A028412 -id:A028412 - cp's OEIS frontend

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

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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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

A001076 Denominators of continued fraction convergents to sqrt(5).

Original entry on oeis.org

0, 1, 4, 17, 72, 305, 1292, 5473, 23184, 98209, 416020, 1762289, 7465176, 31622993, 133957148, 567451585, 2403763488, 10182505537, 43133785636, 182717648081, 774004377960, 3278735159921, 13888945017644, 58834515230497, 249227005939632, 1055742538989025

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A001077(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = -1, a(2*n) with b(2*n) := A001077(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = +1 (cf. Emerson reference).
Bisection: a(2*n+1) = T(2*n+1, sqrt(5))/sqrt(5) = A007805(n), n >= 0 and a(2*n) = 4*S(n-1,18), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. S(n,18)=A049660(n+1). - Wolfdieter Lang, Jan 10 2003
Apart from initial terms, this is the Pisot sequence E(4,17), a(n) = floor(a(n-1)^2/a(n-2) + 1/2).
This is also the Horadam sequence (0,1,1,4), having the recurrence relation a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 4, r = 1. a(n) / a(n-1) converges to 5^1/2 + 2 as n approaches infinity. 5^(1/2) + 2 can also be written as (2 * Phi) + 1 and Phi^2 + Phi. - Ross La Haye, Aug 18 2003
Numerators of continued fraction [4, 4, 4, ...], where the convergents to [4, 4, 4, ...] = (4/1, 17/4, 72/17, ...). Let X = the 2 X 2 matrix [0, 1; 1, 4]; then X^n = [a(n-1), a(n); a(n), a(n+1)]; e.g., X^3 = [4, 17; 17, 72]. Let C = the limit of a(n)/a(n-1) = 2 + sqrt(5) = 4.236067977...; then C^n = a(n+1) + (1/C)*a(n), where (1/C) = 0.236067977... . Example: C^3 = 76.01315556..., = 72 + 17*(0.2360679...). - Gary W. Adamson, Dec 15 2007, corrected by Greg Dresden, Sep 16 2019, corrected by Alex Mark, Jul 21 2020
Sqrt(5) = 4/2 + 4/17 + 4/(17*305) + 4/(305*5473) + 4/(5473*98209) + ... . - Gary W. Adamson, Dec 15 2007
a(p) == 20^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009
a(n) = A167808(3*n). - Reinhard Zumkeller, Nov 12 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 4's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Moreover, a(n) is the second binomial transform of (0,1,0,5,0,25,...) (see also A033887). This fact can be proved similarly like the proof of Paul Barry's remark in A033887 by using the following scaling identity for delta-Fibonacci numbers: y^n b(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) b(k;x) and the fact that b(n;2) = (1-(-1)^n) 5^floor(n/2). - Roman Witula, Jul 12 2012
Binomial transform of 0, 1, 2, 8, 24, 80, 256, ... (A063727 with offset 1). - R. J. Mathar, Feb 05 2014
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,4} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
With offset 1 is the INVERT transform of A006190: (1, 3, 10, 33, 109, 360, ...). - Gary W. Adamson, Jul 24 2015
From Rogério Serôdio, Mar 30 2018: (Start)
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).
gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)
The initial 0 of this sequence is in contradiction with the fact that 0 is no valid denominator and according to all standard references, the first convergent of a continued fraction is p(0)/q(0) = b(0)/1 where b(0) is the first term of the continued fraction, given by the integer part of the number. One may artificially define q(-1) = 0 to have a recurrent relation q(n) = b(n)*q(n-1) + q(n-2), n >= 1, but then its index should be -1. - M. F. Hasler, Nov 01 2019
Number of 4-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 4-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 4 kinds of squares available. (End)
a(n) is the smallest nonnegative integer that is the sum of n, but no fewer, Fibonacci numbers including negative-index Fibonacci numbers (A039834), with that sum being a(n) = Sum_{i=0..n-1} A000045(3*i+1). a(n) is also the smallest nonnegative integer that is the sum of n, but no fewer, terms each of which is either a Fibonacci number or the negative of a Fibonacci number. (See A027941 for negatives disallowed.) - Mike Speciner, Oct 08 2023
From Enrique Navarrete, Dec 16 2023: (Start)
a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n > = 1, where P(k) = A006190(k) is the k-th 3-metallonacci number (see example below).
In general, the number of compositions with k-metallonacci number of parts is counted by the (k+1)-st metallonacci sequence (note k=1 and k=2 are the Fibonacci and the Pell numbers, respectively). (End).
a(n) is the number of tilings of a 2 X n rectangle missing the top right 1 X 1 cell, using 1 X 1 squares, dominoes and right trominoes. Compare to A110679 which is the same problem but without the missing top right cell. - Greg Dresden and Yilin Zhu, Jul 10 2025

