

A001519


a(n) = 3*a(n1)  a(n2) for n >= 2, with a(0) = a(1) = 1.
(Formerly M1439 N0569)


329



1, 1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, 75025, 196418, 514229, 1346269, 3524578, 9227465, 24157817, 63245986, 165580141, 433494437, 1134903170, 2971215073, 7778742049, 20365011074, 53316291173, 139583862445, 365435296162, 956722026041
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OFFSET

0,3


COMMENTS

This is a bisection of the Fibonacci sequence A000045. a(n) = F(2*n1), with F(n) = A000045(n) and F(1) = 1.
Number of ordered trees with n+1 edges and height at most 3 (height=number of edges on a maximal path starting at the root). Number of directed columnconvex polyominoes of area n+1. Number of nondecreasing Dyck paths of length 2n+2.  Emeric Deutsch, Jul 11 2001
Terms for n > 1 are the solutions to: 5x^24 is a square.  Benoit Cloitre, Apr 07 2002
a(0) = a(1) = 1, a(n+1) = smallest Fibonacci number greater than the nth partial sum.  Amarnath Murthy, Oct 21 2002
The fractional part of tau*a(n) decreases monotonically to zero.  Benoit Cloitre, Feb 01 2003
n such that floor(phi^2*n^2)floor(phi*n)^2 = 1 where phi=(1+sqrt(5))/2.  Benoit Cloitre, Mar 16 2003
Number of leftist horizontally convex polyominoes with area n+1.
Number of 31avoiding words of length n on alphabet {1,2,3} which do not end in 3. (E.g., at n=3, we have 111,112,121,122,132,211,212,221,222,232,321,322 and 332.) See A028859.  Jon Perry, Aug 04 2003
Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=phi=(1+sqrt(5))/2.  Benoit Cloitre, Feb 24 2004
a(1) = 1, a(2) = 2, then the least number such that the square of any term is just less than the geometric mean of its neighbors. a(n+1)*a(n1) > a(n)^2.  Amarnath Murthy, Apr 06 2004
All positive integer solutions of Pell equation b(n)^2  5*a(n+1)^2 = 4 together with b(n)=A002878(n), n >= 0.  Wolfdieter Lang, Aug 31 2004
a(n) = L(n,3), where L is defined as in A108299; see also A002878 for L(n,3).  Reinhard Zumkeller, Jun 01 2005
Essentially same as Pisot sequence E(2,5).
Number of permutations of [n+1] avoiding 321 and 3412. E.g., a(3) = 13 because the permutations of [4] avoiding 321 and 3412 are 1234, 2134, 1324, 1243, 3124, 2314, 2143, 1423, 1342, 4123, 3142, 2413, 2341.  Bridget Tenner, Aug 15 2005
Number of 1324avoiding circular permutations on [n+1].
A subset of the Markoff numbers (A002559).  Robert G. Wilson v, Oct 05 2005
(x,y)=(a(n),a(n+1)) are the solutions of x/(yz)+y/(xz)+z/(xy)=3 with z=1.  Floor van Lamoen, Nov 29 2001
Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 5 and s(i)  s(i1) = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 1.  Herbert Kociemba, Jun 10 2004
With interpolated zeros, counts closed walks of length n at the start or end node of P_4. a(n) counts closed walks of length 2n at the start or end node of P_4. The sequence 0,1,0,2,0,5,... counts walks of length n between the start and second node of P_4.  Paul Barry, Jan 26 2005
a(n) is the number of ordered trees on n edges containing exactly one nonleaf vertex all of whose children are leaves (every ordered tree must contain at least one such vertex). For example, a(0) = 1 because the root of the tree with no edges is not considered to be a leaf and the condition "all children are leaves" is vacuously satisfied by the root and a(4) = 13 counts all 14 ordered trees on 4 edges (A000108) except (ignore dots)
..
.\/.
which has two such vertices.  David Callan, Mar 02 2005
