A005248 -id:A005248 - cp's OEIS frontend

A286183 Number of connected induced (non-null) subgraphs of the antiprism graph with 2n nodes.

3, 15, 60, 207, 663, 2038, 6107, 17983, 52272, 150407, 429223, 1216490, 3427635, 9609327, 26821668, 74576703, 206650167, 570877918, 1572754187, 4322192287, 11851474968, 32430381815, 88576465735, 241511251922, 657457204323, 1787147867343, 4851349002252

Offset: 1

Giovanni Resta, May 04 2017

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Antiprism Graph
Eric Weisstein's World of Mathematics, Vertex-Induced Subgraph

Cf. A020873 (wheel), A059020 (ladder), A059525 (grid), A286139 (king), A286182 (prism), A286184 (helm), A286185 (Möbius ladder), A286186 (friendship), A286187 (web), A286188 (gear), A286189 (rook), A285765 (queen).

a[n_] := Block[{g = Graph@ Flatten@ Table[{i <-> Mod[i,n]+1, n+i <-> Mod[i,n] + n+1, i <-> n + Mod[i, n] + 1, i <-> n + Mod[i-1, n] + 1}, {i, n}]}, -1 + ParallelSum[ Boole@ ConnectedGraphQ@ Subgraph[g, s], {s, Subsets@ Range[2 n]}]]; Array[a, 8]

a(n) = 8*a(n-1) - 24*a(n-2) + 34*a(n-3) - 24*a(n-4) + 8*a(n-5) - a(n-6), for n > 6 (conjectured).
a(n) = A005248(n) - 2*n + 2*n*A001906(n) (conjectured). - Eric W. Weisstein, May 08 2017
G.f.: x*(3 - 9*x + 12*x^2 - 15*x^3 + 9*x^4 - 2*x^5) / ((1 - x)^2*(1 - 3*x + x^2)^2) (conjectured). - Colin Barker, May 30 2017

a(17)-a(27) from Andrew Howroyd, May 20 2017

A087215 Lucas(6n): a(n) = 18a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

Original entry on oeis.org

2, 18, 322, 5778, 103682, 1860498, 33385282, 599074578, 10749957122, 192900153618, 3461452808002, 62113250390418, 1114577054219522, 20000273725560978, 358890350005878082, 6440026026380244498, 115561578124838522882, 2073668380220713167378

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 19 2003

a(n+1)/a(n) converges to 9 + sqrt(80) = 17.9442719... a(0)/a(1) = 2/18; a(1)/a(2) = 18/322; a(2)/a(3) = 322/5778; a(3)/a(4) = 5778/103682; etc.
Lim_{n -> oo} a(n)/a(n+1) = 0.05572809000084... = 1/(9 + sqrt(80)) = 9 - sqrt(80).
From Peter Bala, Oct 13 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = (1/2)*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^6) = 1.0555459720... = 1 + 1/(18 + 1/(322 + 1/(5778 + ...))).
Also F(-Phi^6) = 0.9444348576... has the continued fraction representation 1 - 1/(18 - 1/(322 - 1/(5788 - ...))) and the simple continued fraction expansion 1/(1 + 1/((18 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/((5788 - 2) + 1/(1 + ...))))))).
F(Phi^6)*F(-Phi^6) = 0.9968944099... has the simple continued fraction expansion 1/(1 + 1/((18^2 - 4) + 1/(1 + 1/((322^2 - 4) + 1/(1 + 1/((5788^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^6)/F(-Phi^6) = 1.0588241282... has the simple continued fraction expansion 1 + 1/((18 - 2) + 1/(1 + 1/((5778 - 2) + 1/(1 + 1/(1860498 - 2) + 1/(1 + ...))))). (End)

			a(4) = 103682 = 18*a(3) - a(2) = 18*5778 - 322 = (9 + sqrt(80))^4 + (9 - sqrt(80))^4 = 103681.99999035512... + 0.00000964487... = 103682.