			1 2 9 38 161 (A001077)
-,-,-,--,---, ...
0 1 4 17 72 (A001076)
G.f. = x + 4*x^2 + 17*x^3 + 72*x^4 + 305*x^5 + 1292*x^6 + 5473*x^7 + 23184*x^8 + ...
From _Enrique Navarrete_, Dec 16 2023: (Start)
From the comment on compositions with 3-metallonacci sorts of parts, A006190(k), there are A006190(1)=1 type of 1, A006190(2)=3 types of 2, A006190(3)=10 types of 3, A006190(4)=33 types of 4, A006190(5)=109 types of 5 and A006190(6)=360 types of 6. The following table gives the number of compositions of n=6:
Composition, number of such compositions, number of compositions of this type:
 6,              1,      360;
 5+1,            2,      218;
 4+2,            2,      198;
 3+3,            1,      100;
 4+1+1,          3,       99;
 3+2+1,          6,      180;
 2+2+2,          1,       27;
 3+1+1+1,        4,       40;
 2+2+1+1,        6,       54;
 2+1+1+1+1,      5,       15;
 1+1+1+1+1+1,    1,        1;
for a total of a(6)=1292 compositions of n=6. (End)

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 23.
S. Koshkin, Non-classical linear divisibility sequences ..., Fib. Q., 57 (No. 1, 2019), 68-80. See Table 1.
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).
V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 282.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Paraskevas K. Alvanos and Konstantinos A. Draziotis, Integer Solutions of the Equation y^2 = Ax^4 + B, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.4.
Elwyn Berlekamp and Richard K. Guy, Fibonacci Plays Billiards (2003), arXiv:2002.03705 [math.HO], 2020. See p. 5.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
D. W. Boyd, Some integer sequences related to the Pisot sequences, Acta Arithmetica, 34 (1979), 295-305.
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
Latham Boyle and Paul J. Steinhardt, Self-Similar One-Dimensional Quasilattices, arXiv preprint arXiv:1608.08220 [math-ph], 2016.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
M. C. Firengiz and A. Dil, Generalized Euler-Seidel method for second order recurrence relations, Notes on Number Theory and Discrete Mathematics, Vol. 20, 2014, No. 4, 21-32.
Yixing Fu, E. J. König, J. H. Wilson, Yang-Zhi Chou, and J. H. Pixley, Magic-angle semimetals, arXiv:1809.04604 [cond-mat.str-el], 2018.
Juan B. Gil and Aaron Worley, Generalized metallic means, arXiv:1901.02619 [math.NT], 2019.
Brian Hopkins and Stéphane Ouvry, Combinatorics of Multicompositions, arXiv:2008.04937 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 398
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Shaoxiong (Steven) Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Row n=4 of A073133, A172236 and A352361.
Cf. A000045, A001077, A015448, A175183 (Pisano periods).
Cf. A049660, A007805, A110679.
Partial sums of A033887. First differences of A049652. Bisection of A059973.
Third column of array A028412.