Number of directed columnconvex polyominoes of area n. Example: a(2)=2 because we have the 1 X 2 and the 2 X 1 rectangles.  Emeric Deutsch, Jul 31 2006
Same as the number of Kekulé structures in polyphenanthrene in terms of the number of hexagons in extended (1,1)nanotubes. See Table 1 on page 411 of I. Lukovits and D. Janezic.  Parthasarathy Nambi, Aug 22 2006
Number of free generators of degree n of symmetric polynomials in 3noncommuting variables.  Mike Zabrocki, Oct 24 2006
Inverse: With phi = (sqrt(5) + 1)/2, log_phi((sqrt(5) a(n) + sqrt(5 a(n)^2  4))/2) = n for n >= 1.  David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 19 2007
Consider a teacher who teaches one student, then he finds he can teach two students while the original student learns to teach a student. And so on with every generation an individual can teach one more student then he could before. a(n) starting at a(2) gives the total number of new students/teachers (see program).  Ben Paul Thurston, Apr 11 2007
The Diophantine equation a(n)=m has a solution (for m >= 1) iff ceiling(arcsinh(sqrt(5)*m/2)/log(phi))) != ceiling(arccosh(sqrt(5)*m/2)/log(phi))) where phi is the golden ratio. An equivalent condition is A130255(m)=A130256(m).  Hieronymus Fischer, May 24 2007
a(n+1) = B^(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., 2=`0`, 5=`00`, 13=`000`, ..., in Wythoff code.
Bisection of the Fibonacci sequence into oddindexed nonzero terms (1, 2, 5, 13, ...) and evenindexed terms (1, 3, 8, 21, ...) may be represented as row sums of companion triangles A140068 and A140069.  Gary W. Adamson, May 04 2008
a(n) is the number of partitions pi of [n] (in standard increasing form) such that Flatten[pi] is a (213)avoiding permutation. Example: a(4)=13 counts all 15 partitions of [4] except 13/24 and 13/2/4. Here "standard increasing form" means the entries are increasing in each block and the blocks are arranged in increasing order of their first entries. Also number that avoid 312.  David Callan, Jul 22 2008
Let P be the partial sum operator, A000012: (1; 1,1; 1,1,1; ...) and A153463 = M, the partial sum & shift operator. It appears that beginning with any randomly taken sequence S(n), iterates of the operations M * S(n), > M * ANS, > P * ANS, etc. (or starting with P) will rapidly converge upon a twosequence limit cycle of (1, 2, 5, 13, 34, ...) and (1, 1, 3, 8, 21, ...).  Gary W. Adamson, Dec 27 2008
4*a(n) = A153266(n) + A153267(n), apart from initial terms.  Creighton Dement, Jan 02 2009
Sum_{n>=0} atan(1/a(n)) = (3/4)Pi.  Jaume Oliver Lafont, Feb 27 2009
Summation of the squares of fibonacci numbers taken 2 at a time. Offset 1. a(3)=5.  Al Hakanson (hawkuu(AT)gmail.com), May 27 2009
Number of musical compositions of Rhythmmusic over a time period of n1 units. Example: a(4)=13; indeed, denoting by R a rest over a time period of 1 unit and by N[j] a note over a period of j units, we have (writing N for N[1]): NNN, NNR, NRN, RNN, NRR, RNR, RRN, RRR, N[2]R, RN[2], NN[2], N[2]N, N[3]) (see the J. Groh reference, pp. 4348).  Juergen K. Groh (juergen.groh(AT)lhsystems.com), Jan 17 2010
Given an infinite lower triangular matrix M with (1, 2, 3, ...) in every column but the leftmost column shifted upwards one row. Then (1, 2, 5, ...) = lim_{n>inf} M^n. (Cf. A144257.)  Gary W. Adamson, Feb 18 2010
As a fraction: 8/71 = 0.112676 or 98/9701 = 0.010102051334... (fraction 9/71 or 99/9701 for sequence without initial term). 19/71 or 199/9701 for sequence in reverse.  Mark Dols, May 18 2010