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

Colin Barker, Table of n, a(n) for n = 0..750
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence R_4.
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A074919.
Row 2 * 2 of array A188645.
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

[ Lucas(6*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a[0] = 2; a[1] = 18; a[n_] := 18a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
Table[LucasL[6n], {n, 0, 18}]  (* or *) CoefficientList[Series[2*(1 - 9*x)/(1 - 18*x + x^2), {x, 0, 17}], x] (* Indranil Ghosh, Mar 15 2017 *)

Vec(2*(1-9*x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016

a(n) = if(n<2, 17^n + 1, 18*a(n - 1) - a(n - 2));
for(n=0, 17, print1(a(n),", ")) \\ Indranil Ghosh, Mar 15 2017

a(n) = A000032(6*n).
a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.
a(n) = (9 + sqrt(80))^n + (9 - sqrt(80))^n.
G.f.: 2*(1-9*x)/(1-18*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = 2*A023039(n). - R. J. Mathar, Oct 22 2010
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(6*n+6)/F(6) - F(6*n-6)/F(6) = A049660(n+1) - A049660(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^6 = [5, 8; 8, 13].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
16*Sum_{n >= 1} 1/(a(n) - 20/a(n)) = 1: (20 = Lucas(6) + 2 and 16 = Lucas(6) - 2)
20*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 16/a(n)) = 1.
Series acceleration formulas for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/16 - 20*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 20)).
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/20 + 16*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 16)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(9-4*sqrt(5)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(4*sqrt(5)-9))^2 )/4,
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 18*x^2 + 323*x^3 + ... is the o.g.f. for A049660. (End)
E.g.f.: 2*exp(9*x)*cosh(4*sqrt(5)*x). - Stefano Spezia, Oct 18 2019
a(n) = L(2n-1)^2 * F(2n+1) + L(2n+1)^2 * F(2n-1), where F(n) = A000045(n) and L(n) = A000032(n). - Diego Rattaggi, Nov 12 2020
From  Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^3 - 3*Lucas(2*n) = 2*T(3, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind; more generally, for k >= 0, Lucas(2*k*n) = 2*T(k, Lucas(2*n)/2).
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3(9 - sqrt(80))^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3(sqrt(80) - 9)^2), where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

A103997 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2M X 2N Moebius strip.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 11, 7, 1, 1, 41, 71, 18, 1, 1, 153, 769, 539, 47, 1, 1, 571, 8449, 17753, 4271, 123, 1, 1, 2131, 93127, 603126, 434657, 34276, 322, 1, 1, 7953, 1027207, 20721019, 46069729, 10894561, 276119, 843, 1, 1, 29681, 11332097, 714790675, 4974089647, 3625549353, 275770321, 2226851, 2207, 1

Offset: 0

Ralf Stephan, Feb 26 2005

			Array begins:
  1,   1,     1,        1,          1,             1,               1,
  1,   3,     7,       18,         47,           123,             322,
  1,  11,    71,      539,       4271,         34276,          276119,
  1,  41,   769,    17753,     434657,      10894561,       275770321,
  1, 153,  8449,   603126,   46069729,    3625549353,    289625349454,
  1, 571, 93127, 20721019, 4974089647, 1234496016491, 312007855309063,
  ...

Laura Florescu, Daniela Morar, David Perkinson, Nicholas Salter and Tianyuan Xu, Sandpiles and Dominos, El. J. Comb., 22 (2015), P1.66. See Theorem 18.
W. T. Lu and F. Y. Wu, Dimer statistics on the Moebius strip and the Klein bottle, arXiv:cond-mat/9906154 [cond-mat.stat-mech], 1999.
Index entries for sequences related to dominoes

Rows include A005248, A103998.
Columns 1..7 give A001835(n+1), A334135, A334179, A334180, A334181, A334182, A334183.
Main diagonal gives A334124.
Cf. A099390, A103999, A341741, A334178, A256045, A348566.

T[M_, N_] := Product[4Sin[(4n-1)Pi/(4N)]^2 + 4Cos[m Pi/(2M+1)]^2, {n, 1, N}, {m, 1, M}];
Table[T[M - N, N] // Round, {M, 0, 9}, {N, 0, M}] // Flatten (* Jean-François Alcover, Dec 03 2018 *)

T(M, N) = Product_{m=1..M} (Product_{n=1..N} 4*sin(Pi*(4*n-1)/(4*N))^2 + 4*cos(Pi*m/(2*M + 1))^2).

