 2015: (Start)
Sum_{k=0..n} T(n,k)*(k+1)^2 = (2*n+1)*binomial(2*n,n). (A002457)
Sum_{k=0..n} T(n,k)*(k+1)^3 = 4^n*(3*n+2)/2.
Sum_{k=0..n} T(n,k)*(k+1)^4 = (2*n+1)^2*binomial(2*n,n).
Sum_{k=0..n} T(n,k)*(k+1)^5 = 4^n*(15*n^2+15*n+4)/4. (End)
The o.g.f. G(x,t) is such that G(x,t+1) is the o.g.f. for A035324, but with an offset of 0, and G(x,t-1) is the o.g.f. for A033184, again with an offset of 0. - Peter Bala, Sep 20 2015
Denote this lower triangular array by L; then L * transpose(L) is the Cholesky factorization of the Hankel matrix ( 1/(i+j)*binomial(2*i+2*j-2, i+j-1) )A172417%20read%20as%20a%20square%20array.%20See%20Chamberland,%20p.%201669.%20-%20_Peter%20Bala">i,j >= 1 = A172417 read as a square array. See Chamberland, p. 1669. - _Peter Bala, Oct 15 2023

Typo in one entry corrected by Philippe Deléham, Dec 16 2007

A049660 a(n) = Fibonacci(6*n)/8.

Original entry on oeis.org

0, 1, 18, 323, 5796, 104005, 1866294, 33489287, 600940872, 10783446409, 193501094490, 3472236254411, 62306751484908, 1118049290473933, 20062580477045886, 360008399296352015, 6460088606857290384, 115921586524134874897, 2080128468827570457762

Offset: 0

Clark Kimberling

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 18's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,17}. - Milan Janjic, Jan 25 2015
10*a(n)^2 = Tri(4)*S(n-1, 18)^2 is the triangular number Tri((T(n, 9) - 1)/2), with Tri, S and T given in A000217, A049310 and A053120. This is instance k = 4 of the k-family of identities given in a comment on A001109. - Wolfdieter Lang, Feb 01 2016
Possible solutions for y in Pell equation x^2 - 80*y^2 = 1. The values for x are given in A023039. - Herbert Kociemba, Jun 05 2022

			a(3) = F(6 * 3) / 8 = F(18) / 8 = 2584 / 8 = 323. - _Indranil Ghosh_, Feb 06 2017

Indranil Ghosh, Table of n, a(n) for n = 0..796 (terms 0..200 from Vincenzo Librandi)
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Column m=6 of array A028412.
Partial sums of A007805.
Cf. A000045, A001076, A001077, A014445, A014448, A015448, A099843, A134492, A155179.

[Fibonacci(6*n)/8: n in [0..30]]; // G. C. Greubel, Dec 02 2017

with (combinat):seq(fibonacci(2*n,4)/4, n=0..16); # Zerinvary Lajos, Apr 20 2008

Fibonacci[6*Range[0,20]]/8 (* Harvey P. Dale, Nov 23 2011 *)
LinearRecurrence[{18,-1}, {0,1}, 30] (* G. C. Greubel, Dec 02 2017 *)
Table[ChebyshevU[-1 + n, 9], {n, 0, 18}] (* Herbert Kociemba, Jun 05 2022 *)

numlib::fibonacci(6*n)/8 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(6*n)/8  \\ Charles R Greathouse IV, Apr 17 2012