For n >= 1, a(n) is the number of compositions (ordered integer partitions) of 2n1 into an odd number of odd parts. O.g.f.: (xx^3)/(13x^2+x^4) = A(A(x)) where A(x) = 1/(1x)1/(1x^2).
For n > 0, determinant of the n X n tridiagonal matrix with 1's in the super and subdiagonals, (1,3,3,3,...) in the main diagonal, and the rest zeros.  Gary W. Adamson, Jun 27 2011
The Gi3 sums, see A180662, of the triangles A108299 and A065941 equal the terms of this sequence without a(0).  Johannes W. Meijer, Aug 14 2011
The number of permutations for which length equals reflection length.  Bridget Tenner, Feb 22 2012
Number of nonisomorphic graded posets with 0 and 1 and uniform Hasse graph of rank n+1, with exactly 2 elements of each rank between 0 and 1. (Uniform used in the sense of Retakh, Serconek and Wilson. Graded used in R. Stanley's sense that all maximal chains have the same length.)
HANKEL transform of sequence and the sequence omitting a(0) is the sequence A019590(n). This is the unique sequence with that property.  Michael Somos, May 03 2012
a(n) = 2^n a(n;1/2), where a(n;d). n=0,1,...,d, denote the socalled deltaFibonacci number (see Witula et al. papers and comments in A000045).  Roman Witula, Jul 12 2012
The number of Dyck paths of length 2n and height at most 3.  Ira M. Gessel, Aug 06 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
Primes in the sequence are 2, 5, 13, 89, 233, 1597, 28657, ... (apparently A005478 without the 3).  R. J. Mathar, May 09 2013
a(n+1) is the sum of rising diagonal of the Pascal triangle written as a square  cf. comments in A085812. E.g., 13 = 1+5+6+1.  John Molokach, Sep 26 2013
a(n) is the top left entry of the nth power of any of the 3 X 3 matrices [1, 1, 1; 1, 1, 1; 0, 1, 1] or [1, 1, 1; 0, 1, 1; 1, 1, 1] or [1, 1, 0; 1, 1, 1; 1, 1, 1] or [1, 0, 1; 1, 1, 1; 1, 1, 1].  R. J. Mathar, Feb 03 2014
Except for the initial term, positive values of x (or y) satisfying x^2  3xy + y^2 + 1 = 0.  Colin Barker, Feb 04 2014
Except for the initial term, positive values of x (or y) satisfying x^2  18xy + y^2 + 64 = 0.  Colin Barker, Feb 16 2014
Positive values of x such that there is a y satisfying x^2  xy  y^2  1 = 0.  Ralf Stephan, Jun 30 2014
1 = a(n)*a(n+2)  a(n+1)*a(n+1) for all n in Z.  Michael Somos, Jul 08 2014
a(n) is also the number of permutations simultaneously avoiding 231, 312 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n1 nodes. See A245904 for more information on increasing strict binary trees.  Manda Riehl, Aug 07 2014
(1, a(n), a(n+1)), n >= 0, are Markoff triples (see A002559) and Robert G. Wilson v's Oct 05 2005 comment). In the Markoff tree they give one of the outer branches. Proof: a(n)*a(n+1)  1 = A001906(2*n)^2 = (a(n+1)  a(n))^2 = a(n)^2 + a(n+1)^2  2*a(n)*a(n+1), thus 1^2 + a(n)^2 + a(n+1)^2 = 3*a(n)*a(n+1).  Wolfdieter Lang, Jan 30 2015
For n > 0, a(n) is the smallest positive integer not already in the sequence such that a(1) + a(2) + ... + a(n) is a Fibonacci number.  Derek Orr, Jun 01 2015
Number of vertices of degree n2 (n >= 3) in all Fibonacci cubes, see Klavzar, Mollard, & Petkovsek.  Emeric Deutsch, Jun 22 2015
Except for the first term, this sequence can be generated by Corollary 1 (ii) of Azarian's paper in the references for this sequence.  Mohammad K. Azarian, Jul 02 2015
Precisely the numbers F(n)^k + F(n+1)^k that are also Fibonacci numbers with k > 1, see Luca & Oyono.  Charles R Greathouse IV, Aug 06 2015
