A002878 -id:A002878 - cp's OEIS frontend

A098149 a(0)=-1, a(1)=-1, a(n)=-3*a(n-1)-a(n-2) for n>1.

-1, -1, 4, -11, 29, -76, 199, -521, 1364, -3571, 9349, -24476, 64079, -167761, 439204, -1149851, 3010349, -7881196, 20633239, -54018521, 141422324, -370248451, 969323029, -2537720636, 6643838879, -17393796001, 45537549124, -119218851371

Offset: 0

Creighton Dement, Aug 29 2004

Sequence relates bisections of Lucas and Fibonacci numbers.
2*a(n) + A098150(n) = 8*(-1)^(n+1)*A001519(n) - (-1)^(n+1)*A005248(n+1). Apparently, if (z(n)) is any sequence of integers (not all zero) satisfying the formula z(n) = 2(z(n-2) - z(n-1)) + z(n-3) then |z(n+1)/z(n)| -> golden ratio phi + 1 = (3+sqrt(5))/2.
Pisano period lengths: 1, 3, 4, 6, 1, 12, 8, 6, 12, 3, 10, 12, 7, 24, 4, 12, 9, 12, 18, 6, ... . - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Oct 12 2020: (Start)
[X(n) = (-1)^n*(S(n, 3) + S(n-1, 3)), Y(n) = X(n-1)] gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 3*X*Y = +5, for n = -oo..+oo, with Chebyshev S polynomials (A049310), with S(-1, x) = 0,  S(-|n|, x) = - S(|n|-2, x), for |n| >= 2, and S(n,-x) = (-1)^n*S(n, x). The present sequence is a(n) = -X(n-1), for n >= 0. See the formula section.
This binary indefinite quadratic form of discriminant 5, representing 5, has only this family of proper solutions (modulo sign flip), and no improper ones.
This comment is inspired by a paper by Robert K. Moniot (private communication) See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Seong Ju Kim, R. Stees, and L. Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), #16.1.4.
Tanya Khovanova, Recursive Sequences
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Index entries for linear recurrences with constant coefficients, signature (-3,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A098150, A001519, A005248, A000045, A001906, A049310, A027941.

a[0] = a[1] = -1; a[n_] := a[n] = -3a[n - 2] - a[n - 1]; Table[ a[n], {n, 0, 27}] (* Robert G. Wilson v, Sep 01 2004 *)
LinearRecurrence[{-3,-1},{-1,-1},30] (* Harvey P. Dale, Apr 19 2014 *)
CoefficientList[Series[-(1 + 4 x)/(1 + 3 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Apr 19 2014 *)

G.f.: -(1+4*x)/(1+3*x+x^2). - Philippe Deléham, Nov 19 2006
a(n) = (-1)^n*A002878(n-1). - R. J. Mathar, Jan 30 2011
-a(n+1) = Sum_{k, 0<=k<=n}(-5)^k*Binomial(n+k, n-k) = Sum_{k, 0<=k<=n}(-5)^k*A085478(n, k). - Philippe Deléham, Nov 28 2006
a(n) = (-1)^n*(S(n-1, 3) + S(n-2, 3)) = (-1)^n*S(2*(n-1), sqrt(5)), for n >= 0, with Chebyshev S polynomials (A049310), with S(-1, x) = 0 and S(-2, x) = -1. S(n, 3) = A001906(n+1) = F(2*(n+1)), with F = A000045. - Wolfdieter Lang, Oct 12 2020

Simpler definition from Philippe Deléham, Nov 19 2006

A098532 Sum of sixth powers of first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 66, 795, 16420, 278564, 5105373, 90871494, 1635675910, 29316316535, 526297607496, 9442398055752, 169448124595321, 3040546683808010, 54560921044808010, 979052407236876819, 17568407254504944748

Offset: 0

Benoit Cloitre, Sep 12 2004

G. C. Greubel, Table of n, a(n) for n = 0..795
Index entries for linear recurrences with constant coefficients, signature(13,104,-260,-260,104,13,-1).

Cf. A001654, A005968, A005969, A098531, A098533.

Cf. A119287, A000071, A001654, A005968, A005969, A098531, A098533, A128697.