For k > 0, T(n,k) = 2^n * sqrt(Resultant(U_{2*n}(x/2), T_{2*k}(i*x/2))), where T_n(x) is a Chebyshev polynomial of the first kind, U_n(x) is a Chebyshev polynomial of the second kind and i = sqrt(-1). - Seiichi Manyama, Apr 15 2020

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

Christian G. Bower

nonn

			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)

Vaclav Kotesovec, Table of n, a(n) for n = 1..1000
A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31 (11) (2000), 1421-1427.
C. G. Bower, Transforms (2).
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 31.
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
P. Flajolet and M. Soria, The Cycle Construction. [pdf file]
Meghann Moriah Gibson, Combinatorics of compositions, Master of Science, Georgia Southern University, 2017.
Meghann Moriah Gibson, Daniel Gray, and Hua Wang, Combinatorics of n-color compositions, Discrete Mathematics 341 (2018), 3209-3226.
Index entries for sequences related to necklaces

Cf. A000032, A001906, A004146, A005248, A032170, A032287, A088305, A308723, A308747.

Equals A005594(n) - 1.

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 *)

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 */

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)

A056571 Fourth power of Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 1, 16, 81, 625, 4096, 28561, 194481, 1336336, 9150625, 62742241, 429981696, 2947295521, 20200652641, 138458410000, 949005240561, 6504586067281, 44583076827136, 305577005139121, 2094455819300625, 14355614096087056

Offset: 0

Wolfdieter Lang, Jul 10 2000

Divisibility sequence; that is, if n divides m, then a(n) divides a(m).

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/4,3/4)-fences and quarter-squares (1/4 X 1 pieces, always placed so that the shorter sides are horizontal). A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/8,3/8)-fences and (1/8,7/8)-fences. - Michael A. Allen, Jan 11 2022

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 31.
Donald E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 85, (exercise 1.2.8. Nr. 30) and p. 492 (solution).
Hilary I. Okagbue, Muminu O. Adamu, Sheila A. Bishop and Abiodun A. Opanuga, Digit and Iterative Digit Sum of Fibonacci numbers, their identities and powers, International Journal of Applied Engineering Research ISSN 0973-4562 Volume 11, Number 6 (2016) pp. 4623-4627.

Vincenzo Librandi, Table of n, a(n) for n = 0..151
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
Alfred Brousseau, A sequence of power formulas, Fib. Quart., Vol. 6, No. 1 (1968), pp. 81-83.
Andrej Dujella, A bijective proof of Riordan's theorem on powers of Fibonacci numbers, Discrete Math., Vol. 199, No. 1-3 (1999), pp. 217-220. MR1675924 (99k:05016).
Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers cubed, Fib. Q. 58:5 (2020) 128-134.
Hideyuki Ohtsuka, Problem N-1220, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 55, No. 4 (2017), p. 368; Gelin-Cesàro Identity Yields a Telescoping Product, Solution to Problem H-790 by Ramya Dutta, ibid., Vol. 56, No. 4 (2018), p. 372.
John Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J., Vol. 29, No. 1 (1962), pp. 5-12.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000045, A001622, A007598, A056570, A056588, A055870.
First differences of A005969.
Fourth row of array A103323.

[Fibonacci(n)^4: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

A056571 := proc(n)
    combinat[fibonacci](n)^4 ;
end proc:
seq(A056571(n),n=0..10) ; # R. J. Mathar, Jan 23 2022

Fibonacci[Range[0, 25]]^4 (* Wesley Ivan Hurt, Jan 11 2022 *)

a(n)=fibonacci(n)^4 \\ Charles R Greathouse IV, Oct 07 2016

a(n) = F(n)^4 = A007598(n)^2, F(n) = A000045(n).
G.f.: x*p(4, x)/q(4, x) with p(4, x) := sum(A056588(3, m)*x^m, m=0..3) = 1 - 4*x - 4*x^2 + x^3 = (1+x)*(1 - 5*x + x^2) and q(4, x) := Sum_{m=0..5} A055870(5, m)*x^m = 1 - 5*x - 15*x^2 + 15*x^3 + 5*x^4 - x^5 = (1-x)*(1 + 3*x + x^2)*(1 - 7*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): 1*a(n) - 5*a(n-1) - 15*a(n-2) + 15*a(n-3) + 5*a(n-4) - 1*a(n-5) = 0, n >= 5; inputs: a(n), n=0..4.
(1/25)*((-1)^n*(2*F(2*n-2) - 6*F(2*n+1)) + 2*F(4*n-1) + F(4*n) + 6). - Ralf Stephan, May 14 2004
a(n) = F(n-2)*F(n-1)*F(n+1)*F(n+2) + 1 = A244855(n)+1.
Sum_{j=0..n} binomial(n,j)*a(j) = (3^n*A005248(n) - 4*(-1)^n*A000032(n) + 6*2^n)/25. Sum_{j=0..n} (-1)^j*binomial(n,j)*a(j) = -5^((n+1)/2-2)*(A001906(n) + 4*A000045(n)) if n odd. - R. J. Mathar, Oct 16 2006
a(n) = (F(n)*F(3n) - 3*F(n)^2*(-1)^n)/5. - Gary Detlefs, Dec 26 2010
Product_{n>=3} (1 - 1/a(n)) = phi^5/12, where phi is the golden ratio (A001622)(Ohtsuka, 2017). - Amiram Eldar, Dec 02 2021