a:=[0,1];; for n in [3..30] do a[n]:=4*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018

I:=[0,1]; [n le 2 select I[n] else 4*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

A001076:=-1/(-1+4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

Join[{0}, Denominator[Convergents[Sqrt[5], 30]]] (* Harvey P. Dale, Dec 10 2011 *)
a[ n_] := Fibonacci[3*n] / 2; (* Michael Somos, Feb 23 2014 *)
a[ n_] := ((2 + Sqrt[5])^n - (2 - Sqrt[5])^n) /(2 Sqrt[5]) // Simplify; (* Michael Somos, Feb 23 2014 *)
LinearRecurrence[{4, 1}, {0, 1}, 26] (* Jean-François Alcover, Sep 23 2017 *)
a[ n_] := Fibonacci[n, 4]; (* Michael Somos, Nov 02 2021 *)

a(n):=sum(4^(n-1-2*k)*binomial(n-k-1,n-2*k-1),k,0,floor((n)/2));/* Vladimir Kruchinin, Oct 02 2022 */

numlib::fibonacci(3*n)/2 $ n = 0..30; // Zerinvary Lajos, May 09 2008

{a(n) = fibonacci(3*n) / 2}; /* Michael Somos, Aug 11 2009 */

{a(n) = imag( (2 + quadgen(20))^n )}; /* Michael Somos, Feb 23 2014 */

{a(n) = polchebyshev(n-1, 2, 2*I)/I^(n-1)}; /* Michael Somos, Nov 02 2021 */

[lucas_number1(n,4,-1) for n in range(23)] # Zerinvary Lajos, Apr 23 2009

[fibonacci(3*n)/2 for n in range(23)] # Zerinvary Lajos, May 15 2009

a(n) = 4*a(n-1) + a(n-2), n > 1. a(0)=0, a(1)=1.
G.f.: x/(1 - 4*x - x^2).
a(n) = ((2+sqrt(5))^n - (2-sqrt(5))^n)/(2*sqrt(5)).
a(n) = A014445(n)/2 = F(3n)/2.
a(n) = ((-i)^(n-1))*S(n-1, 4*i), with i^2 = -1 and S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x) = 0.
a(n) = Sum_{i=0..n} Sum_{j=0..n} Fibonacci(i+j)*n!/(i!j!(n-i-j)!)/2. - Paul Barry, Feb 06 2004
E.g.f.: exp(2*x)*sinh(sqrt(5)*x)/sqrt(5). - Vladeta Jovovic, Sep 01 2004
a(n) = F(1) + F(4) + F(7) + ... + F(3n-2), for n > 0.
Conjecture: 2a(n+1) = a(n+2) - A001077(n+1). - Creighton Dement, Nov 28 2004
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n, j)*C(j, k)*F(j)/2. - Paul Barry, Feb 14 2005
a(n) = A048876(n) - A048875(n). - Creighton Dement, Mar 19 2005
Let M = {{0, 1}, {1, 4}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = v[n][[1]]. - Roger L. Bagula, May 29 2005
a(n) = F(n, 4), the n-th Fibonacci polynomial evaluated at x=4. - T. D. Noe, Jan 19 2006
[A015448(n), a(n)] = [1,4; 1,3]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = (Sum_{k=0..n} Fibonacci(3*k-2)) + 1. - Gary Detlefs, Dec 26 2010
a(n) = (3*(-1)^n*F(n) + 5*F(n)^3)/2, n >= 0. See the general D. Jennings formula given in a comment on triangle A111125, where also the reference is given. Here the second (k=1) row [3,1] applies. - Wolfdieter Lang, Sep 01 2012
Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (Sum_{k>=1} (-1)^(k-1)/(F_k*F_(k+1)))^3 = phi^(-3), where F_n is the n-th Fibonacci numbers (A000045) and phi is golden ratio (A001622). - Vladimir Shevelev, Feb 23 2013
G.f.: Q(0)*x/(2-4*x), where Q(k) = 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 2/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
a(-n) = -(-1)^n * a(n). - Michael Somos, Feb 23 2014
The o.g.f. A(x) = x/(1 - 4*x - x^2) satisfies A(x) + A(-x) + 8*A(x)*A(-x) = 0 or equivalently (1 + 8*A(x))*(1 + 8*A(-x)) = 1. The o.g.f. for A049660 equals -A(sqrt(x))*A(-sqrt(x)). - Peter Bala, Apr 02 2015
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)*a(n+1) = 4*Sum_{k=1..n} a(k)^2;
