A004146 -id:A004146 - cp's OEIS frontend

A005592 a(n) = F(2n+1) + F(2n-1) - 1.

1, 2, 6, 17, 46, 122, 321, 842, 2206, 5777, 15126, 39602, 103681, 271442, 710646, 1860497, 4870846, 12752042, 33385281, 87403802, 228826126, 599074577, 1568397606, 4106118242, 10749957121, 28143753122, 73681302246, 192900153617, 505019158606, 1322157322202

Offset: 0

N. J. A. Sloane

For any m, the maximum element in the continued fraction of F(2n+m)/F(m) is a(n). - Benoit Cloitre, Jan 10 2006
The continued fraction [a(n);1,a(n)-1,1,a(n)-1,...] = phi^(2n), where phi = 1.618... is the golden ratio, A001622. - Thomas Ordowski, Jun 07 2013
a(n) is the number of labeled subgraphs of the n-cycle C_n. For example, a(3)=17. There are 7 subgraphs of the triangle C_3 with 0 edges, 6 with 1 edge, 3 with 2 edges, and 1 with 3 edges (C_3 itself); here 7+6+3+1 = 17. - John P. McSorley, Oct 31 2016
a(n) equals the sum of the n-th row of triangle A277919. - John P. McSorley, Nov 25 2016

			G.f. = 1 + 2*x + 6*x^2 + 17*x^3 + 46*x^4 + 122*x^5 + 321*x^6 + 842*x^7 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..1000 (terms n = 1..200 from Vincenzo Librandi)
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 41, 56.
M. D. McIlroy, The number of states of a dynamic storage system, Computer J., Vol. 25, No. 3 (1982), pp. 388-392.
M. D. McIlroy, The number of states of a dynamic storage system, Computer J., Vol. 25, No. 3 (1982), pp. 388-392. (Annotated scanned copy)
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.
Jesús Salas and Alan D. Sokal, Transfer Matrices and Partition-Function Zeros for Antiferromagnetic Potts Models. V. Further Results for the Square-Lattice Chromatic Polynomial, J. Stat. Phys., Vol. 135 (2009), pp. 279-373; arXiv preprint, arXiv:0711.1738 [cond-mat.stat-mech], 2007-2009. Mentions this sequence. - N. J. A. Sloane, Mar 14 2014
Robert S. Seamons, Problem B-89, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 4, No. 2 (1966), p. 190; A Close Approximation, Solution to Problem B-89 by Douglas Lind, ibid., Vol. 5, No. 1 (1967), pp. 108-109.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Equals A004146+1 and A005248+1.
Bisection of A014217; the other bisection is A002878, which also bisects A000032.
Cf. A000045, A065034.

a005592 n = a005592_list !! (n-1)
a005592_list = map (subtract 1) $
                   tail $ zipWith (+) a001519_list $ tail a001519_list
-- Reinhard Zumkeller, Aug 09 2013

[Fibonacci(2*n+1)+Fibonacci(2*n-1)-1: n in [1..30]]; // Vincenzo Librandi, Aug 23 2011

A005592:=-(2-2*z+z**2)/(z-1)/(z**2-3*z+1); # conjectured by Simon Plouffe in his 1992 dissertation
# second Maple program:
F:= n-> (<<0|1>, <1|1>>^n)[1,2]:
a:= n-> F(2*n+1)+F(2*n-1)-1:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 04 2016

Table[Fibonacci[2n+1]+Fibonacci[2n-1]-1,{n,30}] (* Harvey P. Dale, Aug 22 2011 *)
a[n_] := LucasL[2n]-1; Array[a, 30] (* Jean-François Alcover, Dec 09 2015 *)

a(n)=fibonacci(2*n+1)+fibonacci(2*n-1)-1 \\ Charles R Greathouse IV, Aug 23 2011

[lucas_number2(n,3,1)-1 for n in range(1,29)] # Zerinvary Lajos, Jul 06 2008

a(n) = Lucas(2*n)-1, with Lucas(n)=A000032(n).
a(n) = floor(r^(2*n)), where r = golden ratio = (1+sqrt(5))/2.
a(n) = floor(Fibonacci(5*n)/Fibonacci(3*n)). - Gary Detlefs, Mar 11 2011
a(n) = +4*a(n-1) -4*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 11 2011
a(n) = A001519(2*n-1) + A001519(2*n+1) - 1. - Reinhard Zumkeller, Aug 09 2013
a(n) = 3*a(n) - a(n-1) + 1; a(n) = A004146(n) + 1, n>0. - Richard R. Forberg, Sep 04 2013
a(n) = 2*cosh(2*n*arcsinh(1/2)) - 1. - Ilya Gutkovskiy, Oct 31 2016
a(n) = floor(sqrt(5)*Fibonacci(2*n)), for n > 0 (Seamons, 1966). - Amiram Eldar, Feb 05 2022

Formulae and comments by Clark Kimberling, Nov 24 2010

a(0)=1 prepended by Alois P. Heinz, Nov 04 2016

A203976 a(n) = 3*a(n-2) - a(n-4), a(0)=0, a(1)=1, a(2)=5, a(3)=4.