[lucas_number1(n,18,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008

[fibonacci(6*n)/8 for n in range(0, 17)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1 - 18*x + x^2).
a(n) = A134492(n)/8.
a(n) ~ (1/40)*sqrt(5)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all terms k of the sequence, 80*k^2 + 1 is a square. Limit_{n->oo} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 14 2002
a(n) = S(n-1, 18) with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n))/(8*sqrt(5)).
a(n) = sqrt((A023039(n)^2 - 1)/80) (cf. Richardson comment).
a(n) = 18*a(n-1) - a(n-2). - Gregory V. Richardson, Oct 14 2002
a(n) = A001076(2n)/4.
a(n) = 17*(a(n-1) + a(n-2)) - a(n-3) = 19*(a(n-1) - a(n-2)) + a(n-3). - Mohamed Bouhamida, May 26 2007
a(n+1) = Sum_{k=0..n} A101950(n,k)*17^k. - Philippe Deléham, Feb 10 2012
Product_{n>=1} (1 + 1/a(n)) = (1/2)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = (2/9)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
a(n) = (1/32)*(F(6*n + 3) - F(6*n - 3)).
Sum_{n>=1} 1/(4*a(n) + 1/(4*a(n))) = 1/4. Compare with A001906 and A049670. - Peter Bala, Nov 29 2013
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -G(x)*G(-x), where G(x) = Sum_{n >= 1} A001076(n)*x^n.
1 + 4*Sum_{n >= 1} a(n)*x^(2*n) = (1 + F(x))*(1 + F(-x)) = (1 + 2*x*G(x))*(1 - 2*x*G(-x)), where F(x) = Sum_{n >= 1} Fibonacci(3*n + 3)*x^n.
1 + 7*Sum_{n >= 1} a(n)*x^(2*n) = (1 + G(x))*(1 + G(-x)) = (1 + 7*G(x))*(1 + 7*G(-x)).
1 + 12*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*G(x))*(1 + 2*G(-x)) = (1 + 6*G(x))*(1 + 6*G(-x)) = (1 + A(x))*(1 + A(-x)), where A(x) = Sum_{n >= 1} Fibonacci(3*n)*x^n is the o.g.f for A014445.
1 + 15*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 5*G(x))*(1 + 5*G(-x)) = (1 + 3*G(x))*(1 + 3*G(-x)) = H(x)*H(-x), where H(x) = Sum_{n >= 0} A155179(n)*x^n.
1 + 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 4*G(x))*(1 + 4*G(-x)) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*(-x)^n) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*(-x)^n).
1 + 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} Lucas(3*n)*x^n)*(1 + Sum_{n >= 1} Lucas(3*n)*(-x)^n).
1 - 5*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} A001077(n+1)*x^n)*(1 + Sum_{n >= 1} A001077(n+1)*(-x)^n).
1 - 9*Sum_{n >= 1} a(n)*x^(2*n) = (1 - G(x))*(1 - G(-x)) = (1 + 9*G(x))*(1 + 9*G(-x)).
1 - 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*Sum_{n >= 1} A099843(n)*x^n)*(1 + 2*Sum_{n >= 1} A099843(n)*(-x)^n).
1 - 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 - 2*G(x))*(1 - 2*G(-x)) = (1 + 10*G(x))*(1 + 10*G(-x)).
(End)

Chebyshev and other comments from Wolfdieter Lang, Nov 08 2002

A049666 a(n) = Fibonacci(5*n)/5.

Original entry on oeis.org

0, 1, 11, 122, 1353, 15005, 166408, 1845493, 20466831, 226980634, 2517253805, 27916772489, 309601751184, 3433536035513, 38078498141827, 422297015595610, 4683345669693537, 51939099382224517, 576013438874163224

Offset: 0

Clark Kimberling

For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 11's along the main diagonal and 1's along the subdiagonal and the superdiagonal. - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,11} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
For n >= 1, a(n) equals the denominator of the continued fraction [11, 11, ..., 11] (with n copies of 11). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 26 2019
From Michael A. Allen, Mar 30 2023: (Start)
Also called the 11-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 11 kinds of squares available. (End)

			G.f. = x + 11*x^2 + 122*x^3 + 1353*x^4 + 15005*x^5 + 166408*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..950
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (11,1).

A column of array A028412.
Cf. A000045, A102312, A243399.
Row n=11 of A073133, A172236 and A352361, and column k=11 of A157103.