A064170 a(1) = 1; a(n+1) = product of numerator and denominator in Sum_{k=1..n} 1/a(k).

Original entry on oeis.org

1, 1, 2, 10, 65, 442, 3026, 20737, 142130, 974170, 6677057, 45765226, 313679522, 2149991425, 14736260450, 101003831722, 692290561601, 4745030099482, 32522920134770, 222915410843905, 1527884955772562, 10472279279564026, 71778070001175617, 491974210728665290

Offset: 1

Leroy Quet, Sep 19 2001

The numerator and denominator in the definition have no common divisors >1.
Also denominators in a system of Egyptian fraction for ratios of consecutive Fibonacci numbers: 1/2 = 1/2, 3/5 = 1/2 + 1/10, 8/13 = 1/2 + 1/10 + 1/65, 21/34 = 1/2 + 1/10 + 1/65 + 1/442 etc. (Rossi and Tout). - Barry Cipra, Jun 06 2002
a(n)-1 is a square. - Sture Sjöstedt, Nov 04 2011
From Wolfdieter Lang, May 26 2020: (Start)
Partial sums of the reciprocals: Sum_{k=1..n} 1/a(k) equal 1 for n=1, and F(2*n - 1)/F(2*n - 3) for n >= 2, with F = A000045. Proof by induction. Hence a(n) = 1 for n=1, and F(2*n - 3)*F(2*n - 5) for n >= 2, with F(-1) = 1 (gcd(F(n), F(n+1)) = 1). See the comment by Barry Cipra.
Thus a(n) = 1, for n = 1, and a(n) = 1 + F(2*(n-2))^2, for n >= 2 (by Cassini-Simson for even index, e.g., Vajda, p. 178 eq.(28)). See the Sture Sjöstedt comment.
The known G.f. of {F(2*n)^2} from A049684 leds then to the conjectured formula by R. J. Mathar below, and this proves also the recurrence given there..
From the partial sums the series Sum_{k>=1} 1/a(k) converges to 1 + phi, with phi = A001622. See the formulas by Gary W. Adamson and Diego Rattaggi below. (End)

			1/a(1) + 1/a(2) + 1/a(3) + 1/a(4) = 1 + 1 + 1/2 + 1/10 = 13/5. So a(5) = 13 * 5 = 65.

S. Vajda, Fibonacci & Lucas Numbers, and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Michael De Vlieger, Table of n, a(n) for n = 1..1199
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
C. Rossi and C. A. Tout, Were the Fibonacci Series and the Golden Section Known in Ancient Egypt?, Historia Mathematica, vol. 29 (2002), 101-113.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A059929, A058038.
Cf. A033890 (first differences). - R. J. Mathar, Jul 03 2009
Cf. A001906, A001622.

A064170[1] := 1; A064170[n_] := A064170[n] = Module[{temp = Sum[1/A064170[i], {i, n - 1}]}, Numerator[temp] Denominator[temp]]; Table[A064170[n], {n, 20}](* Alonso del Arte, Sep 05 2013 *)
Join[{1}, LinearRecurrence[{8, -8, 1}, {1, 2, 10}, 23]] (* Jean-François Alcover, Sep 22 2017 *)

