A002878 -id:A002878 - cp's OEIS frontend

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401

Offset: 0

Ignacio Larrosa Cañestro, Feb 27 2000

Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

			a(1)^2 = 121 is the 5th star number (A003154).

G. C. Greubel, Table of n, a(n) for n = 0..1000
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Star Number
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A003154, A006061, A031138, A007667, A004189.
Cf. A138281. Cf. A100047.
Cf. A142238.

a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019

I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Numerator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A142238 *)
a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A023039 a(n) = 18*a(n-1) - a(n-2).

Original entry on oeis.org

1, 9, 161, 2889, 51841, 930249, 16692641, 299537289, 5374978561, 96450076809, 1730726404001, 31056625195209, 557288527109761, 10000136862780489, 179445175002939041, 3220013013190122249, 57780789062419261441

Offset: 0

David W. Wilson

The primitive Heronian triangle 3*a(n) +- 2, 4*a(n) has the latter side cut into 2*a(n) +- 3 by the corresponding altitude and has area 10*a(n)*A060645(n). - Lekraj Beedassy, Jun 25 2002
Chebyshev polynomials T(n,x) evaluated at x=9.
{a(n)} gives all (unsigned, integer) solutions of Pell equation a(n)^2 - 80*b(n)^2 = +1 with b(n) = A049660(n), n >= 0.
{a(n)} gives all possible solutions for x in Pell equation x^2 - D*y^2 = 1 for D=5, D=20 and D=80. The corresponding values for y are A060645 (D=5), A207832 (D=20) and A049660 (D=80). - Herbert Kociemba, Jun 05 2022
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 24 2004
For all terms x of the sequence, 5*x^2 - 5 is a square, A004292(n)^2.
The a(n) are the x-values in the nonnegative integer solutions of x^2 - 5y^2 = 1, see A060645(n) for the corresponding y-values. - Sture Sjöstedt, Nov 29 2011
Rightmost digits alternate repeatedly: 1 and 9 in fact, a(2) = 18*9 - 1 == 1 (mod 10); a(3) = 18*1 - 9 == 9 (mod 10) therefore a(2n) == 1 (mod 10), a(2n+1) == 9 (mod 10). - Carmine Suriano, Oct 03 2013

			G.f. = 1 + 9*x + 161*x^2 + 2889*x^3 + 51841*x4 + 930249*x^5 + 16692641*x^6 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..750 (terms 0..200 from Vincenzo Librandi)
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A001077, A115032.
Row 2 of array A188645.
Row 4 of A322790.

I:=[1, 9]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 13 2012

a := n -> hypergeom([n, -n], [1/2], -4):
seq(simplify(a(n)), n=0..16); # Peter Luschny, Jul 26 2020

LinearRecurrence[{18, -1}, {1, 9}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
CoefficientList[Series[(1-9*x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

{a(n) = fibonacci(6*n) / 2 + fibonacci(6*n - 1)}; /* Michael Somos, Aug 11 2009 */

x='x+O('x^30); Vec((1-9*x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) ~ (1/2)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = phi^6 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 9) = (S(n, 18) - S(n-2, 18))/2, with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 18)=A049660(n+1).
a(n) = sqrt(80*A049660(n)^2 + 1) (cf. Richardson comment).
a(n) = ((9 + 4*sqrt(5))^n + (9 - 4*sqrt(5))^n)/2.
G.f.: (1 - 9*x)/(1 - 18*x + x^2).
a(n) = cosh(2*n*arcsinh(2)). - Herbert Kociemba, Apr 24 2008
a(n) = A001077(2*n). - Michael Somos, Aug 11 2009
From Johannes W. Meijer, Jul 01 2010: (Start)
a(n) = 2*A167808(6*n+1) - A167808(6*n+3).
Limit_{k->infinity} a(n+k)/a(k) = a(n) + A060645(n)*sqrt(5).
Limit_{n->infinity} a(n)/A060645(n) = sqrt(5).
(End)
a(n) = (1/2)*A087215(n) = (1/2)*(sqrt(5) + 2)^(2*n) + (1/2)*(sqrt(5) - 2)^(2*n).
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1/8. Compare with A005248, A002878 and A075796. - Peter Bala, Nov 29 2013
a(n) = 2*A115032(n-1) - 1 = S(n, 18) - 9*S(n-1, 18), with A115032(-1) = 1, and see the above formula with S(n, 18) using its recurrence. - Wolfdieter Lang, Aug 22 2014
a(n) = A128052(3n). - A.H.M. Smeets, Oct 02 2017
a(n) = A049660(n+1) - 9*A049660(n). - R. J. Mathar, May 24 2018
a(n) = hypergeom([n, -n], [1/2], -4). - Peter Luschny, Jul 26 2020
a(n) = L(6*n)/2 for L(n) the Lucas sequence A000032(n). - Greg Dresden, Dec 07 2021
a(n) = cosh(6*n*arccsch(2)). - Peter Luschny, May 25 2022

