cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A004146 Alternate Lucas numbers - 2.

Original entry on oeis.org

0, 1, 5, 16, 45, 121, 320, 841, 2205, 5776, 15125, 39601, 103680, 271441, 710645, 1860496, 4870845, 12752041, 33385280, 87403801, 228826125, 599074576, 1568397605, 4106118241, 10749957120, 28143753121, 73681302245, 192900153616, 505019158605, 1322157322201
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

This is the r=5 member in the r-family of sequences S_r(n) defined in A092184 where more information can be found.
Number of spanning trees of the wheel W_n on n+1 vertices. - Emeric Deutsch, Mar 27 2005
Also number of spanning trees of the n-helm graph. - Eric W. Weisstein, Jul 16 2011
a(n) is the smallest number requiring n terms when expressed as a sum of Lucas numbers (A000204). - David W. Wilson, Jan 10 2006
This sequence has a primitive prime divisor for all terms beyond the twelfth. - Anthony Flatters (Anthony.Flatters(AT)uea.ac.uk), Aug 17 2007
From Giorgio Balzarotti, Mar 11 2009: (Start)
Determinant of power series of gamma matrix with determinant 1:
a(n) = Determinant(A + A^2 + A^3 + A^4 + A^5 + ... + A^n)
where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant
A = [[1,1,1,1,1,1,...],[1,2,1,2,1,2,...],[1,2,3,1,2,3,...],[1,2,3,4,1,2,...],
[1,2,3,4,5,1,...],[1,2,3,4,5,6,...],...]. Note: Determinant A(1..n,1..n)= (n-1)!.
See A158039, A158040, A158041, A158042, A158043, A158044, for sequences of matrix 2!,3!,... (End)
The previous comment could be rephrased as: a(n) = -det(A^n - I) where I is the 2 X 2 identity matrix and A = [1, 1; 1, 2]. - Peter Bala, Mar 20 2015
a(n) is also the number of points of Arnold's "cat map" that are on orbits of period n-1. This is a map of the two-torus T^2 into itself. If we regard T^2 as R^2 / Z^2, the action of this map on a two vector in R^2 is multiplication by the unit-determinant matrix A = [2, 1;1, 1], with the vector components taken modulo one. As such, an explicit formula for the n-th entry of this sequence is -det(I-A^n). - Bruce Boghosian, Apr 26 2009
7*a(n) gives the total number of vertices in a heptagonal hyperbolic lattice {7,3} with n total levels, in which an open heptagon is centered at the origin. - Robert M. Ziff, Apr 10 2011
The sequence is the case P1 = 5, P2 = 6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014
Determinants of the spiral knots S(3,k,(1,-1)). a(k) = det(S(3,k,(1,-1))). These knots are also the weaving knots W(k,3) and the Turk's Head Links THK(3,k). - Ryan Stees, Dec 14 2014
Even-indexed Fibonacci numbers (1, 3, 8, 21, ...) convolved with (1, 2, 2, 2, 2, ...). - Gary W. Adamson, Aug 09 2016
a(n) is the number of ways to tile a bracelet of length n with 1-color square, 2-color dominos, 3-color trominos, etc. - Yu Xiao, May 23 2020
a(n) is the number of face-labeled unfoldings of a pyramid whose base is a simple n-gon. Cf. A103536. - Rick Mabry, Apr 17 2023

Examples

			For k=3, b(3) = sqrt(5)*b(2) - b(1) = 5 - 1 = 4, so det(S(3,3,(1,-1))) = 4^2 = 16.
G.f. = x + 5*x^2 + 16*x^3 + 45*x^4 + 121*x^5 + 320*x^4 + 841*x^5 + ... - _Michael Somos_, Feb 10 2023

References

  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (p. 193, Problem 3.3.40 (a)).
  • N. Hartsfield and G. Ringel, Pearls in Graph Theory, p. 102. Academic Press: 1990.
  • B. Hasselblatt and A. Katok, "Introduction to the Modern Theory of Dynamical Systems," Cambridge University Press, 1997.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