a(n) = MA(n)  2*(1)^n where MA(n) is exactly the maximum area of a quadrilateral with lengths of sides in order L(n2), L(n2), F(n+1), F(n+1) for n > 1 and L(n)=A000032(n).  J. M. Bergot, Jan 28 2016
a(n) is the number of bargraphs of semiperimeter n+1 having no valleys (i.e., convex bargraphs). Equivalently, number of bargraphs of semiperimeter n+1 having exactly 1 peak. Example: a(5) = 34 because among the 35 (=A082582(6)) bargraphs of semiperimeter 6 only the one corresponding to the composition [2,1,2] has a valley.  Emeric Deutsch, Aug 12 2016
Integers k such that the fractional part of k*phi is less than 1/k. See Byszewski link p. 2.  Michel Marcus, Dec 10 2016
Number of words of length n1 over {0,1,2,3} in which binary subwords appear in the form 10...0.  Milan Janjic, Jan 25 2017
With a(0) = 0 this is the Riordan transform with the Riordan matrix A097805 (of the associated type) of the Fibonacci sequence A000045. See a Feb 17 2017 comment on A097805.  Wolfdieter Lang, Feb 17 2017
Number of sequences (e(1), ..., e(n)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) < e(j) < e(k). [Martinez and Savage, 2.12]  Eric M. Schmidt, Jul 17 2017
Number of permutations of [n] that avoid the patterns 321 and 2341.  Colin Defant, May 11 2018
The sequence solves the following problem: find all the pairs (i,j) such that i divides 1+j^2 and j divides 1+i^2. In fact, the pairs (a(n),a(n+1)), n > 0, are all the solutions.  Tomohiro Yamada, Dec 23 2018
Number of permutations in S_n whose principal order ideals in the Bruhat order are lattices (equivalently, modular, distributive, Boolean lattices).  Bridget Tenner, Jan 16 2020
From Wolfdieter Lang, Mar 30 2020: (Start)
a(n) is the upper left entry of the nth power of the 2 X 2 tridiagonal matrix M_2 = Matrix([1,1], [1,2]) from A322602: a(n) = (M_2)^n)[1,1].
Proof: (M_2)^2 = 3*M + 1_2 (with the 2 X 2 unit matrix 1_2) from the characteristic polynomial of M_2 (see a comment in A322602) and the CayleyHamilton theorem. The recurrence M^n = M*M^(n1) leads to (M_n)^n = S(n, 3)*1_2 + S(na, 3)*(M  3*1_2), for n >= 0, with S(n, 3) = F(2(n+1)) = A001906(n+1). Hence (M_2)^n)[1,1] = S(n, 3)  2*S(n1, 3) = a(n) = F(2*n1) = (1/(2*r+1))*r^(2*n1)*(1 + (1/r^2)^(2*n1)), with r = rho(5) = A001622 (golden ratio) (see the first Aug 31 2004 formula, using the recurrence of S(n, 3), and the Michael Somos Oct 28 2002 formula). This proves a conjecture of Gary W. Adamson in A322602.
The ratio a(n)/a(n1) converges to r^2 = rho(5)^2 = A104457 for n > infinity (see the a(n) formula in terms of r), which is one of the statements by Gary W. Adamson in A322602. (End)
a(n) is the number of ways to stack coins with a bottom row of n coins such that any coin not on the bottom row touches exactly two coins in the row below, and all the coins on any row are contiguous [Wilf, 2.12].  Greg Dresden, Jun 29 2020
a(n) is the upper left entry of the (2*n)th power of the 4 X 4 Jacobi matrix L with L(i,j)=1 if ij = 1 and L(i,j)=0 otherwise.  Michael Shmoish, Aug 29 2020
All positive solutions of the indefinite binary quadratic F(1, 3, 1) := x^2  3*x*y + y^2, of discriminant 5, representing 1 (special Markov triples (1, y=x, z=y) if y <= z) are [x(n), y(n)] = [abs(F(2*n+1)), abs(F(2*n1))], for n = infinity..+infinity. (F(n) = (1)^(n+1)*F(n)). There is only this single family of proper solutions, and there are no improper solutions. [See also the Floor van Lamoen Nov 29 2001 comment, which uses this negative n, and my Jan 30 2015 comment.]  Wolfdieter Lang, Sep 23 2020