[Fibonacci(5*n)/5: n in [0..30]]; // G. C. Greubel, Dec 02 2017

A049666 := proc(n)
    combinat[fibonacci](5*n)/5 ;
end proc: # R. J. Mathar, May 07 2024

Table[Fibonacci[5*n]/5, {n, 0, 100}] (* T. D. Noe, Oct 29 2009 *)
a[ n_] := Fibonacci[n, 11]; (* Michael Somos, May 28 2014 *)

numlib::fibonacci(5*n)/5 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(5*n)/5 \\ Charles R Greathouse IV, Feb 03 2014

from sage.combinat.sloane_functions import recur_gen3
it = recur_gen3(0,1,11,11,1,0)
[next(it) for i in range(1,22)] # Zerinvary Lajos, Jul 09 2008

[lucas_number1(n,11,-1) for n in range(0, 19)] # Zerinvary Lajos, Apr 27 2009

[fibonacci(5*n)/5 for n in range(0, 19)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1 - 11*x - x^2).
a(n) = A102312(n)/5.
a(n) = 11*a(n-1) + a(n-2) for n > 1, a(0)=0, a(1)=1. With a=golden ratio and b=1-a, a(n) = (a^(5n)-b^(5n))/(5*sqrt(5)). - Mario Catalani (mario.catalani(AT)unito.it), Jul 24 2003
a(n) = F(n, 11), the n-th Fibonacci polynomial evaluated at x=11. - T. D. Noe, Jan 19 2006
a(n) = ((11+sqrt(125))^n-(11-sqrt(125))^n)/(2^n*sqrt(125)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n) = 11*A049670(n), a(2n+1) = A097843(n).
a(3n+1) = A041227(5n), a(3n+2) = A041227(5n+3), a(3n+3) = 2*A041227(5n+4).
Limit_{k->oo} a(n+k)/a(k) = (A001946(n) + A049666(n)*sqrt(125))/2.
Limit_{n->oo} A001946(n)/A049666(n) = sqrt(125).
(End)
a(n) = F(n) + (-1)^n*5*F(n)^3 + 5*F(n)^5, n >= 0. See the D. Jennings formula given in a comment on A111125, where also the reference is given. - Wolfdieter Lang, Aug 31 2012
a(-n) = -(-1)^n * a(n). - Michael Somos, May 28 2014
E.g.f.: (exp((1/2)*(11-5*sqrt(5))*x)*(-1 + exp(5*sqrt(5)*x)))/(5*sqrt(5)). - Stefano Spezia, Aug 02 2019

A028412 Rectangular array of numbers Fibonacci(m(n+1))/Fibonacci(m), m >= 1, n >= 0, read by downward antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 4, 8, 3, 1, 7, 17, 21, 5, 1, 11, 48, 72, 55, 8, 1, 18, 122, 329, 305, 144, 13, 1, 29, 323, 1353, 2255, 1292, 377, 21, 1, 47, 842, 5796, 15005, 15456, 5473, 987, 34, 1, 76, 2208, 24447, 104005, 166408, 105937, 23184, 2584, 55, 1, 123, 5777

Offset: 0

N. J. A. Sloane

Every integer-valued quotient of two Fibonacci numbers is in this array. - Clark Kimberling, Aug 28 2008

Not only does 5 divide row 5, but 50 divides (-5 + row 5), as in A214984. - Clark Kimberling, Nov 02 2012

			   1   1    1      1       1        1
   1   3    4      7      11       18
   2   8   17     48     122      323
   3  21   72    329    1353     5796
   5  55  305   2255   15005   104005
   8 144 1292  15456  166408  1866294
  13 377 5473 105937 1845493 33489287
  ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 142.

Clark Kimberling, Table of n, a(n) for n = 0..1829
I. Strazdins, Lucas factors and a Fibonomial generating function, in Applications of Fibonacci numbers, Vol. 7 (Graz, 1996), 401-404, Kluwer Acad. Publ., Dordrecht, 1998.