This is the r=5 member of the family S_r(n) defined in A092184.
Cf. A005248. Partial sums of A002878. Pairwise sums of A027941. Bisection of A074392.
Sequence A032170, the Möbius transform of this sequence, is then the number of prime periodic orbits of Arnold's cat map. - Bruce Boghosian, Apr 26 2009
Cf. A000032, A100047, A001350.
Cf. also A158039, A158040, A158041, A158042, A158043, A158044.
Cf. A103536 for the number of geometrically distinct edge-unfoldings of the regular pyramid. - Rick Mabry, Apr 17 2023

Programs

  • Magma
    [Lucas(n)-2: n in [0..60 by 2]]; // Vincenzo Librandi, Mar 20 2015
  • Mathematica
    Table[LucasL[2*n] - 2, {n, 0, 20}]
(* Second program: *)
LinearRecurrence[{4, -4, 1}, {0, 1, 5}, 30] (* Jean-François Alcover, Jan 08 2019 *)
  • PARI
    a(n) = { we = quadgen(5);((1+we)^n) + ((2-we)^n) - 2;} /* Michel Marcus, Aug 18 2012 */

Formula

a(n) = A005248(n) - 2 = A000032(2*n) - 2.
a(n+1) = 3*a(n) - a(n-1) + 2.
G.f.: x*(1+x)/(1-4*x+4*x^2-x^3) = x*(1+x)/((1-x)*(1-3*x+x^2)).
a(n) = 2*(T(n, 3/2)-1) with Chebyshev's polynomials T(n, x) of the first kind. See their coefficient triangle A053120.
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n>=3, a(0)=0, a(1)=1, a(2)=5.
a(n) = 2*T(n, 3/2) - 2, with twice the Chebyshev polynomials of the first kind, 2*T(n, x=3/2) = A005248(n).
a(n) = b(n) + b(n-1), n>=1, with b(n):=A027941(n-1), n>=1, b(-1):=0, the partial sums of S(n, 3) = U(n, 3/2) = A001906(n+1), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind.
a(2n) = A000204(2n)^2 - 4 = 5*A000045(2n)^2; a(2n+1) = A000204(2n+1)^2. - David W. Wilson, Jan 10 2006
a(n) = ((3+sqrt(5))/2)^n + ((3-sqrt(5))/2)^n - 2. - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jun 09 2001
a(n) = b(n-1) + b(n-2), n>=1, with b(n):=A027941(n), b(-1):=0, partial sums of S(n, 3) = U(n, 3/2) = A001906(n+1), Chebyshev's polynomials of the second kind.
a(n) = n*Sum_{k=1..n} binomial(n+k-1,2*k-1)/k, n > 0. - Vladimir Kruchinin, Sep 03 2010
a(n) = floor(tau^(2*n)*(tau^(2*n) - floor(tau^(2*n)))), where tau = (1+sqrt(5))/2. - L. Edson Jeffery, Aug 26 2013
From Peter Bala, Apr 03 2014: (Start)
a(n) = U(n-1,sqrt(5)/2)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -3/2; 1, 5/2] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(k) = det(S(3,k,(1,-1))) = b(k)^2, where b(1)=1, b(2)=sqrt(5), b(k)=sqrt(5)*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + Sum_{n >= 1} Fibonacci(2*n)*x^n. Cf. A001350. - Peter Bala, Mar 19 2015
E.g.f.: exp(phi^2*x) + exp(x/phi^2) - 2*exp(x), where phi = (1 + sqrt(5))/2. - G. C. Greubel, Aug 24 2015
a(n) = a(-n) for all n in Z. - Michael Somos, Aug 27 2015
From Peter Bala, Jun 03 2016: (Start)
a(n) = Lucas(2*n) - Lucas(0*n);
a(n)^2 = Lucas(4*n) - 3*Lucas(2*n) + 3*Lucas(0*n) - Lucas(-2*n);
a(n)^3 = Lucas(6*n) - 5*Lucas(4*n) + 10*Lucas(2*n) - 10*Lucas(0*n) + 5*Lucas(-2*n) - Lucas(-4*n) and so on (follows from Binet's formula for Lucas(2*n) and the algebraic identity (x + 1/x - 2)^m = f(x) + f(1/x) where f(x) = (x - 1)^(2*m - 1)/x^(m-1) ). (End)
Limit_{n->infinity} a(n+1)/a(n) = (3 + sqrt(5))/2 = A104457. - Ilya Gutkovskiy, Jun 03 2016
a(n) = (phi^n - phi^(-n))^2, where phi = A001622 = (1 + sqrt(5))/2. - Diego Rattaggi, Jun 10 2020
a(n) = 4*sinh(n*A002390)^2, where A002390 = arcsinh(1/2). - Gleb Koloskov, Sep 18 2021
a(n) = 5*F(n)^2 = L(n)^2 - 4 if n even and a(n) = L(n)^2 = 5*F(n)^2 - 4 if n odd. - Michael Somos, Feb 10 2023
a(n) = n^2 + Sum_{i=1..n-1} i*a(n-i). - Fern Gossow, Dec 03 2024