Original entry on oeis.org

0, 1, 5, 4, 15, 11, 40, 29, 105, 76, 275, 199, 720, 521, 1885, 1364, 4935, 3571, 12920, 9349, 33825, 24476, 88555, 64079, 231840, 167761, 606965, 439204, 1589055, 1149851, 4160200, 3010349, 10891545, 7881196, 28514435, 20633239, 74651760, 54018521, 195440845

Offset: 0

Michael Somos, Jan 08 2012

a(n+1) = p(n+2) where p(x) is the unique degree-n polynomial such that p(k) = Lucas(k) for k = 1, ..., n+1.
This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m).
a(n) = row sums of triangle A226377(n), based on differences among Lucas Numbers.  - Richard R. Forberg, Aug 01 2013
A strong divisibility sequence, i.e., gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. The sequence of convergents of the 2-periodic continued fraction [0; 1, -5, 1, -5, ...] = 1/(1 - 1/(5 - 1/(1 - 1/(5 - ...)))) = 1/2*(5 - sqrt(5)) begins [0/1, 1/1, 5/4, 4/3, 15/11, 11/8, 40/29,...]. The present sequence is the sequence of numerators; the sequence of denominators [1, 1, 4, 3, 11, 8, 29,...] is A005013. - Peter Bala, May 19 2014
It appears that the first homology group of the branched n-th cyclic covering of the group of figure-eight knot is the direct sum of cyclic groups of orders a(n) and A005013(n), so the order of that group is the product of these numbers, i. e. A004146(n); see the table on p. 156 of the paper by Fox. - Andrey Zabolotskiy, Mar 16 2023

			a(3) = 4 since p(x) = (-x^2 + 7*x - 4) / 2 interpolates p(1) = 1, p(2) = 3, p(3) = 4, and p(4) = 4.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
R. H. Fox, A quick trip through knot theory, pages 120-167 in: Topology of 3-manifolds and related topics (Proceedings of The University of Georgia Institute, 1961), Prentice-Hall, 1962.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).

Cf. A000032, A000045, A201157 (bisection), A002878 (bisection). A005013.

a203976 n = a203976_list !! n
a203976_list = 0 : 1 : 5 : 4 : zipWith (-)
   (map (* 3) $ drop 2 a203976_list) a203976_list
-- Reinhard Zumkeller, Jan 10 2012

I:=[0,1,5,4]; [n le 4 select I[n] else 3*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Mar 29 2016

LinearRecurrence[{0,3,0,-1},{0,1,5,4},40] (* Harvey P. Dale, Apr 06 2013 *)

{a(n) = if( n%2, fibonacci(n+1) + fibonacci(n-1), 5 * fibonacci(n))}

{a(n) = if( n<0, -a(-n), polcoeff( x * (1 + 5*x + x^2) / (1 - 3*x^2 + x^4) + x * O(x^n), n))}

{a(n) = if( n<0, -a(-n), subst( polinterpolate( vector( n, k, fibonacci(k-1) + fibonacci(k+1) )), x, n + 1))}

a(1) = 1, a(2) = 5, a(3) = 4, a(n) * a(n-3) = a(n-1) * a(n-2) - 5. a(-n) = -a(n).
G.f.: x * (1 + 5*x + x^2) /  ( (x^2+x-1)*(x^2-x-1) ).
a(2*n) = 5 * A000045(2*n) (Fibonacci). a(2*n+1) = A000032(2*n+1) (Lucas).
a(A004277(n)) = A054888(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = A000032(n+1) - A061084(n). - R. J. Mathar, Jun 23 2013
a(2n) = a(2n-1) + a(2n+1), for n>0. - Richard R. Forberg, Aug 01 2013
a(n) = (2^(-1-n)*((-5-sqrt(5)+(-1)^n*(-5+sqrt(5)))*((-1+sqrt(5))^n-(1+sqrt(5))^n)))/sqrt(5). - Colin Barker, Mar 28 2016
E.g.f.: exp(-phi*x)*(exp(x) - 1)*(phi*exp(sqrt(5)*x) - 1/phi), where phi = (1 + sqrt(5))/2. - G. C. Greubel, Mar 28 2016

A032170 "CHK" (necklace, identity, unlabeled) transform of 1, 2, 3, 4, ...