Columns include A000045, A001906, A001076, A004187, A049666, A049660, A049667, A049668, A049669, A049670.
Rows include (essentially) A000032, A047946, A083564, A103226.
Main diagonal is A051294.
Transpose is A214978.

max = 11; col[m_] := CoefficientList[ Series[ 1/(1 - LucasL[m]*x + (-1)^m*x^2), {x, 0, max}], x]; t = Transpose[ Table[ col[m], {m, 1, max}]] ; Flatten[ Table[ t[[n - m + 1, m]], {n, 1, max }, {m, n, 1, -1}]] (* Jean-François Alcover, Feb 21 2012, after Paul D. Hanna *)
f[n_] := Fibonacci[n]; t[m_, n_] := f[m*n]/f[n]
TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]] (* array *)
t = Flatten[Table[t[k, n + 1 - k], {n, 1, 120}, {k, 1, n}]] (* sequence *) (* Clark Kimberling, Nov 02 2012 *)

{T(n,m)=polcoeff(1/(1 - Lucas(m)*x + (-1)^m*x^2 +x*O(x^n)),n)}

T(n, m) = Sum_{i_1>=0} Sum_{i_2>=0} ... Sum_{i_m>=0} C(n-i_m, i_1)*C(n-i_1, i_2)*C(n-i_2, i_3)*...*C(n-i_{m-1}, i_m).

G.f. for column m >= 1: 1/(1 - Lucas(m)*x + (-1)^m*x^2), where Lucas(m) = A000204(m). - Paul D. Hanna, Jan 28 2012

More terms from Erich Friedman, Jun 03 2001
Edited by Ralf Stephan, Feb 03 2005
Better description from Clark Kimberling, Aug 28 2008

A065705 a(n) = Lucas(10*n).

Original entry on oeis.org

2, 123, 15127, 1860498, 228826127, 28143753123, 3461452808002, 425730551631123, 52361396397820127, 6440026026380244498, 792070839848372253127, 97418273275323406890123, 11981655542024930675232002, 1473646213395791149646646123, 181246502592140286475862241127

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 25 2003

Lim_{n->infinity} a(n+1)/a(n) = (123 + sqrt(15125))/2 = 122.9918693812...
Lim_{n->infinity} a(n)/a(n+1) = (123 - sqrt(15125))/2 = 0.00813061875578...
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^10) = 1.0081300769... = 1 + 1/(123 + 1/(15127 + 1/(1860498 + ...))).
Also F(-Phi^10) = 0.9918699143... has the continued fraction representation 1 - 1/(123 - 1/(15127 - 1/(1860498 - ...))) and the simple continued fraction expansion 1/(1 + 1/((123 - 2) + 1/(1 + 1/((15127 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + ...))))))).
F(Phi^10)*F(-Phi^10) = 0.9999338930... has the simple continued fraction expansion 1/(1 + 1/((123^2 - 4) + 1/(1 + 1/((15127^2 - 4) + 1/(1 + 1/((1860498^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^10)/F(-Phi^10) = 1.0081967213... has the simple continued fraction expansion 1 + 1/((123 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + 1/(28143753123 - 2) + 1/(1 + ...))))). (End)

			a(4) = 228826127 = 123*a(3) - a(2) = 123*1860498 - 15127=((123+sqrt(15125))/2)^4 + ( (123-sqrt(15125))/2)^4 =228826126.99999999562986 + 0.00000000437013 = 228826127.
a(4) = L(10 * 4) = L(40) = 228826127. - _Indranil Ghosh_, Feb 08 2017

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Indranil Ghosh, Table of n, a(n) for n = 0..477
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem,Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (123,-1).

Cf. A000032: a(n) = A000032(10*n).

Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(10*n): n in [0..90]]; // Vincenzo Librandi, Apr 14 2011

LucasL[10*Range[0, 20]] (* Paolo Xausa, Mar 04 2024 *)

a(n) = 123*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 123.
a(n) = ((123 + sqrt(15125))/2)^n + ((123 - sqrt(15125))/2)^n.
a(n)^2 = a(2*n) + 2.
G.f.: (2 - 123*x)/(1 - 123*x + x^2). -  Philippe Deléham, Nov 18 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(10*n+10)/F(10) - F(10*n-10)/F(10) = A049670(n+1) - A049670(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^10 = [34, 55; 55, 89].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
121*Sum_{n >= 1} 1/(a(n) - 125/a(n)) = 1: (125 = Lucas(10) + 2 and 121 = Lucas(10) - 2)
125*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 121/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 123*x^2 + 15128*x^3 + ... is the o.g.f. for A049670. (End)
E.g.f.: exp((1/2)*(123 - 55*sqrt(5))*x)*(1 + exp(55*sqrt(5)*x)). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^5 - 5*Lucas(2*n)^3 + 5*Lucas(2*n) = 2*T(5, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (123 - sqrt(15125))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(15125) - 123)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

A097842 Chebyshev polynomials S(n,123) + S(n-1,123) with Diophantine property.

Original entry on oeis.org

1, 124, 15251, 1875749, 230701876, 28374454999, 3489827263001, 429220378894124, 52790616776714251, 6492816643156958749, 798563656491529211876, 98216836931814936101999, 12079872378956745611334001, 1485726085774747895257980124, 182732228677915034371120221251, 22474578401297774479752529233749

Offset: 0

Wolfdieter Lang, Sep 10 2004

(11*a(n))^2 - 5*(5*b(n))^2 = -4 with b(n)=A097843(n) give all positive solutions of this Pell equation.

			All positive solutions of Pell equation x^2 - 125*y^2 = -4 are (11=11*1,1), (1364=11*124,122), (167761=11*15251,15005), (20633239=11*1875749,1845493), ...

Indranil Ghosh, Table of n, a(n) for n = 0..477
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (123,-1).
Index entries for sequences related to Chebyshev polynomials.

a:=[1,124];; for n in [3..20] do a[n]:=123*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019

m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-123*x+x^2) )); // G. C. Greubel, Jan 13 2019

CoefficientList[Series[(1+x)/(1-123x+x^2), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)
LinearRecurrence[{123,-1}, {1,124}, 20] (* G. C. Greubel, Jan 13 2019 *)

a(n)=polchebyshev(n, 2, 123/2) + polchebyshev(n - 1, 2, 123/2); \\ Michel Marcus, Aug 04 2017

my(x='x+O('x^20)); Vec((1+x)/(1-123*x+x^2)) \\ G. C. Greubel, Jan 13 2019

((1+x)/(1-123*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019

a(n) = S(n, 123) + S(n-1, 123) = S(2*n, 5*sqrt(5)), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x). S(n, 123) = A049670(n+1).
a(n) = (-2/11)*i*((-1)^n)*T(2*n+1, 11*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-123*x+x^2).
a(n) = 123*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=124. - Philippe Deléham, Nov 18 2008
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(10*n + 10 - 2*k) + Fibonacci(10*n + 2*k) )/( Fibonacci(10 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(10*n + 10 - 2*k - 1) - Fibonacci(10*n + 2*k + 1) )/( Fibonacci(10 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer, k != 2.
The aerated sequence (b(n))n>=1 = [1, 0, 124, 0, 15251, 0, 1875749, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -121, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = Lucas(10*n + 5)/11. - Ehren Metcalfe, Jul 29 2017

A056568 Fibonomial coefficients.

Original entry on oeis.org

1, 89, 12816, 1493064, 187628376, 22890661872, 2824135408458, 346934172869802, 42689423937884208, 5249543573067466872, 645693859487298425256, 79413089729752455762384, 9767258556969762111163771, 1201288963378036364032704659, 147748983166877427393815516256

Offset: 0

Wolfdieter Lang, Jul 10 2000

Alois P. Heinz, Table of n, a(n) for n = 0..150

Cf. A010048, A000045, A001654-8, A056565-7, A001906, A004187 (signed), A049660, A049668 (signed), A049670.