Extensions

Correction to formula from Nephi Noble (nephi(AT)math.byu.edu), Apr 09 2002
Chebyshev comments from Wolfdieter Lang, Sep 10 2004

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

Original entry on oeis.org

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

Views

Author

Christian G. Bower

Keywords

Examples

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

Links

Crossrefs

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

Programs

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

Formula

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

A203853 a(n) = (1/n) * Sum_{d|n} moebius(n/d) * Lucas(d)^2, where Lucas(n) = A000204(n).

Original entry on oeis.org

1, 4, 5, 10, 24, 50, 120, 270, 640, 1500, 3600, 8610, 20880, 50700, 124024, 304290, 750120, 1854400, 4600200, 11440548, 28527320, 71289000, 178526880, 447910470, 1125750120, 2833885800, 7144449920, 18036373140, 45591631800, 115381697740, 292329067800, 741410800830
Offset: 1

Views

Author

Paul D. Hanna, Jan 07 2012

Keywords

Comments

Apparently the same as A032170, if n > 2. - R. J. Mathar, Jan 11 2012

Examples

			G.f.: F(x) = 1/((1-x-x^2) * (1-3*x^2+x^4)^4 * (1-4*x^3-x^6)^5 * (1-7*x^4+x^8)^10 * (1-11*x^5-x^10)^24 * (1-18*x^6+x^12)^50 * (1-29*x^7-x^14)^120 * ... * (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^a(n) * ...)
where F(x) = exp( Sum_{n>=1} Lucas(n)^3 * x^n/n ) = g.f. of A203803:
F(x) = 1 + x + 14*x^2 + 35*x^3 + 205*x^4 + 744*x^5 + 3414*x^6 + ...
where
log(F(x)) = x + 3^3*x^2/2 + 4^3*x^3/3 + 7^3*x^4/4 + 11^3*x^5/5 + 18^3*x^6/6 + 29^3*x^7/7 + 47^3*x^8/8 + ... + Lucas(n)^3*x^n/n + ...

Links

Crossrefs

Cf. A001254, A203803, A006206, A203854, A203855, A203856, A203857, A203858, A203859, A203800.

Programs

  • Mathematica
    a[n_]:= 1/n DivisorSum[n, MoebiusMu[n/#] LucasL[#]^2 &]; Array[a, 30] (* G. C. Greubel, Dec 25 2017 *)
  • PARI
    {a(n)=if(n<1, 0, sumdiv(n, d, moebius(n/d)*(fibonacci(d-1)+fibonacci(d+1))^2)/n)}
  • PARI
    {Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=local(F=exp(sum(m=1, n, Lucas(m)^3*x^m/m)+x*O(x^n)));if(n==1,1,polcoeff(F*prod(k=1,n-1,(1 - Lucas(k)*x^k + (-1)^k*x^(2*k) +x*O(x^n))^a(k)),n)/Lucas(n))}

Formula

G.f.: 1/Product_{n>=1} (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^a(n) = exp(Sum_{n>=1} Lucas(n)^3 * x^n/n), which is the g.f. of A203803.
a(n) ~ phi^(2*n) / n, where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Sep 02 2017

A180363 a(n) = Lucas(prime(n)).