a(n) = Fibonacci(2*n-5)*Fibonacci(2*n-3), for n >= 3. - Barry Cipra, Jun 06 2002
Sum_{n>=3} 1/a(n) = 2/(1+sqrt(5)) = phi - 1, with phi = A001622. - Gary W. Adamson, Jun 07 2003
Conjecture: a(n) = 8*a(n-1)-8*a(n-2)+a(n-3), n>4. G.f.: -x*(2*x^2+x^3-7*x+1)/((x-1)*(x^2-7*x+1)). - R. J. Mathar, Jul 03 2009 [For a proof see the W. Lang comment above.]
a(n+1) = (A005248(n)^2 - A001906(n)^2)/4, for n => 0. - Richard R. Forberg, Sep 05 2013
From Diego Rattaggi, Apr 21 2020: (Start)
a(n) = 1 + A049684(n-2) for n>1.
Sum_{n>=2} 1/a(n) = phi = (1+sqrt(5))/2 = A001622.
Sum_{n>=1} 1/a(n) = phi^2 = 1 + phi. (End) [See a comment above for the proof]
a(n) = F(2*n - 3)*F(2*n - 5) = 1 + F(2*(n - 2))^2, for n >= 2, with F(-1) = 1. See the W. Lang comments above. - Wolfdieter Lang, May 26 2020

A075796 Numbers k such that 5*k^2 + 5 is a square.

Original entry on oeis.org

2, 38, 682, 12238, 219602, 3940598, 70711162, 1268860318, 22768774562, 408569081798, 7331474697802, 131557975478638, 2360712083917682, 42361259535039638, 760141959546795802, 13640194012307284798, 244763350261984330562, 4392100110703410665318, 78813038642399407645162

Offset: 1

Gregory V. Richardson, Oct 13 2002

Bisection of A001077; a(n) = A001077(2*n-1). - Greg Dresden, Jun 08 2021
From Peter Bala, Aug 25 2022: (Start)
The aerated sequence (b(n))n>=1 = [2, 0, 38, 0, 682, 0, 1238, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). The sequence (1/2)*(b(n))n>=1 is the case P1 = 0, P2 = -16, Q = -1  of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. (End)

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A000290, A306380, A001077.

I:=[2,38]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 30 2011

[Lucas(6*n-3)/2: n in [1..20]]; // G. C. Greubel, Feb 13 2019

with(combinat); A075796:=n->fibonacci(6*n+3)+fibonacci(6*n)/2; seq(A075796(n), n=1..50); # Wesley Ivan Hurt, Nov 29 2013

LinearRecurrence[{18, -1}, {2, 38}, 50] (* Sture Sjöstedt, Nov 29 2011; typo fixed by Vincenzo Librandi, Nov 30 2011 *)
LucasL[6*Range[20]-3]/2 (* G. C. Greubel, Feb 13 2019 *)
CoefficientList[Series[2*(1+x)/( 1-18*x+x^2 ), {x,0,20}],x] (* Stefano Spezia, Mar 02 2019 *)

vector(20, n, (fibonacci(6*n-2) + fibonacci(6*n-4))/2) \\ G. C. Greubel, Feb 13 2019

[(fibonacci(6*n-2) + fibonacci(6*n-4))/2 for n in (1..20)] # G. C. Greubel, Feb 13 2019

a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n) + ((9 + 4*sqrt(5))^(n-1) - (9 - 4*sqrt(5))^(n-1)))/(4*sqrt(5)).
a(n) = 18*a(n-1) - a(n-2).
a(n) = 2*A049629(n-1).
Limit_{n->oo} a(n)/a(n-1) = 8*phi + 1 = 9 + 4*sqrt(5).
a(n+1) = 9*a(n) + 4*sqrt(5)*sqrt((a(n)^2+1)). - Richard Choulet, Aug 30 2007
G.f.: 2*x*(1 + x)/(1 - 18*x + x^2). - Richard Choulet, Oct 09 2007
From Johannes W. Meijer, Jul 01 2010: (Start)
a(n) = A000045(6*n+3) + A000045(6*n)/2.
a(n) = 2*A167808(6*n+4) - A167808(6*n+6).
Limit_{k->oo} a(n+k)/a(k) = A023039(n)*A060645(n)*sqrt(5).
(End)
5*A007805(n)^2 - 1 = a(n+1)^2. - Sture Sjöstedt, Nov 29 2011
From Peter Bala, Nov 29 2013: (Start)
a(n) = Lucas(6*n - 3)/2.
Sum_{n >= 1} 1/(a(n) + 5/a(n)) = 1/4. Compare with A002878, A005248, A023039. (End)
Limit_{n->oo} a(n)/A007805(n-1) = sqrt(5). - A.H.M. Smeets, May 29 2017
E.g.f.: (exp((9 - 4*sqrt(5))*x)*(- 5 + 2*sqrt(5) + (5 + 2*sqrt(5))*exp(8*sqrt(5)*x)))/(2*sqrt(5)). - Stefano Spezia, Feb 13 2019
Sum_{n > 0} 1/a(n) = (1/log(9 - 4*sqrt(5)))*(- 17 - 38/sqrt(5))*sqrt(5*(9 - 4*sqrt(5)))*(- 9 + 4*sqrt(5))*(psi_{9 - 4*sqrt(5)}(1/2) - psi_{9 - 4*sqrt(5)}(1/2 - (I*Pi)/log(9 - 4*sqrt(5)))) approximately equal to 0.527868600269500798938265500122302016..., where psi_q(x) is the q-digamma function. - Stefano Spezia, Feb 25 2019
a(n) = sinh((6*n - 3)*arccsch(2)). - Peter Luschny, May 25 2022