Original entry on oeis.org

1, 2, 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: 1

Christian G. Bower

nonn

Apparently, for n > 2, the same as A072337. - Ralf Stephan, Feb 01 2004
a(n) is the number of prime period-n periodic orbits of Arnold's cat map. - Bruce Boghosian, Apr 26 2009
From Petros Hadjicostas, Nov 17 2017: (Start)
A first proof of the g.f., given below, can be obtained using the first of Vladeta Jovovic's formulae. If b(n) = A004146(n), then B(x) = Sum_{n >= 1} b(n)*x^n = x*(1 + x)/((1 - x)*(1 - 3*x + x^2)) (see the documentation for sequence A004146). From Jovovich's first formula, A(x) = Sum_{n >= 1} a(n)*x^n = Sum_{n >= 1} (1/n)*Sum_{d | n} mu(d)*b(n/d)*x^n. Letting m = n/d, we get A(x) = Sum_{d >= 1} (mu(d)/d)*Sum_{m >= 1} b(m)*(x^d)^m/m = Sum_{d >= 1} (mu(d)/d)*f(x^d), where f(y) = Sum_{m >= 1} b(m)*y^m/m = int(B(w)/w, w = 0..y) = int((1 + w)/((1 - w)*(1 - 3*w + w^2)), w = 0..y) = log((1 - y)^2/(1 - 3*y + y^2)) for |y| < (3 - sqrt(5))/2.
A second proof of the g.f. can be obtained using C. G. Bower's definition of the CHK transform of a sequence (e(n): n>=1) with g.f. E(x) (see the links below). If (c_k(n): n >= 1) = CHK_k(e(n): n >= 1), then (c_k(n): n >= 1) = (1/k)*(MOEBIUS*AIK)k (e_n: n >= 1) = (1/k)*Sum{d | gcd(n,k)} mu(d)*AIK_{k/d}(e(n/d): n multiple of d), where the * between MOEBIUS and AIK denotes Dirichlet convolution and (d_k(n): n >= 1) = AIK_k(e(n): n >= 1) has g.f. E(x)^k. (There is a typo in the given definition of CHK in the link.)
If C(x) is the g.f. of CHK(e(n): n >= 1) = Sum_{k = 1..n} CHK_k(e(n): n >= 1), then C(x) = Sum_{n>=1} Sum_{k = 1..n} c_k(n)*x^n = Sum_{k >= 1} (1/k) Sum_{n >= k} Sum_{d | gcd(n,k)} mu(d)*d_{k/d}(n/d)*x^n. Letting m = n/d and s = k/d and using the fact that E(0) = 0, we get C(x) = Sum_{d >= 1} (mu(d)/d)*Sum_{s >= 1} (1/s)*Sum_{m >= s} d_s(m)*(x^d)^m = Sum_{d >= 1} (mu(d)/d)*Sum_{s >= 1} E(x^d)^s. Thus, C(x) = -Sum_{d >= 1} (mu(d)/d)*log(1 - E(x^d)).
For the sequence (e(n): n >= 1) = (n: n >= 1), we have E(x) = Sum_{n>=1} n*x^n = x/(1 - x)^2, and thus A(x) = C(x) = -Sum_{d >= 1} (mu(d)/d)*log(1 - x/(1-x)^2), from which we can easily get the g.f. given in the formula section.
Apparently, for this sequence and for sequences A032165, A032166, A032167, the author assumes that C(0) = 0 (i.e., he assumes the CHK transform has no constant term), while for sequences A032164, A108529, and possibly others, he assumes that the CHK transform starts with the constant term 1 (i.e., he assumes C(x) = 1 - Sum_{d >= 1} (mu(d)/d)*log(1 - E(x^d))). (End)
From Petros Hadjicostas, Jul 13 2020: (Start)
We elaborate further on Michel Marcus's claim below. Consider his sequence (b(n): n >= 1) with b(1) = 3 and b(n) = a(n) for n >= 2.
Using the identity -Sum_{k >= 1} (mu(k)/k)*log(1 - x^k) = x for |x| < 1 and the g.f. of (a(n): n >= 1) below, we see that Sum_{n >= 1} b(n)*x^n = 3*x - a(1)*x + Sum_{n >= 1} a(n)*x^n = 2*x + Sum_{k >= 1} (mu(k)/k)*(2*log(1 - x^k) - log(1 - 3*x^k + x^(2*k))) = -Sum_{k >= 1} (mu(k)/k)*log(1 - 3*x^k + x^(2*k)).
Following Kam Cheong Au (2020), let d(w,N) be the dimension of the Q-span of weight w and level N of colored multiple zeta values (CMZV). Here Q are the rational numbers.
Deligne's bound says that d(w,N) <= D(w,N), where 1 + Sum_{w >= 1} D(w,N)*t^w = (1 - a*t + b*t^2)^(-1) when N >= 3, where a = phi(N)/2 + omega(N) and b = omega(N) - 1 (with omega(N) being the number of distinct primes of N).
For N = 6, a = phi(6)/2 + omega(6) = 2/2 + 2 = 3 and b = omega(6) - 1 = 1. It follows that D(w, N=6) = A001906(w+1) = Fibonacci(2*(w+1)).
For some reason, Kam Cheong Au (2020) assumes Deligne's bound is tight, i.e., d(w,N) = D(w,N). He sets Sum_{w >= 1} c(w,N)*t^w = log(1 + Sum_{w >= 1} d(w,N)*t^w) = log(1 + Sum_{w >= 1} D(w,N)*t^w) = -log(1 - a*t + b*t^2) for N >= 3.
For N = 6, we get that c(w, N=6) = A005248(w)/w.
He defines d*(w,N) = Sum_{k | w} (mu(k)/k)*c(w/k,N) to be the "number of primitive constants of weight w and level N". (Using the terminology of A113788, we may perhaps call d*(w,N) the number of irreducible colored multiple zeta values at weight w and level N.)
Using standard techniques of the theory of g.f.'s, we can prove that Sum_{w >= 1} d*(w,N)*t^w = Sum_{s >= 1} (mu(s)/s) Sum_{k >= 1} c(k,N)*(t^s)^k = -Sum_{s >= 1} (mu(s)/s)*log(1 - a*t^s + b*t^(2*s)).
For N = 6, we saw that a = 3 and b = 1, and hence d*(w, N=6) = b(w) for w >= 1 (as claimed by Michel Marcus below). See Table 1 on p. 6 in Kam Cheong Au (2020). (End)