[&*[Fibonacci(n+i): i in [0..9]]/122522400: n in [1..15]]; // Vincenzo Librandi, Oct 31 2014

F:= combinat[fibonacci]: a:= n-> mul(F(n+i), i=0..9)/122522400: seq(a(n), n=1..18); # Zerinvary Lajos, Oct 07 2007
a:= n-> (Matrix(11, (i,j)-> if (i=j-1) then 1 elif j=1 then [1514513, -582505, -83215, 4895, 89, -1][abs(i-11/2)+1/2] else 0 fi)^n)[1, 1]; seq(a(n), n=0..18);  # Alois P. Heinz, Aug 15 2008

Times@@@Partition[Fibonacci[Range[30]],10,1]/122522400 (* Harvey P. Dale, Jul 27 2019 *)

a(n)=prod(k=0,9,fibonacci(n+k))/122522400; \\ Joerg Arndt, Oct 31 2014

a(n) = A010048(n+10,10) =: Fibonomial(n+10,10).
G.f.: 1/p(11,n) with p(11,n) = 1-89*x -4895*x^2 +83215*x^3 +582505*x^4 -1514513*x^5 -1514513*x^6 +582505*x^7 +83215*x^8 -4895*x^9 -89*x^10 +x^11 = (1+x) *(1-3*x+x^2) *(1+7*x+x^2) *(1-18*x+x^2) *(1+47*x+x^2) *(1-123*x+x^2) (n=8 row polynomial of signed Fibonomial triangle A055870; see this entry for Knuth and Riordan references).
Recursion: a(n)=123*a(n-1)-a(n-2)+((-1)^n)*A056566(n), n >= 2, a(0)=1, a(1)=89.
G.f.: exp( Sum_{k>=1} F(11*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

A305413 a(n) = Fibonacci(11*n)/89.

Original entry on oeis.org

0, 1, 199, 39602, 7880997, 1568358005, 312111123992, 62111682032413, 12360536835574179, 2459808941961294034, 489514339987133086945, 97415813466381445596089, 19386236394149894806708656, 3857958458249295447980618633, 767753119428003944042949816623

Offset: 0

Vincenzo Librandi, Jun 05 2018

S. Falcon, Generalized Fibonacci Sequences Generated from a k-Fibonacci Sequence, Journal of Mathematics Research, Vol. 4, No. 2 (2012), 97-100.
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (199,1).

Cf. A000045, A167398.

Cf. similar sequences: F(3*n)/2 (A001076), F(4*n)/3 (A004187), F(5*n)/5 (A049666), F(6*n)/8 (A049660), F(7*n)/13 (A049667), F(8*n)/21 (A049668), F(9*n)/34 (A049669), F(10*n)/55 (A049670), F(11*n)/89 (this sequence), F(12*n)/144 (A253368).

Magma
```
[Fibonacci(11*n)/89: n in [0..30]];
```

Fibonacci[11 Range[0, 20]]/89
LinearRecurrence[{199,1},{0,1},20] (* Harvey P. Dale, Aug 03 2024 *)

a(n) = fibonacci(11*n)/89 \\ Felix Fröhlich, Jul 30 2019

G.f.: x/(1 - 199*x - x^2).
a(n) = 199*a(n-1) + a(n-2) for n>1, a(0)=0, a(1)=1.
a(n) = A167398(n)/89.
For n >= 1, a(n) equals the denominator of the continued fraction [199, 199, ..., 199] (with n copies of 199). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 29 2019

cp's OEIS Frontend

A097843 First differences of Chebyshev polynomials S(n,123) = A049670(n+1) with Diophantine property.

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

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A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

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A049660 a(n) = Fibonacci(6*n)/8.

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A049666 a(n) = Fibonacci(5*n)/5.

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