[(Fibonacci(n)^5*Fibonacci(n+3) + Fibonacci(2*n))/4: n in [0..30]]; // G. C. Greubel, Jan 17 2018

Table[(Fibonacci[n]^5*Fibonacci[n+3] + Fibonacci[2*n])/4, {n,0,30}] (* G. C. Greubel, Jan 17 2018 *)

PARI
```
a(n)=sum(i=0,n,fibonacci(i)^6);
```

for(n=0,30, print1((fibonacci(n)^5*fibonacci(n+3) + fibonacci(2*n))/4, ", ")) \\ G. C. Greubel, Jan 17 2018

a(n) = (1/500)*(F(6*n+1) +3*F(6*n+2) -(-1)^n*(16*F(4*n+1)+8*F(4*n+2))-60*F(2*n+1) +120*F(2*n+2) -(-1)^n*40 ) where F(n)=A000045(n).
G.f.: x*(1-11*x-64*x^2-11*x^3+x^4)/((x+1)*(1-18*x+x^2)*(1-3*x+x^2)*(1+7*x+x^2)). - R. J. Mathar, Feb 26 2012
a(n) = -6*(-1)^n*A049685(n)/125 +3*A002878(n)/25 +A049629(n)/125 -2*(-1)^n/25. - R. J. Mathar, Feb 26 2012
a(n)= (F(n)^5 * F(n+3) + F(2*n))/4. - Gary Detlefs, Jan 05 2013

A110391 a(n) = L(3*n)/L(n), where L(n) = Lucas number.

Original entry on oeis.org

1, 4, 6, 19, 46, 124, 321, 844, 2206, 5779, 15126, 39604, 103681, 271444, 710646, 1860499, 4870846, 12752044, 33385281, 87403804, 228826126, 599074579, 1568397606, 4106118244, 10749957121, 28143753124, 73681302246, 192900153619, 505019158606, 1322157322204

Offset: 0

Amarnath Murthy, Jul 27 2005

Subsidiary sequences: a(n) = L((2k+1)*n)/L(n) for k = 2,3, etc. This is the sequence for k = 1.

			a(1) = L(3)/L(1) = 4/1 = 4.

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

Cf. A000204, A000032, A002878, A114525, A153173, A153175, A153177, A153179, A153180.

[Lucas(3*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 17 2017

with(combinat): L:=n->fibonacci(n+2)-fibonacci(n-2): seq(L(3*n)/L(n),n=0..30); # Emeric Deutsch, Jul 31 2005

Table[LucasL[3 n]/LucasL[n], {n, 0, 27}] (* Michael De Vlieger, Mar 18 2015 *)
LinearRecurrence[{2,2,-1},{1,4,6},40] (* Harvey P. Dale, Aug 20 2020 *)

Vec((1+2*x-4*x^2)/((1+x)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Jun 03 2016

for(n=0,30, print1((fibonacci(3*n+1) + fibonacci(3*n-1))/( fibonacci(n+1) + fibonacci(n-1)), ", ")) \\ G. C. Greubel, Dec 17 2017

From R. J. Mathar, Oct 18 2010: (Start)
a(n) = A005248(n) - (-1)^n.
a(n) = +2*a(n-1) +2*a(n-2) -a(n-3).
G.f.: ( 1+2*x-4*x^2 ) / ( (1+x)*(x^2-3*x+1) ). (End)
Exp( Sum_{n >= 1} a(n)*t^n/n ) = 1 + 4*t + 11*t^2 + 29*t^3 + ... is the o.g.f. for A002878. This is the case x = 1 of the general result exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n, where L(n,x) is the n-th Lucas polynomial of A114525. - Peter Bala, Mar 18 2015
a(n) = 2^(-n)*(-(-2)^n+(3-sqrt(5))^n+(3+sqrt(5))^n). - Colin Barker, Jun 03 2016

Corrected and extended by Emeric Deutsch and Erich Friedman, Jul 31 2005

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

Original entry on oeis.org

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

A013946 Least d for which the number with continued fraction [n,n,n,n...] is in Q(sqrt(d)).

Original entry on oeis.org

5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 5, 37, 173, 2, 229, 65, 293, 82, 365, 101, 445, 122, 533, 145, 629, 170, 733, 197, 5, 226, 965, 257, 1093, 290, 1229, 13, 1373, 362, 61, 401, 1685, 442, 1853, 485, 2029, 530, 2213, 577, 2405, 626, 2605, 677, 2813, 730, 3029, 785, 3253

Offset: 1

Clark Kimberling

nonn

Square roots of a(n) are found in the limiting ratios of A000045, A001333, A003688, A015448, A015449, A015451 and so on. I.e., the limiting ratios are the golden ratio, silver mean, bronze ratio and so on. - Mats Granvik, Oct 20 2010

Clark Kimberling, Table of n, a(n) for n = 1..1000
Robin James Spivey, Close encounters of the golden and silver ratios, Notes on Number Theory and Discrete Mathematics (2019) Vol. 25, No. 3, 170-184.
Ofer Yifrach-Stav, Fast and Private Pool Testing and Contributions to Experimental Mathematics, Doctoral thesis, École normale supérieure (Paris, France), HAL Science [math.cs] 2024, Art. No. tel-04513104. See p. 104.