Michael De Vlieger, Table of n, a(n) for n = 1..2400
Kam Cheong Au, Evaluation of one-dimensional polylogarithmic integral, with applications to infinite series, arXiv:2007.03957 [math.NT], 2020. See line N = 6 in Table 1 (p. 6).
C. G. Bower, Transforms (2).
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Eric Weisstein's World of Mathematics, Arnold's cat map.
Wikipedia, Arnold's cat map.
Index entries for sequences related to Lyndon words

Cf. A000010, A001221, A001906, A004146, A032164, A032165, A032166, A032167, A032198, A005248, A108529, A113788.

Table[DivisorSum[n, MoebiusMu[n/#] (LucasL[2 #] - 2) &]/n, {n, 31}] (* Michael De Vlieger, Nov 18 2017 *)

a(n) = (1/n)*Sum_{d | n} mu(n/d)*A004146(d). - Vladeta Jovovic, Feb 15 2003
Inverse EULER transform of Fibonacci(2*n). - Vladeta Jovovic, May 04 2006
G.f.: Sum_{n >= 1} (mu(n)/n)*f(x^n), where f(y) = log((1 - y)^2/(1 - 3*y + y^2)). - Petros Hadjicostas, Nov 17 2017
It appears that the sequence b(1) = 3, b(n) = a(n) for n >= 2 is related to the rational sequence (c(w, N=6): w >= 1) = (A005248(w)/w: w >= 1) whose g.f. is log(1/(1 - a*t + b*t^2)), where a = phi(N)/2 + omega(N) and b = omega(N) - 1 when N = 6, where phi is A000010 and omega is A001221. See Kam Cheong Au (2020). - Michel Marcus, Jul 13 2020 [Edited by Petros Hadjicostas, Jul 13 2020]

A049682 a(n) = (Lucas(8*n) - 2)/45.