Chebyshev and Pell comments from Wolfdieter Lang, Nov 08 2002

Sture Sjöstedt's comment corrected and reformulated by Wolfdieter Lang, Aug 24 2014

A057080 Even-indexed Chebyshev U-polynomials evaluated at sqrt(10)/2.

Original entry on oeis.org

1, 9, 71, 559, 4401, 34649, 272791, 2147679, 16908641, 133121449, 1048062951, 8251382159, 64962994321, 511452572409, 4026657584951, 31701808107199, 249587807272641, 1965000650073929, 15470417393318791, 121798338496476399

Offset: 0

Wolfdieter Lang, Aug 04 2000

a(n) = L(n,-8)*(-1)^n, where L is defined as in A108299; see also A070997 for L(n,+8). - Reinhard Zumkeller, Jun 01 2005
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim n->infinity a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 71, 34649, 16908641, 8251382159, 31701808107199,... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 9, 0, 71, 0, 559, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -6, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
From Klaus Purath, May 06 2025: (Start)
Nonnegative solutions to the Diophantine equation 3*a(n)^2 - 5*b(n)^2 = -2. The corresponding b(n) are A070997(n). Note that (a(n)*a(n+2) - a(n+1)^2)/2 = -5 and (b(n)*b(n+2) - b(n+1)^2)/2 = 3.
(a(n) + b(n))/2 = (a(n+1) - b(n+1))/2 = A001090(n+1) = Lucas U(8,1). Also a(n)*b(n+1) - a(n+1)*b(n) = -2.
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)) as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0 where (t) is a sequence satisfying t(i+3) = 9*t(i+2) - 9*t(i+1) + t(i) or t(i+2) = 8*t(i+1) - t(i), regardless of the initial values and including this sequence itself. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=10.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A033890, A100047.

a:=[1,9];; for n in [3..30] do a[n]:=8*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

I:=[1,9]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

A057080 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,9]);
    else
        8*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1+x)/(1-8x+x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)

Vec((1+x)/(1-8*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,8,1)-lucas_number2(n-1,8,1))/6 for n in range(1, 21)] # Zerinvary Lajos, Nov 10 2009

For all elements x of the sequence, 15*x^2 + 10 is a square. Lim. n-> Inf. a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 13 2002
a(n) = 8*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 8) + S(n-1, 8) = S(2*n, sqrt(10)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 8) = A001090(n).
G.f.: (1+x)/(1-8*x+x^2).
a(n) = ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)). - Gregory V. Richardson, Oct 13 2002
a(n) = sqrt((5*A070997(n)^2 - 2)/3) (cf. Richardson comment).
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i) then a(n) = (-1)^n*q(n,-10). - Benoit Cloitre, Nov 10 2002
a(n) = Jacobi_P(n,1/2,-1/2,4)/Jacobi_P(n,-1/2,1/2,1); - Paul Barry, Feb 03 2006
a(n+1) = 4*a(n) + sqrt(5*(3*a(n)^2 + 2)). - Richard Choulet, Aug 30 2007
In addition to the first formula above: In general, the following applies to all recurrences (a(n)) of the form (8,-1) with a(0) = 1 and arbritrary a(1): 15*a(n)^2 + y = b^2 where y = x^2 + 8*x + 1 and x = a(1) - 8. Also y = a(k+1)^2 - a(k)*a(k+1) for any k >=0. - Klaus Purath, May 06 2025
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 4), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 8*a(n)*a(n+1) + a(n+1)^2 = 10.
More generally, for arbitrary x, a(n+x)^2 - 8*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 10 with a(n) := ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)) as given above.
a(n+1/2) = sqrt(10) * A001090(n+1).
a(n+3/4) + a(n+1/4) = sqrt(10)*sqrt(sqrt(10) + 2) * A001090(n+1).
a(n+3/4) - a(n+1/4) = sqrt((sqrt(40) - 4)/3) * A001091(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/10 (telescoping series: for n >= 1, 10/(a(n) - 1/a(n)) = 1/A001090(n) + 1/A001090(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5/3) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 5/3 * (1 - 2/(1 + A001091(k+1)))). (End)