a(n) = 2 is equivalent to "n is in the sequence A077444", a(n) = 5 is equivalent to "n is in the sequence A002878".

z = 5000; u = Table[{p, e} = Transpose[FactorInteger[n]];
Times @@ (p^Mod[e, 2]), {n, z}]; Table[u[[n^2 + 4]], {n, 1, Sqrt[z - 4]}]  (* Clark Kimberling, Jul 20 2015, based on T. D. Noe's program at A007913 *)

A013946(n)=core(n^2+4)  \\ M. F. Hasler, Dec 08 2010

a(n) = A007913(n^2+4). - David W. Wilson, Dec 08 2010

More terms from David W. Wilson

A061171 One half of second column of Lucas bisection triangle (odd part).

Original entry on oeis.org

3, 19, 79, 283, 940, 2982, 9171, 27581, 81557, 237995, 687158, 1966764, 5588259, 15780103, 44323195, 123920827, 345062176, 957403026, 2647935987, 7302634865, 20087869313, 55128445259, 150971982314

Offset: 0

Wolfdieter Lang, Apr 20 2001

Numerator of g.f. is on half of row polynomial Sum_{m=0..2} A061187(1,m) * x^m.

G. C. Greubel, Table of n, a(n) for n = 0..1000
É. Czabarka, R. Flórez, L. Junes, A Discrete Convolution on the Generalized Hosoya Triangle, Journal of Integer Sequences, 18 (2015), #15.1.6.
Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).

Cf. A000032, A000045, A001906, A002878, A005248.

I:=[3,19,79,283]; [n le 4 select I[n] else 6*Self(n-1) - 11*Self(n-2) + 6*Self(n-3) - Self(n-4): n in [1..30]]; // G. C. Greubel, Dec 21 2017

CoefficientList[Series[(1+x)(3-2x)/(1-3x+x^2)^2,{x,0,30}],x] (* or *) LinearRecurrence[{6,-11,6,-1},{3,19,79,283},30] (* Harvey P. Dale, Oct 11 2012 *)

my(x='x+O('x^30)); Vec((1+x)*(3-2*x)/(1-3*x+x^2)^2) \\ G. C. Greubel, Dec 21 2017

2*a(n) = A060924(n+1, 1).
G.f.: (1+x)*(3-2*x)/(1-3*x+x^2)^2.
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4), with a(0)=3, a(1)=19, a(2)=79, a(3)=283. - Harvey P. Dale, Oct 11 2012
a(n) = Fibonacci(2*n+4) + n*Lucas(2*n+3). - Lechoslaw Ratajczak, May 06 2020

A065034 a(n) = Lucas(2*n) + 1.

Original entry on oeis.org

3, 4, 8, 19, 48, 124, 323, 844, 2208, 5779, 15128, 39604, 103683, 271444, 710648, 1860499, 4870848, 12752044, 33385283, 87403804, 228826128, 599074579, 1568397608, 4106118244, 10749957123, 28143753124, 73681302248, 192900153619, 505019158608, 1322157322204

Offset: 0

N. J. A. Sloane, Nov 04 2001

Harry J. Smith, Table of n, a(n) for n = 0..200
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 8.
Markus Grassl and Andrew J. Scott, Fibonacci-Lucas SIC-POVMs, arXiv:1707.02944 [quant-ph], 2017.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000032, A000045, A005248, A005592.

Cf. A002878 (first differences). - R. J. Mathar, Jul 18 2009

[ Lucas(2*n) + 1: n in [0..210]]; // Vincenzo Librandi, Apr 15 2011

a:= n-> (<<0|1>, <1|1>>^(2*n). <<2,1>>)[1, 1]+1:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 01 2016

LucasL[2 Range[30]]+1 (* Harvey P. Dale, Oct 21 2011 *)
LinearRecurrence[{4, -4, 1}, {3, 4, 8}, 30] (* Jean-François Alcover, Jan 08 2019 *)

a(n) = { fibonacci(2*n + 1) + fibonacci(2*n - 1) + 1 } \\ Harry J. Smith, Oct 03 2009

Vec((3-2*x)*(1-2*x)/((1-x)*(1-3*x+x^2)) + O(x^40)) \\ Colin Barker, Nov 01 2016

a(n) = F(2*n+1) + F(2*n-1) + 1 = A005248(n) + 1.
From R. J. Mathar, Jul 18 2009: (Start)
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
G.f.: 1/(1-x) + (2-3*x)/(1-3*x+x^2). (End)
a(n) = 1 + ((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n. - Colin Barker, Nov 01 2016

a(0)=3 prepended by Joerg Arndt, Nov 01 2016

A082762 Trinomial transform of Lucas numbers (A000032).