Original entry on oeis.org

0, 1, 49, 2304, 108241, 5085025, 238887936, 11222647969, 527225566609, 24768378982656, 1163586586618225, 54663801192073921, 2568035069440856064, 120642984462528161089, 5667652234669382715121, 266259012044998459449600, 12508505913880258211416081

Offset: 0

Clark Kimberling

This is a divisibility sequence.

			G.f. = x + 49*x^2 + 2304*x^3 + 108241*x^4 + 5085025*x^5 + 238887936*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..595
Index to divisibility sequences.
Index entries for linear recurrences with constant coefficients, signature (48,-48,1).

Cf. A000032, A000045, A004146, A049683, A049684.

List([0..20], n-> (Fibonacci(4*n)/3)^2 ); # G. C. Greubel, Dec 14 2019

[(Fibonacci(4*n)/3)^2: n in [0..20]]; // G. C. Greubel, Dec 02 2017

with(combinat); seq( fibonacci(4*n)^2/9, n=0..20); # G. C. Greubel, Dec 14 2019

LinearRecurrence[{48,-48,1},{0,1,49},20] (* or *) CoefficientList[Series[ (-x-x^2)/ (x^3-48x^2+48x-1),{x,0,20}],x] (* Harvey P. Dale, Apr 22 2011 *)

numlib::fibonacci(4*n)^2/9 $ n = 0..25; // Zerinvary Lajos, May 09 2008

vector(21, n, (fibonacci(4*(n-1))/3)^2) \\ G. C. Greubel, Dec 02 2017

[(fibonacci(4*n)/3)^2 for n in (0..20)] # G. C. Greubel, Dec 14 2019

a(n) = (1/45)*(-2 + ((47 + 7*sqrt(45))/2)^n + ((47 - 7*sqrt(45))/2)^n). - Ralf Stephan, Apr 14 2004
From R. J. Mathar, Jun 03 2009: (Start)
a(n) = (A004187(n))^2.
a(n) = 48*a(n-1) - 48*a(n-2) + a(n-3).
G.f.: x*(1 + x)/((1 - x)*(1 - 47*x + x^2)). (End)
From R. K. Guy, Feb 24 2010: (Start)
a(n) = F(4*n)^2/9.
a(n) - a(n-1) = A004187(2n-1). (End)
From Peter Bala, Jun 03 2016: (Start)
exp( Sum_{n >= 1} 45*a(n)*x^n/n ) = 1 + 15/7*Sum_{n >= 1} Fibonacci(8*n)*x^n.
This is the particular case k = 4 of the relation exp( Sum_{n >= 1} 5*F(k*n)^2*x^n/n ) = 1 + 5*Fibonacci(k)/Lucas(k) * ( Sum_{n >= 1} F(2*k*n)*x^n ). (End)
Lim_{n->infinity} a(n+1)/a(n) = (47 + 21*sqrt(5))/2 = phi^8, where phi is the golden ratio (A001622). - Ilya Gutkovskiy, Jun 06 2016
a(n) = a(-n) for all n in Z. - Michael Somos, Jun 12 2016
0 = a(n)*(+a(n) -98*a(n+1) -2*a(n+2)) + a(n+1)*(+2401*a(n+1) -98*a(n+2)) + a(n+2)^2 for all integer n. - Michael Somos, Jun 12 2016

More terms from N. J. A. Sloane, Feb 26 2010

A049683 a(n) = (Lucas(6*n) - 2)/16.

Original entry on oeis.org

0, 1, 20, 361, 6480, 116281, 2086580, 37442161, 671872320, 12056259601, 216340800500, 3882078149401, 69661065888720, 1250017107847561, 22430646875367380, 402501626648765281, 7222598632802407680, 129604273763794572961, 2325654329115499905620

Offset: 0

Clark Kimberling

This is the r = 20 member of the r-family of sequences S_r(n), n >= 1, defined in A092184 where more information can be found.

Colin Barker, Table of n, a(n) for n = 0..750
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A000032, A004146, A023039, A049660, A049682, A049684, A092184.