A026998 Triangular array T read by rows: T(n, k) = t(n, 2k), t given by A027960, 0 <= k <= n, n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 4, 8, 1, 1, 4, 11, 13, 1, 1, 4, 11, 26, 19, 1, 1, 4, 11, 29, 54, 26, 1, 1, 4, 11, 29, 73, 101, 34, 1, 1, 4, 11, 29, 76, 171, 174, 43, 1, 1, 4, 11, 29, 76, 196, 370, 281, 53, 1, 1, 4, 11, 29, 76, 199, 487, 743, 431, 64, 1

Offset: 0

Clark Kimberling

Right-edge columns are polynomials approximating Lucas(2n+1).

			  .................................... 1;
  ................................. 1, 1;
  ............................. 1,  4, 1;
  ........................ 1,   4,  8, 1;
  ................... 1,   4,  11, 13, 1;
  .............. 1,   4,  11,  26, 19, 1;
  .......... 1,  4,  11,  29,  54, 26, 1;
  ...... 1,  4, 11,  29,  73, 101, 34, 1;
  .. 1,  4, 11, 29,  76, 171, 174, 43, 1;
  1, 4, 11, 29, 76, 196, 370, 281, 53, 1;

G. C. Greubel, Table of n, a(n) for n = 0..1325

This is a bisection of the "Lucas array" A027960, see A027011 for the other bisection.
Row sums give A095121.
Signed row sums give A090132.
Diagonal sums give A027010.
Right-edge columns include A034856, A027966, A027968, A027970, A027972.
Cf. A000032.

function t(n, k) // t = A027960
      if k le n then return Lucas(k+1);
      elif k gt 2*n then return 0;
      else return t(n-1, k-2) + t(n-1, k-1);
      end if;
end function;
A026998:= func< n,k | t(n, 2*k) >;
[A026998(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 09 2025

f[n_, k_]:= f[n, k]= Sum[Binomial[2*n-k+j,j]*LucasL[2*(k-n-j)], {j,0,k-n-1}];
A027960[n_, k_]:= LucasL[k+1] - f[n,k]*Boole[k>n];
A026998[n_, k_]:= A027960[n,2*k];
Table[A026998[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 09 2025 *)

@CachedFunction
def t(n, k): # t = A027960
    if (k>2*n): return 0
    elif (kA026998(n,k): return t(n, 2*k)
print(flatten([[A026998(n, k) for k in (0..n)] for n in (0..12)])) # G. C. Greubel, Jul 09 2025

T(n, k) = Lucas(2*n+1) = A002878(n) for 2*k <= n, otherwise the (2*n-2*k)-th coefficient of the power series for (1+2*x)/( (1-x-x^2)*(1-x)^(2*k-n) ).

Edited by Ralf Stephan, May 05 2005

A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

Original entry on oeis.org

1, 10, 89, 791, 7030, 62479, 555281, 4935050, 43860169, 389806471, 3464398070, 30789776159, 273643587361, 2432002510090, 21614379003449, 192097408520951, 1707262297685110, 15173263270645039, 134852107138120241, 1198495700972437130, 10651609201613813929

Offset: 0

Wolfdieter Lang, Aug 04 2000

This is the m=11 member of the m-family of sequences S(n,m-2)+S(n-1,m-2) = S(2*n,sqrt(m)) (for S(n,x) see Formula). The m=4..10 instances are A005408, A002878, A001834, A030221, A002315, A033890 and A057080, resp. The m=1..3 (signed) sequences are: A057078, A057077 and A057079, resp.
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim_{n->oo} a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 89, 389806471, 192097408520951, 7477414486269626733119, ... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 10, 0, 89, 0, 791, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -7, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=11.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A005408, A002878, A001834, A030221, A002315, A033890, A057080.
Cf. A057078, A057077, A057079.
Cf. A018913, A049310.
Cf. A100047.