A060924 Bisection of Lucas triangle A060922: odd-indexed members of column sequences of A060922 (not counting leading zeros).

Original entry on oeis.org

3, 7, 6, 18, 38, 9, 47, 158, 120, 12, 123, 566, 753, 280, 15, 322, 1880, 3612, 2568, 545, 18, 843, 5964, 15040, 16220, 7043, 942, 21, 2207, 18342, 57366, 83780, 57560, 16536, 1498, 24, 5778, 55162, 206115

Offset: 0

Wolfdieter Lang, Apr 20 2001

Row sums give A060927. Column sequences (without leading zeros) are, for m=0..5: A005248(n+1), 2*A061171, A061172, 4*A061173, A061174, 2*A061175.

Companion triangle A060923 (even part).

			{3}; {7,6}; {18,38,9}; {47,158,120,12}; ...; pLo(2,x)= 2*(3+x-2*x^2).

Cf. A005248.

a(n, m) = A060922(2*n+1-m, m).
a(n, m) = ((2*n-m+1)*A060923(n, m-1) + 2*(2*(2*n+1)-3*m)*a(n-1, m-1) + 4*(2*n-m)*A060923(n-1, m-1))/(5*m), m >= n >= 1; a(n, 0) = A005248(n); otherwise 0.
G.f. for column m >= 0: x^m*pLo(m+1, x)/(1-3*x+x^2)^(m+1), where pLo(n, x) := Sum_{m=0..n+floor((n-1)/2)} A061187(n-1, m)*x^m are the row polynomials of the (signed) staircase A061187.
T(n,k) = 3*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) + 2*T(n-2,k-1) - T(n-2,k-2) + 4*T(n-3,k-2), T(0,0) = 3, T(1,0) = 7, T(1,1) = 6, T(2,0) = 18, T(2,1) = 38, T(2,2) = 9, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 21 2014

A087265 Lucas numbers L(8*n).

Original entry on oeis.org

2, 47, 2207, 103682, 4870847, 228826127, 10749957122, 505019158607, 23725150497407, 1114577054219522, 52361396397820127, 2459871053643326447, 115561578124838522882, 5428934300813767249007, 255044350560122222180447

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 19 2003

a(n+1)/a(n) converges to (47+sqrt(2205))/2 = 46.9787137... a(0)/a(1)=2/47; a(1)/a(2)=47/2207; a(2)/a(3)=2207/103682; a(3)/a(4)=103682/4870847; etc. Lim_{n->infinity} a(n)/a(n+1) = 0.02128623625... = 2/(47+sqrt(2205)) = (47-sqrt(2205))/2.
a(n) = a(-n). - Alois P. Heinz, Aug 07 2008
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^8) = 1.0212763906... = 1 + 1/(47 + 1/(2207 + 1/(103682 + ...))).
Also F(-Phi^8) = 0.9787231991... has the continued fraction representation 1 - 1/(47 - 1/(2207 - 1/(103682 - ...))) and the simple continued fraction expansion 1/(1 + 1/((47 - 2) + 1/(1 + 1/((2207 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + ...))))))).
F(Phi^8)*F(-Phi^8) = 0.9995468962... has the simple continued fraction expansion 1/(1 + 1/((47^2 - 4) + 1/(1 + 1/((2207^2 - 4) + 1/(1 + 1/((103682^2 - 4) + 1/(1 + ...))))))).
1/2 + 1/2*F(Phi^8)/F(-Phi^8) = 1.0217391349... has the simple continued fraction expansion 1 + 1/((47 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + 1/(228826127 - 2) + 1/(1 + ...))))). (End)

			a(4) = 4870847 = 47*a(3) - a(2) = 47*103682 - 2207=((47+sqrt(2205))/2)^4 + ( (47-sqrt(2205))/2)^4 =4870846.999999794696 + 0.000000205303 = 4870847.