List([0..30], n-> (Lucas(1,-1,6*n)[2] - 2)/16 ); # G. C. Greubel, Dec 14 2019

[(Lucas(6*n) -2)/16: n in [0..30]]; // G. C. Greubel, Dec 02 2017

with(combinat); seq( (5*fibonacci(3*n)^2 -2*(1-(-1)^n))/16, n=0..30); # G. C. Greubel, Dec 14 2019

LinearRecurrence[{19,-19,1}, {0,1,30}, 20] (* or *) Table[(LucasL[6*n] -2)/16, {n,0,30}] (* G. C. Greubel, Dec 02 2017 *)

concat(0, Vec(x*(1+x)/((1-x)*(1-18*x+x^2)) + O(x^30))) \\ Colin Barker, Jun 03 2016

vector(31, n, (5*fibonacci(3*n-3)^2 -2*(1+(-1)^n))/16 ) \\ G. C. Greubel, Dec 14 2019

[(lucas_number2(6*n,1,-1) -2)/16 for n in (0..30)] # G. C. Greubel, Dec 14 2019

a(n) = (-2 + (9 + 4*sqrt(5))^n + (9 - 4*sqrt(5))^n)/16. - Ralf Stephan, Apr 14 2004
a(n) = (T(n, 9) - 1)/8 with Chebyshev's polynomials of the first kind evaluated at x = 9: T(n, 9) = A023039(n). Wolfdieter Lang, Oct 18 2004
G.f.: x*(1 + x)/((1 - x)*(1 - 18*x + x^2)) = x*(1 + x)/(1 - 19*x + 19*x^2 - x^3). (from the Stephan link, see A092184).
exp( Sum_{n >= 1} 16*a(n)*x^n/n ) = 1 + 2*Sum_{n >= 1} Fibonacci(6*n)*x^n. - Peter Bala, Jun 03 2016
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3) for n>2. - Colin Barker, Jun 03 2016

A127595 a(n) = F(4n) - 2F(2n) where F(n) = Fibonacci numbers A000045.

Original entry on oeis.org

0, 1, 15, 128, 945, 6655, 46080, 317057, 2176335, 14925184, 102320625, 701373311, 4807434240, 32951037313, 225850798095, 1548007091840, 10610205501105, 72723448842367, 498453982018560, 3416454544730369, 23416728143799375

Offset: 0

Peter Bala, Apr 10 2007

a(n) is a divisibility sequence; that is, if h|k then a(h)|a(k).

			G.f. = x + 15*x^2 + 128*x^3 + 945*x^4 + 6655*x^5 + ... - _Michael Somos_, Dec 30 2022

Michael De Vlieger, Table of n, a(n) for n = 0..1196
E. L. Roettger and H. C. Williams, Appearance of Primes in Fourth-Order Odd Divisibility Sequences, J. Int. Seq., Vol. 24 (2021), Article 21.7.5.
Hugh Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory vol. 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Odd and even linear divisibility sequences of order 4, INTEGERS, 2015, #A33.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (10,-23,10,-1).

Cf. A000032, A000045, A001906, A004146.

With[{r = 3}, CoefficientList[Series[x (1 + (r + 2) x + x^2)/((1 - r x + x^2)*(1 - (r^2 - 2)*x + x^2)), {x, 0, 20}], x]] (* Michael De Vlieger, Nov 09 2021 *)

{a(n) = my(w = quadgen(5)^(2*n)); imag(w^2 - 2*w)}; /* Michael Somos, Dec 30 2022 */

a(n) = F(2n)*(L(2n)-2) = A001906(n)*A004146(n), where L(n) are the Lucas numbers A000032.
a(2n) = 5*(F(2n))^3*L(2n), a(2n+1) = F(2n+1)*L(2n+1)^3.
a(n) = [(Phi^(2n))-1]^2*[(Phi^(4n))-1]/[sqrt(5)*(Phi^(4n))].
G.f.: A(x)=x*(1+(r+2)*x+x^2)/((1-r*x+x^2)*(1-(r^2-2)*x+x^2)) at r=3. The case r=2 is A000578.
a(n) = -a(-n) for all n in Z. - Michael Somos, Dec 30 2022

A259913 Discriminant of the number field containing the number with periodic continued fraction [1,n,1,n,1,n,...].

Original entry on oeis.org

5, 12, 21, 8, 5, 60, 77, 24, 13, 140, 165, 12, 221, 28, 285, 5, 357, 44, 437, 120, 21, 572, 69, 168, 29, 780, 93, 56, 957, 1020, 1085, 8, 1221, 1292, 1365, 40, 1517, 1596, 1677, 440, 205, 1932, 2021, 33, 5, 92, 2397, 156, 53, 12, 2805, 728, 3021, 348, 3245

Offset: 1

Clark Kimberling, Jul 20 2015

a(n) is the first term in row n of the triangle at A259911.

It appears that a(n) = 5 if n is a nonzero term of A004146.