REFERENCES

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J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 39.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. Stanley, Enumerative combinatorics, Vol. 1. Cambridge University Press, Cambridge, 1997, pp. 96100.
H. S. Wilf, Generatingfunctionology, 3rd ed., A K Peters Ltd., Wellesley, MA, 2006, p. 41.


LINKS

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Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,1).
Index entries for "core" sequences


FORMULA

G.f.: (12*x)/(13*x+x^2).
G.f.: 1 / (1  x / (1  x / (1  x))).  Michael Somos, May 03 2012
a(n) = A001906(n+1)  2*A001906(n).
a(n) = a(1n) for all n in Z.
a(n+2) = (a(n+1)^2+1)/a(n) with a(1)=1, a(2)=2.  Benoit Cloitre, Aug 29 2002
a(n) = (phi^(2*n1) + phi^(12*n))/sqrt(5) where phi=(1+sqrt(5))/2.  Michael Somos, Oct 28 2002
a(n) = A007598(n1) + A007598(n) = A000045(n1)^2 + A000045(n)^2 = F(n)^2 + F(n+1)^2.  Henry Bottomley, Feb 09 2001
a(n) = Sum_{k=0..n} binomial(n+k, 2*k).  Len Smiley, Dec 09 2001
a(n) ~ (1/5)*sqrt(5)*phi^(2*n+1).  Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = Sum_{k=0..n} C(n, k)*F(k+1).  Benoit Cloitre, Sep 03 2002
Let q(n, x) = Sum_{i=0..n} x^(ni)*binomial(2*ni, i); then q(n, 1)=a(n) (this comment is essentially the same as that of L. Smiley).  Benoit Cloitre, Nov 10 2002
a(n) = (1/2)*(3*a(n1) + sqrt(5*a(n1)^24)).  Benoit Cloitre, Apr 12 2003
Main diagonal of array defined by T(i, 1) = T(1, j) = 1, T(i, j) = max(T(i1, j) + T(i1, j1); T(i1, j1) + T(i, j1)).  Benoit Cloitre, Aug 05 2003
Hankel transform of A002212. E.g., Det([1, 1, 3;1, 3, 10;3, 10, 36]) = 5.  Philippe Deléham, Jan 25 2004
Solutions x > 0 to equation floor(x*r*floor(x/r)) = floor(x/r*floor(x*r)) when r=phi.  Benoit Cloitre, Feb 15 2004
a(n) = Sum_{i=0..n} binomial(n+i, ni).  Jon Perry, Mar 08 2004
a(n) = S(n1, 3)  S(n2, 3) = T(2*n1, sqrt(5)/2)/(sqrt(5)/2) with S(n, x) = U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first kind. See triangle A049310, resp. A053120.  Wolfdieter Lang, Aug 31 2004
a(n) = ((1)^(n1))*S(2*(n1), i), with the imaginary unit i and S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310.  Wolfdieter Lang, Aug 31 2004
a(n) = Sum_{0<=i_1<=i_2<=n} binomial(i_2, i_1)*binomial(n, i_1+i_2).  Benoit Cloitre, Oct 14 2004
a(n) = a(n1) + Sum_{i=0..n1} a(i)*a(n) = F(2*n+1)*Sum_{i=0..n1} a(i) = F(2*n).  Andras Erszegi (erszegi.andras(AT)chello.hu), Jun 28 2005
The ith term of the sequence is the entry (1, 1) of the ith power of the 2 X 2 matrix M = ((1, 1), (1, 2)).  Simone Severini, Oct 15 2005
a(n1) = (1/n)*Sum_{k=0..n} B(2*k)*F(2*n2*k)*binomial(2*n, 2*k) where B(2*k) is the (2*k)th Bernoulli number.  Benoit Cloitre, Nov 02 2005
a(n) = A055105(n,1) + A055105(n,2) + A055105(n,3) = A055106(n,1) + A055106(n,2).  Mike Zabrocki, Oct 24 2006
a(n) = (2/sqrt(5))*cosh((2n1)*psi), where psi=log(phi) and phi=(1+sqrt(5))/2.  Hieronymus Fischer, Apr 24 2007