A057081 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,10]);
    else
        9*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1 + x)/(1 - 9*x + x^2), {x,0,50}], x] (* or *) LinearRecurrence[{9,-1}, {1,10}, 50] (* G. C. Greubel, Apr 12 2017 *)

Vec((1+x)/(1-9*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,9,1)-lucas_number2(n-1,9,1))/7 for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

a(n) = 9*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 9) + S(n-1, 9) = S(2*n, sqrt(11)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 9) = A018913(n).
G.f.: (1+x)/(1-9*x+x^2).
Let q(n, x) = Sum{i=0..n} x^(n-i)*binomial(2*n-i, i), a(n) = (-1)^n*q(n, -11). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-9)*(-1)^n, where L is defined as in A108299; see also A070998 for L(n,+9). - Reinhard Zumkeller, Jun 01 2005
From Peter Bala, Jun 08 2025: (Start)
a(n) = (1/sqrt(7)) * ( ((sqrt(11) + sqrt(7))/2)^(2*n+1) - ((sqrt(11) - sqrt(7))/2)^(2*n+1) ).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/11 (telescoping series: 11/(a(n) - 1/a(n)) = 1/A018913(n+1) + 1/A018913(n)).
Conjecture: for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(11/7) [telescoping product: ((a(n) + 1)/(a(n) - 1))^2 = (1 - 4/b(n+1))/(1 - 4/b(n)), where b(n) = 2 + A056918(n)]. (End)

A054888 Layer counting sequence for hyperbolic tessellation by regular pentagons of angle Pi/2.

Original entry on oeis.org

1, 5, 15, 40, 105, 275, 720, 1885, 4935, 12920, 33825, 88555, 231840, 606965, 1589055, 4160200, 10891545, 28514435, 74651760, 195440845, 511670775, 1339571480, 3507043665, 9181559515, 24037634880, 62931345125

Offset: 0

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

The layer sequence is the sequence of the cardinalities of the layers accumulating around a (finite-sided) polygon of the tessellation under successive side-reflections.

Reinhard Zumkeller, Table of n, a(n) for n = 0..999 (indices corrected to start at zero by _Sidney Cadot_, Jan 07 2022)
Paolo Dominici, Illustration
Index entries for Coordination Sequences [A layer sequence is a kind of coordination sequence. - _N. J. A. Sloane_, Nov 20 2022]
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000045, A001906, A002878, A004277, A203976.

Cf. A054886, A054887, A054889, A054890.

a054888 n = a054888_list !! (n-1)
a054888_list = 1 : zipWith (+) (tail a002878_list) a002878_list
-- Reinhard Zumkeller, Jan 11 2012

[n eq 0 select 1 else 5*Fibonacci(2*n): n in [0..40]]; // G. C. Greubel, Feb 08 2023

LinearRecurrence[{3,-1},{1,5,15},30] (* Harvey P. Dale, Jan 15 2023 *)
Join[{1}, 5*Fibonacci[2*Range[40]]] (* G. C. Greubel, Feb 08 2023 *)

{a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k)^2*x^k/k)+x*O(x^n)), n)} /* Paul D. Hanna, Feb 21 2012 */

[5*fibonacci(2*n) + int(n==0) for n in range (41)] # G. C. Greubel, Feb 08 2023

a(n) = 5*A001906(n) + [n=0].
G.f.: (1+x)^2/(1-3*x+x^2).
G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^2 * x^n/n ). - Paul D. Hanna, Feb 21 2012
a(n) = A001906(n-1) + 2*A001906(n) + A001906(n+1). - R. J. Mathar, Nov 28 2011
a(n) = A203976(A004277(n-1)). - Reinhard Zumkeller, Jan 11 2012
a(n) = 5*A000045(2*n) for n >= 1. - Robert Israel, Jun 01 2015
a(n) = A002878(n-1)+A002878(n). - R. J. Mathar, Jul 09 2024

Offset changed to 0 by N. J. A. Sloane, Jan 03 2022 at the suggestion of Michel Marcus

A005013 a(n) = 3*a(n-2) - a(n-4), a(0)=0, a(1)=1, a(2)=1, a(3)=4. Alternates Fibonacci (A000045) and Lucas (A000032) sequences for even and odd n.