Original entry on oeis.org

3, 4, 11, 29, 199, 521, 3571, 9349, 64079, 1149851, 3010349, 54018521, 370248451, 969323029, 6643838879, 119218851371, 2139295485799, 5600748293801, 100501350283429, 688846502588399, 1803423556807921, 32361122672259149, 221806434537978679
Offset: 1

Views

Author

Jonathan Vos Post, Aug 31 2010

Keywords

Comments

This is to A030426, Fibonacci(prime(n)), as A000032 (Lucas numbers beginning at 2) is to A000045.

Examples

			a(1) = 3 because the 1st prime is 2, and the 2nd Lucas number is A000032(2) = 3.
a(2) = 4 because the 2nd prime is 3, and the 3rd Lucas number is A000032(3) = 4.
a(3) = 11 because the 3rd prime is 5, and the 5th Lucas number is A000032(5) = 11.

Links

Crossrefs

Cf. A000032, A000040, A000045, A030426.

Programs

  • Magma
    [Lucas(NthPrime(n)): n in [1..30]]; // Vincenzo Librandi, Dec 01 2015
  • Maple
    A180363 := proc(n) A000032(ithprime(n)) ; end proc: seq(A180363(n),n=1..30) ; # R. J. Mathar, Sep 01 2010
# second Maple program:
a:= n-> (<<1|1>, <1|0>>^ithprime(n). <<2, -1>>)[1, 1]:
seq(a(n), n=1..50);  # Alois P. Heinz, Jan 03 2022
  • Mathematica
    LucasL[Prime[Range[30]]] (* Vincenzo Librandi, Dec 01 2015 *)
  • Python
    from sympy import lucas, prime
def a(n): return lucas(prime(n))
print([a(n) for n in range(1, 24)]) # Michael S. Branicky, Dec 30 2021

Formula

a(n) = A000032(A000040(n)) = Lucas(prime(n)).
a(n) = A032170(A000040(n)) / A064723(n-1) - 1 for n>1. - Flávio V. Fernandes, Dec 30 2021

Extensions

Entries checked by R. J. Mathar, Sep 01 2010
Edited by N. J. A. Sloane, Nov 28 2011

A072337 Inverse EULER transform of A064831 (with its initial 1 omitted).

Original entry on oeis.org

1, 3, 3, 5, 10, 24, 50, 120, 270, 640, 1500, 3600, 8610, 20880, 50700, 124024, 304290, 750120, 1854400, 4600200, 11440548, 28527320, 71289000, 178526880, 447910470, 1125750120, 2833885800, 7144449920, 18036373140, 45591631800, 115381697740, 292329067800
Offset: 0