a(n) = (phi+1)^n  phi*A001906(n) with phi=(1+sqrt(5))/2.  Reinhard Zumkeller, Nov 22 2007
a(n) = 2*a(n1) + 2*a(n2)  a(n3); a(n) = ((sqrt(5) + 5)/10)*(3/2 + sqrt(5)/2)^(n2) + ((sqrt(5) + 5)/10)*(3/2  sqrt(5)/2)^(n2).  Antonio Alberto Olivares, Mar 21 2008
a(n) = A147703(n,0).  Philippe Deléham, Nov 29 2008
a(n) = a(n1) + 11*a(n2)  4*a(n3), this formula (which is one of two 3rdorder linear recurrence relations given for this sequence) is a result of the generating floretion Z = X*Y with X = 1.5'i + 0.5i' + .25(ii + jj + kk + ee) and Y = 0.5'i + 1.5i' + .25(ii + jj + kk + ee).  Creighton Dement, Jan 02 2009
With X,Y defined as X = ( F(n) F(n+1) ), Y = ( F(n+2) F(n+3) ), where F(n) is the nth Fibonacci number (A000045), it follows a(n+2) = X.Y', where Y' is the transpose of Y (n >= 0).  K.V.Iyer, Apr 24 2009
a(n) = Fibonacci(2*n+2) mod Fibonacci(2*n), n > 1.  Gary Detlefs, Nov 22 2010
a(n) = (Fibonacci(n1)^2 + Fibonacci(n)^2 + Fibonacci(2*n1))/2.  Gary Detlefs, Nov 22 2010
INVERT transform is A007051. First difference is A001906. Partial sums is A055588. Binomial transform is A093129. Binomial transform of A000045(n1).  Michael Somos, May 03 2012
a(n) = (Fibonacci(n+2)^2 + Fibonacci(n3)^2)/5.  Gary Detlefs, Dec 14 2012
G.f.: 1 + x/( Q(0)  x ) where Q(k) = 1  x/(x*k + 1 )/Q(k+1); (recursively defined continued fraction).  Sergei N. Gladkovskii, Feb 23 2013
G.f.: (12*x)*G(0)/(23*x), where G(k) = 1 + 1/( 1  x*(5*k9)/(x*(5*k4)  6/G(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Jul 19 2013
G.f.: 1 + x*(1x^2)*Q(0)/2, where Q(k) = 1 + 1/(1  x*(4*k+2 + 2*x  x^2)/( x*(4*k+4 + 2*x  x^2 ) + 1/Q(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Sep 11 2013
G.f.: Q(0,u), where u=x/(1x), Q(k,u) = 1 + u^2 + (k+2)*u  u*(k+1 + u)/Q(k+1); (continued fraction).  Sergei N. Gladkovskii, Oct 07 2013
Sum_{n>=2} 1/(a(n)  1/a(n)) = 1. Compare with A001906, A007805 and A097843.  Peter Bala, Nov 29 2013
Let F(n) be the nth Fibonacci number, A000045(n), and L(n) be the nth Lucas number, A000032(n). Then for n > 0, a(n) = F(n)*L(n1) + (1)^n.  Charlie Marion, Jan 01 2014
a(n) = A238731(n,0).  Philippe Deléham, Mar 05 2014
a(n) = (L(2*n+4) + L(2*n6))/25 for L(n)=A000032(n).  J. M. Bergot, Dec 30 2014
a(n) = (L(n1)^2 + L(n)^2)/5 with L(n)=A000032(n).  J. M. Bergot, Dec 31 2014
a(n) = (L(n2)^2 + L(n+1)^2)/10 with L(n)=A000032(n).  J. M. Bergot, Oct 23 2015
a(n) = 3*F(n1)^2 + F(n3)*F(n)  2*(1)^n.  J. M. Bergot, Feb 17 2016
a(n) = (F(n1)*L(n) + F(n)*L(n1))/2.  J. M. Bergot, Mar 22 2016
E.g.f.: (2*exp(sqrt(5)*x) + 3 + sqrt(5))*exp(x*(sqrt(5)3)/2)/(5 + sqrt(5)).  Ilya Gutkovskiy, Jul 04 2016
a(n) = (M_2)^n)[1,1] = S(n, 3)  2*S(n1, 3), with the 2 X 2 tridiagonal matrix M_2 = Matrix([1,1], [1,2]) from A322602. For a proof see the Mar 30 2020 comment above.  Wolfdieter Lang, Mar 30 2020
Sum_{n>=1} 1/a(n) = A153387.  Amiram Eldar, Oct 05 2020