Original entry on oeis.org

0, 1, 1, 4, 3, 11, 8, 29, 21, 76, 55, 199, 144, 521, 377, 1364, 987, 3571, 2584, 9349, 6765, 24476, 17711, 64079, 46368, 167761, 121393, 439204, 317811, 1149851, 832040, 3010349, 2178309, 7881196, 5702887, 20633239, 14930352, 54018521, 39088169, 141422324

Offset: 0

N. J. A. Sloane

S(n,sqrt(5)), with the Chebyshev polynomials A049310, is an integer sequence in the real quadratic number field Q(sqrt(5)) with basis numbers <1,phi>, phi:=(1+sqrt(5))/2. S(n,sqrt(5)) = A(n) + 2*B(n)*phi, with A(n)= a(n+1)*(-1)^n and B(n)= A147600(n-1), n>=0, with A147600(-1):=0.
a(n) = p(n+1) where p(x) is the unique degree-(n-1) polynomial such that p(k) = Fibonacci(k) for k = 1, ..., n. - Michael Somos, Jan 08 2012
Row sums of A227431. - Richard R. Forberg, Jul 29 2013
This is the sequence of Lehmer numbers u_n(sqrt(R),Q) with the parameters R = 5 and Q = 1. It is a strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n,m)) for all natural numbers n and m. The sequence satisfies a linear recurrence of order four. - Peter Bala, Apr 18 2014
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 denominators give the present sequence. The sequence of numerators [0, 1, 5, 4, 15, 11, 40, ...] is A203976. Cf. A108412 and A026741. - Peter Bala, May 19 2014
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. We have (1/2)*a(2*n + 1) = 1/2 o 1/2 o ... o 1/2 (2*n + 1 terms) and (1/2)*sqrt(5)* a(2*n) = 1/2 o 1/2 o ... o 1/2 (2*n terms). Cf. A084068 and A049629. - Peter Bala, Mar 23 2018

			G.f. = x + x^2 + 4*x^3 + 3*x^4 + 11*x^5 + 8*x^6 + 29*x^7 + 21*x^8 + 76*x^9 + ...
a(3) = 4 since p(x) = (x^2 - 3*x + 4) / 2 interpolates p(1) = 1, p(2) = 1, p(3) = 2, and p(4) = 4. - _Michael Somos_, Jan 08 2012

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

G. C. Greubel, Table of n, a(n) for n = 0..2000 (terms 0..500 from T. D. Noe)
A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979. [Annotated scanned copy]
Seong Ju Kim, R. Stees and L. Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), #16.1.4.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
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.
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.
E. W. Weisstein, MathWorld: Lehmer Number
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 (0,3,0,-1).

Cf. A000032, A000045, A001906, A002878, A005247, A096140, A026741, A108412, A084068, A049629.

a:=[0,1,1,4];; for n in [5..40] do a[n]:=3*a[n-2]-a[n-4]; od; a; # Muniru A Asiru, Oct 21 2018

a005013 n = a005013_list !! n
a005013_list = alt a000045_list a000032_list where
   alt (f::fs) (:l:ls) = f : l : alt fs ls
-- Reinhard Zumkeller, Jan 10 2012

I:=[0,1,1,4]; [n le 4 select I[n]  else 3*Self(n-2) - Self(n-4): n in [1..40]]; // Vincenzo Librandi, Feb 09 2016

with(combinat): A005013 := n-> if n mod 2 = 0 then fibonacci(n) else fibonacci(n+1)+fibonacci(n-1); fi;
A005013:=z*(z**2+z+1)/((z**2+z-1)*(z**2-z-1)); # Simon Plouffe in his 1992 dissertation

CoefficientList[Series[(x + x^2 + x^3)/(1 - 3x^2 + x^4), {x, 0, 40}], x]
f[n_] = Product[(1 + 4*Sin[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}]; a = Table[f[n], {n, 0, 30}]; Round[a]; FullSimplify[ExpandAll[a]] (* Roger L. Bagula and Gary W. Adamson, Nov 26 2008 *)
LinearRecurrence[{0, 3, 0, -1}, {0, 1, 1, 4}, 100] (* G. C. Greubel, Feb 08 2016 *)

{a(n) = if( n%2, fibonacci(n+1) + fibonacci(n-1), fibonacci(n))}; /* Michael Somos, Jan 08 2012 */