Original entry on oeis.org

1, 8, 44, 232, 1216, 6368, 33344, 174592, 914176, 4786688, 25063424, 131233792, 687149056, 3597959168, 18839158784, 98643116032, 516502061056, 2704439902208, 14160631169024, 74146027405312, 388233639755776, 2032817728913408, 10643971814457344

Offset: 0

Emanuele Munarini, May 21 2003

G. C. Greubel, Table of n, a(n) for n = 0..1000
Yuhan Jiang, The doubly asymmetric simple exclusion process, the colored Boolean process, and the restricted random growth model, arXiv:2312.09427 [math.CO], 2023.
Index entries for linear recurrences with constant coefficients, signature (6,-4).

Cf. A000032, A027907, A091042, A292277.

I:=[1, 8]; [n le 2 select I[n] else 6*Self(n-1)-4*Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 21 2017

a[n_]:=(MatrixPower[{{2,2},{2,4}},n].{{2},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
f[n_] := Block[{s = Sqrt@ 5}, Simplify[((1 + s)(3 + s)^n + (1 - s)(3 - s)^n)/2]]; Array[f, 21, 0] (* Robert G. Wilson v, Mar 07 2011 *)
LinearRecurrence[{6,-4}, {1, 8}, 30] (* G. C. Greubel, Dec 21 2017 *)

x='x+O('x^30); Vec((1 + 2*x)/(1 - 6*x + 4*x^2)) \\ G. C. Greubel, Dec 21 2017

a(n) = Sum_{k=0..2*n} Trinomial(n,k)*Lucas(k+1), where Trinomial(n,k) = trinomial coefficients (A027907).
a(n) = 2^n*Lucas(2*n+1), where Lucas = A000032.
From Philippe Deléham, Mar 01 2004: (Start)
a(n) = 2^n*A002878(n) = 2^(-n)*Sum_{k>=0} C(2*n+1,2*k)*5^k; see A091042.
a(0) = 1, a(1) = 8, a(n+1) = 6*a(n) - 4*a(n-1). (End)
From Al Hakanson (hawkuu(AT)gmail.com), Jul 13 2009: (Start)
a(n) = ((1+sqrt(5))*(3+sqrt(5))^n + (1-sqrt(5))*(3-sqrt(5))^n)/2.
Third binomial transform of 1, 5, 5, 25, 25, 125. (End)
G.f.: (1 + 2*x)/(1 - 6*x + 4*x^2). - Colin Barker, Mar 23 2012

A110035 Row sums of an unsigned characteristic triangle for the Fibonacci numbers.

Original entry on oeis.org

1, 2, 5, 12, 31, 80, 209, 546, 1429, 3740, 9791, 25632, 67105, 175682, 459941, 1204140, 3152479, 8253296, 21607409, 56568930, 148099381, 387729212, 1015088255, 2657535552, 6957518401, 18215019650, 47687540549, 124847601996

Offset: 0

Paul Barry, Jul 08 2005

Rows sums of abs(A110033).

			G.f. = 1 + 2*x + 5*x^2 + 12*x^3 + 31*x^4 + 80*x^5 + 209*x^6 + ... - _Michael Somos_, Mar 03 2023

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A110034, A236438.

LinearRecurrence[{3,0,-3,1},{1,2,5,12},50] (* Harvey P. Dale, May 01 2022 *)
a[ n_] := With[{F = Fibonacci}, (1 + F[n+1]*F[n+2] + F[n+n])/2]; (* Michael Somos, Mar 03 2023 *)

{a(n) = my(F = fibonacci); (1 + F(n+1)*F(n+2) + F(n+n))/2}; /* Michael Somos, Mar 03 2023 */

G.f.: (1-x-x^2)/((1-x^2)(1-3x+x^2));
a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4);
a(n) = F(2n) + 1 + Sum_{k=0..n-1} F(k)*F(k+1).
From R. J. Mathar, Jul 22 2010: (Start)
a(n) = Sum_{i=0..n} A061646(i).
a(n) = (5 + (-1)^n + 4*A002878(n))/10. (End)
a(n) = A110034(-n) = 1 - A110034(1+n) = A236438(n) + (n mod 2) = (1 + F(n+1)*F(n+2) + F(2*n))/2 for all n in Z. - Michael Somos, Mar 03 2023

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A013946 Least d for which the number with continued fraction [n,n,n,n...] is in Q(sqrt(d)).

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