EXAMPLE

a(3)=13: there are 14 ordered trees with 4 edges; all of them, except for the path with 4 edges, have height at most 3.
G.f. = 1 + x + 2*x^2 + 5*x^3 + 13*x^4 + 34*x^5 + 89*x^6 + 233*x^7 + ...


MAPLE

A001519:=(1+z)/(13*z+z**2); # Simon Plouffe in his 1992 dissertation; gives sequence without an initial 1
A001519 := proc(n) option remember: if n=0 then 1 elif n=1 then 1 elif n>=2 then 3*procname(n1)procname(n2) fi: end: seq(A001519(n), n=0..28); # Johannes W. Meijer, Aug 14 2011


MATHEMATICA

Fibonacci /@ (2Range[29]  1) (* Robert G. Wilson v, Oct 05 2005 *)
LinearRecurrence[{3, 1}, {1, 1}, 29] (* Robert G. Wilson v, Jun 28 2012 *)
a[ n_] := With[{c = Sqrt[5]/2}, ChebyshevT[2 n  1, c]/c]; (* Michael Somos, Jul 08 2014 *)
CoefficientList[ Series[(1  2x)/(1  3x + x^2), {x, 0, 30}], x] (* Robert G. Wilson v, Feb 01 2015 *)


PROG

(PARI) {a(n) = fibonacci(2*n  1)}; /* Michael Somos, Jul 19 2003 */
(PARI) {a(n) = real( quadgen(5) ^ (2*n))}; /* Michael Somos, Jul 19 2003 */
(PARI) {a(n) = subst( poltchebi(n) + poltchebi(n  1), x, 3/2) * 2/5}; /* Michael Somos, Jul 19 2003 */
(Sage) [lucas_number1(n, 3, 1)lucas_number1(n1, 3, 1) for n in range(30)] # Zerinvary Lajos, Apr 29 2009
(Haskell)
a001519 n = a001519_list !! n
a001519_list = 1 : zipWith () (tail a001906_list) a001906_list
 Reinhard Zumkeller, Jan 11 2012
a001519_list = 1 : f a000045_list where f (_:x:xs) = x : f xs
 Reinhard Zumkeller, Aug 09 2013
(Maxima) a[0]:1$ a[1]:1$ a[n]:=3*a[n1]a[n2]$ makelist(a[n], n, 0, 30); /* Martin Ettl, Nov 15 2012 */
(MAGMA) [1] cat [(Lucas(2*n)  Fibonacci(2*n))/2: n in [1..50]]; // Vincenzo Librandi, Jul 02 2014
(GAP)
a:=[1, 1];; for n in [3..10^2] do a[n]:=3*a[n1]a[n2]; od; a; # Muniru A Asiru, Sep 27 2017


CROSSREFS

Fibonacci A000045 = union of this sequence and A001906.
Cf. A001653, A055105, A055106, A055107, A074664, A101368, A124292, A124293, A124294, A124295, A140069, A153463, A153266, A153267, A144257, A211216, A002559, A082582.
a(n)= A060920(n, 0).
Row 3 of array A094954.
Equals A001654(n+1)  A001654(n1), n > 0.
A122367 is another version. Inverse sequences A130255 and A130256. Row sums of A140068, A152251, A153342, A179806, A179745, A213948.
Cf. A001622, A001906, A104457, A153387, A322602.
Sequence in context: A027933 A141448 A011783 * A048575 A099496 A122367
Adjacent sequences: A001516 A001517 A001518 * A001520 A001521 A001522


KEYWORD

nonn,nice,easy,core,changed


AUTHOR

N. J. A. Sloane


EXTENSIONS

Entry revised by N. J. A. Sloane, Aug 24 2006, May 13 2008


STATUS

approved