{a(n) = if( n<0, -a(-n), subst( polinterpolate( vector( n, k, fibonacci(k))), x, n+1))}; /* Michael Somos, Jan 08 2012 */

a(1) = a(2) = 1, a(3) = 4, a(n) = (a(n-1) * a(n-2) - 1) / a(n-3), unless n=3. a(-n) = -a(n).
a(2n) = A001906(n), a(2n+1) = A002878(n). a(n)=F(n+1)+(-1)^(n+1)F(n-1). - Mario Catalani (mario.catalani(AT)unito.it), Sep 20 2002
G.f.: x*(1+x+x^2)/((1-x-x^2)*(1+x-x^2)).
a(n) = Product_{k=1..floor((n-1)/2)} (1 + 4*sin(k*Pi/n)^2). - Roger L. Bagula and Gary W. Adamson, Nov 26 2008
Binomial transform is A096140. - Michael Somos, Apr 13 2012
From Peter Bala, Apr 18 2014: (Start)
a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, and a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even, where alpha = (1/2)*(sqrt(5) + 1) and beta = (1/2)*(sqrt(5) - 1). Equivalently, a(n) = U(n-1, sqrt(5)/2) for n odd and a(n) = (1/sqrt(5))*U(n-1, sqrt(5)/2) for n even, where U(n,x) is the Chebyshev polynomial of the second kind. (End)
E.g.f.: (Phi/sqrt(5))*exp(-Phi*x)*(exp(x)-1)*(exp(sqrt(5)*x) - 1/(Phi)^2), where Phi = (1+sqrt(5))/2. - G. C. Greubel, Feb 08 2016
a(n) = (5^floor((n-1)/2)/2^(n-1))*Sum_{k=0..n-1} binomial(n-1,k)/5^floor(k/2). - Tony Foster III, Oct 21 2018
a(n) = hypergeom([(1 - n)/2, (n + 1) mod 2 - n/2], [1 - n], -4) for n >= 2. - Peter Luschny, Sep 03 2019

Additional comments from Michael Somos, Jun 01 2000

A153387 Decimal expansion of Sum_{n>=1} 1/Fibonacci(2*n-1).

Original entry on oeis.org

1, 8, 2, 4, 5, 1, 5, 1, 5, 7, 4, 0, 6, 9, 2, 4, 5, 6, 8, 1, 4, 2, 1, 5, 8, 4, 0, 6, 2, 6, 7, 3, 2, 8, 1, 7, 3, 3, 2, 1, 8, 9, 3, 5, 4, 2, 6, 6, 0, 8, 2, 9, 9, 2, 3, 2, 6, 0, 2, 9, 0, 1, 5, 0, 1, 9, 4, 0, 8, 3, 0, 4, 0, 3, 6, 7, 7, 7, 3, 9, 6, 7, 5, 9, 8, 9, 1, 3, 8, 9, 9, 8, 1, 9, 8, 2, 0, 7, 5, 0, 7, 6, 4, 2, 4

Offset: 1

Eric W. Weisstein, Dec 25 2008

Borwein et al. express the sum in terms of theta functions. - N. J. A. Sloane, May 16 2011

Duverney et al. (1997) proved that this constant is transcendental. - Amiram Eldar, Oct 30 2020

			1.8245151574069245681...

J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See pp. 202-203.

Joerg Arndt, Table of n, a(n) for n = 1..1000
Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
L. Carlitz, Problem B-110, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; An Infinite Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
Daniel Duverney, Keiji Nishioka, Kumiko Nishioka and Iekata Shiokawa, Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers, Proceedings of the Japan Academy, Series A, Mathematical Sciences, Vol. 73, No. 7 (1997), pp. 140-142.
Index entries for transcendental numbers

Cf. A000032, A002878, A079586, A153386, A190649.

sumpos(n=1, 1/fibonacci(2*n-1)) \\ Michel Marcus, Feb 05 2022

Equals sqrt(5)/4 * (T(b^2)^2 - T(b)^2) where T(q) = 1 + 2*Sum_{n>=1} q^(n^2) and b = 1/2*(1-sqrt(5)); see the Arndt reference and the references cited there. - Joerg Arndt, Feb 01 2014

Equals sqrt(5) * Sum_{n>=0} (-1)^n/Lucas(2*n+1) (Carlitz, 1967). - Amiram Eldar, Feb 05 2022