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

Indranil Ghosh, Table of n, a(n) for n = 0..596
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (47,-1).

Cf. A000032. Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

a(n) = A000032(8n).

[ Lucas(8*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a:= n-> (Matrix([[2,47]]). Matrix([[47,1],[ -1,0]])^(n))[1,1]:
seq(a(n), n=0..14);  # Alois P. Heinz, Aug 07 2008

LucasL[8*Range[0,20]] (* or *) LinearRecurrence[{47,-1},{2,47},20] (* Harvey P. Dale, Oct 23 2017 *)

a(n) = 47*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 47.
a(n) = ((47+sqrt(2205))/2)^n + ((47-sqrt(2205))/2)^n
(a(n))^2 = a(2n)+2.
G.f.: (2-47*x)/(1-47*x+x^2). - Alois P. Heinz, Aug 07 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(8*n+8)/F(8) - F(8*n-8)/F(8) = A049668(n+1) - A049668(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^8 = [13, 21; 21, 34].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
45*Sum_{n >= 1} 1/(a(n) - 49/a(n)) = 1: (49 = Lucas(8) + 2 and 45 = Lucas(8) - 2)
49*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 45/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 47*x^2 + 2208*x^3 + ... is the o.g.f. for A049668. (End)
E.g.f.: 2*exp(47*x/2)*cosh(21*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^4 - 4*Lucas(2*n)^2 + 2 = 2*T(4, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind; more generally, for k >= 0, Lucas(2*k*n) = 2*T(k, Lucas(2*n)/2).
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (47 - sqrt(2205))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(2205) - 47)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

Terms a(22)-a(27) from John W. Layman, Jun 14 2004

A098648 Expansion of (1-3x)/(1 - 6x + 4*x^2).

Original entry on oeis.org

1, 3, 14, 72, 376, 1968, 10304, 53952, 282496, 1479168, 7745024, 40553472, 212340736, 1111830528, 5821620224, 30482399232, 159607914496, 835717890048, 4375875682304, 22912382533632, 119970792472576, 628175224700928, 3289168178315264, 17222308171087872

Offset: 0

Paul Barry, Sep 18 2004

Binomial transform of A001077. Second binomial transform of A084057. Third binomial transform of 1/(1-5*x^2). Let A=[1,1,1,1;3,1,-1,-3;3,-1,-1,3;1,-1,1,-1], the 4 X 4 Krawtchouk matrix. Then a(n)=trace((16(A*A`)^(-1))^n)/4.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (6,-4).

Cf. A098647.

a[n_]:=(MatrixPower[{{5,1},{1,1}},n].{{2},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
CoefficientList[Series[(1-3x)/(1-6x+4x^2),{x,0,30}],x] (* or *) LinearRecurrence[{6,-4},{1,3},31] (* Harvey P. Dale, Jun 06 2011 *)
Table[2^(n - 1) LucasL[2 n], {n, 0, 20}] (* Eric W. Weisstein, Mar 31 2017 *)

Vec((1-3*x)/(1 - 6*x + 4*x^2) + O(x^25)) \\ Jinyuan Wang, Jul 24 2021

E.g.f.: exp(3*x)*cosh(sqrt(5)*x).
a(n) = ((3-sqrt(5))^n + (3+sqrt(5))^n)/2.
a(n) = 2*(3*a(n-1) - 2*a(n-2)). - Lekraj Beedassy, Oct 22 2004
a(n) = A084326(n+1) - 3*A084326(n). - R. J. Mathar, Nov 10 2013
a(n) = 2^(n-1)*Lucas(2*n) = 2^(n-1)*A005248(n), n>0. - Eric W. Weisstein, Mar 31 2017

cp's OEIS Frontend

A286183 Number of connected induced (non-null) subgraphs of the antiprism graph with 2n nodes.

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A087215 Lucas(6*n): a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

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A103997 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2*M X 2*N Moebius strip.

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

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A056571 Fourth power of Fibonacci numbers A000045.

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A064170 a(1) = 1; a(n+1) = product of numerator and denominator in Sum_{k=1..n} 1/a(k).

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A075796 Numbers k such that 5*k^2 + 5 is a square.

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A087215 Lucas(6n): a(n) = 18a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

A103997 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2M X 2N Moebius strip.

A098648 Expansion of (1-3x)/(1 - 6x + 4*x^2).