(2) a(n)^2 + a(n+1)^2 = a(2*n+1);
(3) a(n)^2 - a(n-2)^2 = 4*a(n-1)*(a(n) + a(n-2));
(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);
(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(6) a(n-1)*a(n+1) = a(n)^2 + (-1)^n (particular case of (5)!);
(7) a(2*n) = 2*a(n)*(2*a(n) + a(n-1));
(8) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;
(9) a(n) - a(n-2*k+1) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = (2+sqrt(5))^(2*k-1) + (2-sqrt(5))^(2*k-1);
(10) 31|Sum_{k=n..n+9} a(k), for all positive n. (End)
O.g.f.: x*exp(Sum_{n >= 1} Lucas(3*n)*x^n/n) = x + 4*x^2 + 17*x^3 + .... - Peter Bala, Oct 11 2019
a(n) = Sum_{k=0..floor(n/2)} 4^(n-2*k-1)*C(n-k-1,n-2*k-1). - Vladimir Kruchinin, Oct 02 2022
a(n) = i^(n-1)*S(n-1, -4*i), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. - Gary Detlefs and Wolfdieter Lang, Mar 06 2023
G.f.: x/(1 - 4*x - x^2) = Sum_{n >= 0} x^(n+1) * ( Product_{k = 1..n} (m*k + 4 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
a(n) = 4^(n-1)*hypergeom([(1-n)/2, 1-n/2], [1-n], -1/4) for n > 0. - Peter Luschny, Mar 30 2025
a(n) = a(n-1) + A110679(n-1) + A110679(n-2) = a(n-1) + Fibonacci(3*n-2). - Greg Dresden and Yilin Zhu, Jul 10 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)

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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

A004187 a(n) = 7*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 7, 48, 329, 2255, 15456, 105937, 726103, 4976784, 34111385, 233802911, 1602508992, 10983760033, 75283811239, 516002918640, 3536736619241, 24241153416047, 166151337293088, 1138818207635569, 7805576116155895, 53500214605455696, 366695926122033977

Offset: 0

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

Define the sequence T(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0 . A004187 (with initial 0 omitted) is T(1,7).
This is a divisibility sequence.
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 7'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
a(n) and b(n) := A056854(n) are the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 5*(3*a(n))^2 = +4. see the cross-reference to A056854 below. - Wolfdieter Lang, Jun 26 2013
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4,5,6}. - Milan Janjic, Jan 25 2015
The digital root is A253298, which shares its digital root with A253368. - Peter M. Chema, Jul 04 2016
Lim_{n->oo} a(n+1)/a(n) = 2 + 3*phi = 1+ A090550 = 6.854101...  - Wolfdieter Lang, Nov 16 2023

			a(2) = 7*a(1) - a(0) = 7*7 - 1 = 48. - _Michael B. Porter_, Jul 04 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3 , example 12
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993
Zvonko Cerin, Some alternating sums of Lucas numbers, Centr. Eur. J. Math. vol 3 no 1 (2005) 1-13.
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).
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=7, q=-1.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=9.
Index entries for sequences related to Chebyshev polynomials.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).
Index entries for Pisot sequences

Cf. A000027, A001906, A001353, A004254, A001109, A049685, A033888. a(n)=sqrt((A056854(n)^2 - 4)/45).

Second column of array A028412.