Definition reconciled to sequence and example by Clark Kimberling, Aug 06 2013

A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))n^(k-j), n >= 1, k >= 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 5, -1, 1, 5, 11, 7, -2, 1, 6, 19, 29, 9, -1, 1, 7, 29, 71, 76, 11, 1, 1, 8, 41, 139, 265, 199, 13, 2, 1, 9, 55, 239, 666, 989, 521, 15, 1, 1, 10, 71, 377, 1393, 3191, 3691, 1364, 17, -1, 1, 11, 89, 559, 2584, 8119, 15289, 13775, 3571, 19, -2

Offset: 1

L. Edson Jeffery, Bob Selcoe and Andrew Hone, Oct 22 2017

This array is used to compute A269254: A269254(n) = least k such that A(n,k) is a prime, or -1 if no such k exists.
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

			Array begins:
  1   2    1    -1     -2      -1        1         2          1          -1
  1   3    5     7      9      11       13        15         17          19
  1   4   11    29     76     199      521      1364       3571        9349
  1   5   19    71    265     989     3691     13775      51409      191861
  1   6   29   139    666    3191    15289     73254     350981     1681651
  1   7   41   239   1393    8119    47321    275807    1607521     9369319
  1   8   55   377   2584   17711   121393    832040    5702887    39088169
  1   9   71   559   4401   34649   272791   2147679   16908641   133121449
  1  10   89   791   7030   62479   555281   4935050   43860169   389806471
  1  11  109  1079  10681  105731  1046629  10360559  102558961  1015229051

Robert Price, Table of n, a(n) for n = 1..5050
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.

Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A269253, A269254, A294099, A298675, A298677, A298878, A299045, A299071.
Rows: A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, ...
Columns: A000012, A000027, A028387, ...

(* Array: *)
Grid[Table[LinearRecurrence[{n, -1}, {1, 1 + n}, 10], {n, 10}]]
(* Array antidiagonals flattened (gives this sequence): *)
A294099[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] n^(k - j), {j, 0, k}]; Flatten[Table[A294099[n - k, k], {n, 11}, {k, 0, n - 1}]]

{A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*n^(k-j))}; /* Michael Somos, Jun 19 2023 */

A(n,0) = 1, A(n,1) = n + 1, A(n,k) = n*A(n,k-1) - A(n,k-2), n >= 1, k >= 2.
G.f. for row n: (1 + x)/(1 - n*x + x^2), n >= 1.
A(n, k) = B(-n, k) where B = A299045. - Michael Somos, Jun 19 2023

A079472 Number of perfect matchings on an n X n L-shaped graph.

Original entry on oeis.org

0, 2, 4, 12, 30, 80, 208, 546, 1428, 3740, 9790, 25632, 67104, 175682, 459940, 1204140, 3152478, 8253296, 21607408, 56568930, 148099380, 387729212, 1015088254, 2657535552, 6957518400, 18215019650, 47687540548, 124847601996, 326855265438, 855718194320

Offset: 1

Helen King (h.king(AT)uea.ac.uk), Jan 15 2003

a(n+1) = 2*F(n)*F(n+1) appears as the second component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, with F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), a(n+1), A059929(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014
a(n+1) is the numerator of the continued fraction [1,...,1,2,1,...,1] with n 1's to the left of the central 2, and n 1's to the right of the central 2. For the denominators, see A061646. - Greg Dresden and Max Liu, Jun 25 2023
For n >= 3, a(n) equals the sum of the sides of the right triangle with side lengths [F(n)*F(n-3), 2*F(n-1)*F(n-2), F(2*n-3)] (n = 4 corresponds to the 3-4-5 right triangle). - Peter Bala, Nov 03 2023

			a(7) = 2*13*8 = 208 = number of matchings. F(7) = 13 F(6) = 8
a(3) = 4 because in the graph with vertex set {(0,0), (1,0), (2,0), (0,1), (1,1), (2,1), (0,2), (1,2)} and edge set {h(0,0), h(1,0), h(0,1), h(1,1), h(0,2), v(0,0), v(0,1), v(1,0), v(1,1), v(2,0)}, where h(i,j) (v(i,j)) is a horizontal (vertical) edge of unit length starting from vertex (i,j), we have the following four perfect matchings: {h(0,0), h(0,1), h(0,2), v(2,0)}, {h(0,0), v(0,1), v(1,1), v(2,0)}, {v(0,0), v(1,0), v(2,0), h(0,2)} and {v(0,0), h(1,0), h(1,1), h(0,2)}. - _Emeric Deutsch_, Dec 30 2004
G.f. = 2*x^2 + 4*x^3 + 12*x^4 + 30*x^5 + 80*x^6 + 208*x^7 + 546*x^8 + ...