Views

Author

N. J. A. Sloane, Jul 16 2002

Keywords

Comments

Also, dimensions of the graded Yang-Mills algebra ym(3) [Herscovich and Solotar]. - N. J. A. Sloane, Jan 02 2013
Apparently the same as A032170 for n > 2. - Ralf Stephan, Feb 01 2004
From Petros Hadjicostas, Dec 03 2017: (Start)
We clarify the name of the sequence and provide a sketch of a proof of R. Stephan's statement above.
For n>=0, let b(n) = A064831(n+1). Then B(x) = Sum_{n>=0} b(n)*x^n = Sum_{n>=0} A064831(n+1)*x^n = 1/((1-x^2)*(1-3*x+x^2)) (see the formula section for sequence A064831). Define a(0) = 1, and (a(n): n>=1) = invEULER(b(n): n>=0), i.e., we give a new (more correct) definition of the current sequence. Using Bernstein and Sloane (1995), we define the sequence (d(n): n>=1) via Sum_{n>=1} (d(n)/n)*x^n = log B(x). Since EULER(a(n): n>=1) = (b(n): n>=0), we have a(n) = (1/n)*Sum_{s|n} mu(s)*d(n/s). Using the change of indexes m=n/s, we get A(x) = 1 + Sum_{n>=1} a(n)*x^n = 1 + Sum_{s>=1} (mu(s)/s)*Sum_{m>=1} (d(m)/m)*(x^s)^m = 1 + Sum_{s>=1} (mu(s)/s)*log B(x^s).
From the documentation of sequence A032170, we see that its g.f. is Sum_{s>=1} (mu(s)/s)*log C(x^s), where C(x) = (1-x)^2/(1-3*x+x^2). To prove R. Stephan's claim above, we need to prove that Sum_{s>=1} (mu(s)/s)*log C(x^s) - x - 2*x^2 = Sum_{s>=1} (mu(s)/s)*log B(x^s) - 3*x - 3*x^2. The last equality is equivalent to Sum_{s>=1} (mu(s)/s)*log(B(x^s)/C(x^s)) = 2*x + x^2, which in turn is equivalent to -Sum_{s>=1} (mu(s)/s) log((1-x^s)^2*(1-x^{2*s})) = 2*x + x^2. The last equality follows from the identity -Sum_{s>=1} (mu(s)/s)*log(1-y^s) = y, which follows from the Lambert series Sum_{s>=1} mu(s)*y^s/(1-y^s) = y.
(End)

Links

Crossrefs

Cf. A032170, A064831.

Programs

  • Mathematica
    mob[m_, n_] := If[Mod[m, n] == 0, MoebiusMu[m/n], 0];
EULERi[b_] := Module[{a, c, i, d}, c = {}; For[i = 1, i <= Length[b], i++, c = Append[c, i*b[[i]] - Sum[c[[d]]*b[[i - d]], {d, 1, i - 1}]]]; a = {}; For[i = 1, i <= Length[b], i++, a = Append[a, (1/i)*Sum[mob[i, d]*c[[d]], {d, 1, i}]]]; Return[a]];
Join[{1}, EULERi[LinearRecurrence[{3, 0, -3, 1}, {3, 9, 24, 64}, 31]]] (* Jean-François Alcover, Aug 19 2018 *)

Formula

G.f.: 1 + Sum_{s>=1} (mu(s)/s)*log B(x^s), where B(x) = 1/((1-x^2) * (1-3*x+x^2)). - Petros Hadjicostas, Dec 03 2017
a(n) ~ ((1 + sqrt(5))/2)^(2*n) / n. - Vaclav Kotesovec, Aug 19 2018

A308446 Expansion of Product_{k>=1} 1/(1 - x^k)^Fibonacci(2*k).

Original entry on oeis.org

1, 1, 4, 12, 39, 118, 371, 1129, 3468, 10524, 31910, 96155, 289016, 865000, 2581577, 7679762, 22784896, 67418329, 199004329, 586052299, 1722165404, 5050349249, 14781877481, 43185726143, 125949155473, 366716549379, 1066057177765, 3094398005409, 8969054893842
Offset: 0

Views

Author

Ilya Gutkovskiy, May 27 2019

Keywords

Comments

Euler transform of A001906.

Crossrefs

Cf. A000045, A001906, A032170, A034691, A088305, A166861.

Programs

  • Mathematica
    nmax = 28; CoefficientList[Series[Product[1/(1 - x^k)^Fibonacci[2 k], {k, 1, nmax}], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d Fibonacci[2 d], {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 28}]

Formula

a(n) ~ phi^(2*n) * exp(2*sqrt(n)/5^(1/4) - 3/10 + S) / (2 * 5^(1/8) * sqrt(Pi) * n^(3/4)), where S = Sum_{k>=2} 1/((phi^(2*k) - 3 + 1/phi^(2*k))*k) = 0.155349347463140787939176213528043741704916536093946010733676987281... and phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, May 28 2019
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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