[Fibonacci(4*n)/3 : n in [0..30]]; // Vincenzo Librandi, Jun 07 2011

/* By definition: */ [n le 2 select n-1 else 7*Self(n-1)-Self(n-2): n in [1..23]]; // Bruno Berselli, Dec 24 2012

seq(combinat:-fibonacci(4*n)/3, n = 0 .. 30); # Robert Israel, Jan 26 2015

LinearRecurrence[{7,-1},{0,1},30] (* Harvey P. Dale, Jul 13 2011 *)
CoefficientList[Series[x/(1 - 7*x + x^2), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 23 2012 *)

a[0]:0$ a[1]:1$ a[n]:=7*a[n-1] - a[n-2]$ A004187(n):=a[n]$ makelist(A004187(n),n,0,30); /* Martin Ettl, Nov 11 2012 */

numlib::fibonacci(4*n)/3 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(4*n)/3 \\ Charles R Greathouse IV, Mar 09 2012

concat(0, Vec(x/(1-7*x+x^2) + O(x^99))) \\ Altug Alkan, Jul 03 2016

[lucas_number1(n,7,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

[fibonacci(4*n)/3 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1-7*x+x^2).
a(n) = F(4*n)/3 = A033888(n)/3, where F=A000045 (the Fibonacci sequence).
a(n) = S(2*n-1, sqrt(9))/sqrt(9) = S(n-1, 7); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = Sum_{i = 0..n-1} C(2*n-1-i, i)*5^(n-i-1). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
[A049685(n-1), a(n)] = [1,5; 1,6]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167816(4*n). - Reinhard Zumkeller, Nov 13 2009
a(n) = (((7+sqrt(45))/2)^n-((7-sqrt(45))/2)^n)/sqrt(45). - Noureddine Chair, Aug 31 2011
a(n+1) = Sum_{k = 0..n} A101950(n,k)*6^k. - Philippe Deléham, Feb 10 2012
a(n) = (A081072(n)/3)-1. - Martin Ettl, Nov 11 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = (1/5)*(5 + 3*sqrt(5)).
Product {n >= 2} (1 - 1/a(n)) = (1/14)*(5 + 3*sqrt(5)). (End)
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -A(x)*A(-x), where A(x) = Sum_{n >= 1} Fibonacci(2*n)* x^n.
1 + 5*Sum_{n >= 1} a(n)*x^(2*n) = F(x)*F(-x) = G(x)*G(-x), where F(x) = 1 + A(x) and G(x) = 1 + 5*A(x).
1 + Sum_{n >= 1} a(n)*x^(2*n) = H(x)*H(-x) = I(x)*I(-x), where H(x) = 1 + Sum_{n >= 1} Fibonacci(2*n + 3)*x^n and I(x) = 1 + x + x*Sum_{n >= 1} Fibonacci(2*n - 1)*x^n. (End)
E.g.f.: 2*exp(7*x/2)*sinh(3*sqrt(5)*x/2)/(3*sqrt(5)). - Ilya Gutkovskiy, Jul 03 2016
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*9^k*binomial(n+k, 2*k+1). - Peter Bala, Jul 17 2023
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*7^(n-2*k)*binomial(n-k, k). - Greg Dresden, Aug 03 2024
From Peter Bala, Jul 22 2025: (Start)
The following products telescope:
Product {n >= 2} (1 + (-1)^n/a(n)) = (3/14)*(3 + sqrt(5)).
Product {n >= 1} (1 - (-1)^n/a(n)) = (1/3)*(3 + sqrt(5)).
Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = (3/5)*sqrt(5). (End)

Entry improved by comments from Michael Somos and Wolfdieter Lang, Aug 02 2000

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

cp's OEIS Frontend

A214978 Array T(m,n) = Fibonacci(m*n)/Fibonacci(m), by antidiagonals; transpose of A028412.

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A214983 Antidiagonal sums of the array A028412.

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

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A001076 Denominators of continued fraction convergents to sqrt(5).

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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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