Daniele Corradetti, La Metafisica del Numero, 2008.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. pp. 178, 255.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
P. F. F. Espinosa, J. F. González, J. P. Herrán, A. M. Cañadas, and J. L. Ramírez, On some relationships between snake graphs and Brauer configuration algebras, Algebra Disc. Math. (2022) Vol. 33, No. 2, 29-59.
I. Gutman and S. J. Cyvin, A result on 1-factors related to Fibonacci numbers, The Fibonacci Quarterly, 28 (1990), pp. 81-84.
C.-A. Laisant, Observations sur les triangles rectangles en nombres entiers et les suites de Fibonacci, Nouvelles Annales de Math. (1919, in French) Series 4, Vol. 19, 391-397.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001654, A121646.

List([1..30], n -> 2*Fibonacci(n-1)*Fibonacci(n)); # G. C. Greubel, Jan 07 2019

[2*Fibonacci(n)*Fibonacci(n-1): n in [1..30]]; // Vincenzo Librandi, Jun 29 2014

with(combinat,fibonacci):seq(2*fibonacci(n)*fibonacci(n-1),n=1..30);

LinearRecurrence[{2,2,-1}, {0,2,4}, 30] (* Arkadiusz Wesolowski, Sep 15 2012 *)
Table[(2*Fibonacci[n]*Fibonacci[n-1]), {n,30}] (* Vincenzo Librandi, Jun 29 2014 *)

{a(n) = 2 * fibonacci(n) * fibonacci(n-1)}; \\ Michael Somos, Jun 28 2014

concat(0, Vec(2*x^2/((x+1)*(x^2-3*x+1)) + O(x^40))) \\ Colin Barker, Sep 27 2016

[2*fibonacci(n-1)*fibonacci(n) for n in (1..30)] # G. C. Greubel, Jan 07 2019

a(n) = 2*F(n)*F(n-1) where F(n) are the Fibonacci numbers (A000045).
From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003: (Start)
a(n) = 2*A001654(n) = F(2*n) - F(n)^2 = A001906(n) - A007598(n).
a(n) = (F(n+1)^2 - F(n-2)^2)/2 = (A007598(n+1) - A007598(n-2))/2.
a(n) = 2*(L(2*n-1) + (-1)^n)/5 = (2/5)*(A002878(n-1) + A033999(n)), where L(n) = A000032(n).
a(n+1) = a(n) + 2*F(n)^2.
G.f.: 2*x^2/((1+x)*(1-3*x+x^2)). (End)
a(n) = Im( (F(n) + i*F(n+1))^2 ) (cf. A121646). - Daniele Corradetti (d.corradetti(AT)gmail.com), May 02 2008
From Michael Somos, Jun 28 2014: (Start)
a(n) = F(n+1)^2 - F(n)^2 - F(n-1)^2.
a(1 - n) = -a(n). (End)
a(n) = ( 2*(-1)^n - (1+sqrt(5))*((3-sqrt(5))/2)^n - (1-sqrt(5))*((3+sqrt(5))/2)^n )/5. - Colin Barker, Sep 27 2016
From Rigoberto Florez, May 06 2020: (Start)
a(n) = F(2n-2) + F(n-1)^2, where F(n) is the n-th Fibonacci number.
a(n) = M^(n+1)[2,1], for n>0 where M=[0,0,1;0,1,2;1,1,1]. (End)
a(n) = F(n)^2 + F(n-1)^2 - F(n-2)^2. - Michael Somos, Mar 02 2023

More terms from Benoit Cloitre and Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jan 18 2003

cp's OEIS Frontend

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

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A023039 a(n) = 18*a(n-1) - a(n-2).

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A057080 Even-indexed Chebyshev U-polynomials evaluated at sqrt(10)/2.

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A026998 Triangular array T read by rows: T(n, k) = t(n, 2k), t given by A027960, 0 <= k <= n, n >= 0.

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A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

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A054888 Layer counting sequence for hyperbolic tessellation by regular pentagons of angle Pi/2.

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A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))n^(k-j), n >= 1, k >= 0.