			[1,3,1,3,1,3,...] = (1/6)(3 + sqrt(21)), so that a(3) = 21.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A259911.

v = Table[FromContinuedFraction[{1, {n, 1}}], {n, 1, 60}];
Flatten[NumberFieldDiscriminant[v]]

A316659 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial x(x^2 - 2) + x(((v - w)/2)^n + ((v + w)/2)^n), where v = 3 + 2x and w = sqrt(5 + 4x).

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 2, 1, 0, 5, 8, 3, 0, 16, 30, 16, 2, 0, 45, 104, 81, 24, 2, 0, 121, 340, 356, 170, 35, 2, 0, 320, 1068, 1411, 932, 315, 48, 2, 0, 841, 3262, 5209, 4396, 2079, 532, 63, 2, 0, 2205, 9760, 18281, 18784, 11440, 4144, 840, 80, 2, 0, 5776, 28746

Offset: 0

Franck Maminirina Ramaharo, Jul 09 2018

The triangle is related to the Kauffman bracket polynomial for the Turk's Head Knot ((3,n)-torus knot). Column 1 matches the determinant of the Turk's Head Knots THK(3,k) A004146.

			The triangle T(n,k) begins:
n\k: 0      1      2       3       4       5       6      7      8    9  10 11
0:   0      0      0       1
1:   0      1      2       1
2:   0      5      8       3
3:   0     16     30      16       2
4:   0     45    104      81      24       2
5:   0    121    340     356     170      35       2
6:   0    320   1068    1411     932     315      48      2
7:   0    841   3262    5209    4396    2079     532     63      2
8:   0   2205   9760   18281   18784   11440    4144    840     80    2
9:   0   5776  28746   61786   74838   55809   26226   7602   1260   99   2
10:  0  15125  83620  202841  282980  249815  144488  54690  13080 1815 120  2
...

Louis H. Kauffman, An invariant of regular isotopy, Trans. Amer. Math. Soc., Vol. 318 (1990), 417-471.
Seong Ju Kim, R. Stees, and L. Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), #16.1.4.
Alexander Stoimenow, Square numbers, spanning trees and invariants of achiral knots, Communications in Analysis and Geometry, Vol. 13 (2005), 591-631.

Row sums: A000302 (Powers of 4).
Row 1: row 1 of A300184, A300192 and row 0 of A300454.
Row 2: row 2 of A300454.
Cf. A137396, A299989, A300453.

v = 3 + 2*x; w = Sqrt[5 + 4*x];
row[n_] := CoefficientList[x*(x^2 - 2) + x*(((v - w)/2)^n + ((v + w)/2)^n), x];
Array[row, 15, 0] // Flatten

v : 3 + 2*x$ w : sqrt(5 + 4*x)$
p(n, x) := expand(x*(x^2 - 2) + x*(((v - w)/2)^n + ((v + w)/2)^n))$
for n:0 thru 15 do print(makelist(ratcoef(p(n, x), x, k), k, 0, max(3, n + 1)));

T(n,1) = A004146(n).
T(n,2) = A122076(n,1) = A099920(2*n-1).
G.f.: (x^3 - 2*x)/(1 - y) + (2*x - 3*x*y - 2*x^2*y)/(1 - 3*y - 2*x*y + y^2 + 2*x*y^2 + x^2*y^2).

A074392 a(n) = Lucas(n+1) + (3*(-1)^n - 1)/2.

Original entry on oeis.org

2, 1, 5, 5, 12, 16, 30, 45, 77, 121, 200, 320, 522, 841, 1365, 2205, 3572, 5776, 9350, 15125, 24477, 39601, 64080, 103680, 167762, 271441, 439205, 710645, 1149852, 1860496, 3010350, 4870845, 7881197, 12752041, 20633240, 33385280, 54018522

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 22 2002

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1)

Cf. A000032, A074331.

Cf. A004146.

CoefficientList[Series[(2-x)/(1-x-2*x^2+x^3+x^4), {x, 0, 40}], x]

Vec((2-x) / ((x-1)*(1+x)*(x^2+x-1)) + O(x^50)) \\ Colin Barker, Jul 12 2017

a(n) = Sum (L(2i+e), (i=0, 1, .., Floor(n/2))), where L(n) are Lucas numbers and e=2(n/2 - Floor(n/2)).
Convolution of L(n) with the sequence (1, 0, 1, 0, 1, 0, ...)
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4) for n>3.
G.f.: ( 2-x ) / ( (x-1)*(1+x)*(x^2+x-1) ).
a(n) = 2*A052952(n)-A052952(n-1). - R. J. Mathar, Oct 04 2013
a(n) = 2^(-1-n) * (3*(-2)^n - 2^n + (1-sqrt(5))^(1+n) + (1+sqrt(5))^(1+n)). - Colin Barker, Jul 12 2017

A117202 Binomial transform of n*F(n).

Original entry on oeis.org

0, 1, 4, 15, 52, 170, 534, 1631, 4880, 14373, 41810, 120406, 343884, 975325, 2749852, 7713435, 21540304, 59917826, 166094370, 458998523, 1264919720, 3477182961, 9536877614, 26102772910, 71309161752, 194468551225, 529490287924

Offset: 0

Paul Barry, Mar 02 2006

Binomial transform of A045925.
Number of acyclic subgraphs of the wheel graph W_n (on n+1 vertices) with exactly n-1 edges. - Emil R. Vaughan, Jun 12 2007
Equivalently, number of two-component spanning forests of the wheel graph W_n (on n+1 vertices). - Harry Richman, Jul 31 2023
Starting (1, 4, 15, 52, ...) = binomial transform of A136376. - Gary W. Adamson, Sep 03 2008

Indranil Ghosh, Table of n, a(n) for n = 0..1000
Harry Richman, Farbod Shokrieh, and Chenxi Wu, Counting two-forests and random cut size via potential theory, arXiv:2308.03859 [math.CO], 2023. See p. 18.
J. Salas and A. D. Sokal, Transfer Matrices and Partition-Function Zeros for Antiferromagnetic Potts Models. V. Further Results for the Square-Lattice Chromatic Polynomial, J. Stat. Phys. 135 (2009) 279-373; arXiv:0711.1738 [cond-mat.stat-mech], 2007-2009. Mentions this sequence. - _N. J. A. Sloane_, Mar 14 2014
Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).

Cf. A001519, A111262.
Cf. A136376.
Cf. A004146 (number of spanning trees of wheel graph).

Table[n Fibonacci[2n-1],{n,0,26}] (* or *) Table[Sum[Fibonacci[2k]*BernoulliB[2n-2k]*Binomial[2n,2k],{k,1,n}],{n,0,26}] (* or *) CoefficientList[Series[x(1-2x+2x^2)/(1-3x+x^2)^2 ,{x,0,26}],x] (* Indranil Ghosh, Feb 26 2017 *)

a(n) = n*fibonacci(2*n-1); \\ Indranil Ghosh, Feb 26 2017

concat(0, Vec(x*(1-2*x+2*x^2) / (1-3*x+x^2)^2 + O(x^30))) \\ Colin Barker, Feb 26 2017

G.f.: x*(1-2x+2x^2)/(1-3x+x^2)^2.
a(n) = 6*a(n-1)-11*a(n-2)+6*a(n-3)-a(n-4).
a(n) = Sum_{k=0..n} C(n,k)*k*F(k).
From Benoit Cloitre, Nov 29 2006: (Start)
a(n) = Sum_{k=1..n} F(2k)*B(2n-2k)*binomial(2n,2k) where F=Fibonacci numbers and B=Bernoulli numbers;
a(n) = n*F(2n-1). (End)
a(n) = (2^(-1-n)*(-(-5+sqrt(5))*(3+sqrt(5))^n + (3-sqrt(5))^n*(5+sqrt(5)))*n) / 5. - Colin Barker, Feb 26 2017
a(n) = (1/sqrt(5)) * n * (((1 + sqrt(5)) / 2)^(2*n-1) - ((1 - sqrt(5)) / 2)^(2*n-1)). - Harry Richman, Jul 31 2023
a(n) = round((1/sqrt(5)) * n * phi^(2n-1)), where phi = (1+sqrt(5))/2 is the golden ratio A001622. - Harry Richman, Jul 31 2023

cp's OEIS Frontend

A005592 a(n) = F(2n+1) + F(2n-1) - 1.

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A203976 a(n) = 3*a(n-2) - a(n-4), a(0)=0, a(1)=1, a(2)=5, a(3)=4.

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A032170 "CHK" (necklace, identity, unlabeled) transform of 1, 2, 3, 4, ...

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A049682 a(n) = (Lucas(8*n) - 2)/45.

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A049683 a(n) = (Lucas(6*n) - 2)/16.

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A127595 a(n) = F(4n) - 2F(2n) where F(n) = Fibonacci numbers A000045.

A316659 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial x(x^2 - 2) + x(((v - w)/2)^n + ((v + w)/2)^n), where v = 3 + 2x and w = sqrt(5 + 4x).