A033891 -id:A033891 - cp's OEIS frontend

A103134 a(n) = Fibonacci(6n+4).

3, 55, 987, 17711, 317811, 5702887, 102334155, 1836311903, 32951280099, 591286729879, 10610209857723, 190392490709135, 3416454622906707, 61305790721611591, 1100087778366101931, 19740274219868223167, 354224848179261915075, 6356306993006846248183

Offset: 0

Creighton Dement, Jan 24 2005

Gives those numbers which are Fibonacci numbers in A103135.
Generally, for any sequence where a(0)= Fibonacci(p), a(1) = F(p+q) and Lucas(q)*a(1) +- a(0) = F(p+2q), then a(n) = L(q)*a(n-1) +- a(n-2) generates the following Fibonacci sequence: a(n) = F(q(n)+p). So for this sequence, a(n) = 18*a(n-1) - a(n-2) = F(6n+4): q=6, because 18 is the 6th Lucas number (L(0) = 2, L(1)=1); F(4)=3, F(10)=55 and F(16)=987 (F(0)=0 and F(1)=1). See Lucas sequence A000032. This is a special case where a(0) and a(1) are increasing Fibonacci numbers and Lucas(m)*a(1) +- a(0) is another Fibonacci. - Bob Selcoe, Jul 08 2013
a(n) = x + y where x and y are solutions to x^2 = 5*y^2 - 1. (See related sequences with formula below.) - Richard R. Forberg, Sep 05 2013

Colin Barker, Table of n, a(n) for n = 0..750
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Subsequence of A033887.
Cf. A000032, A000045, A001906, A001519, A015448, A014445, A033888, A033889, A033890, A033891, A049310, A049660, A102312, A099100, A134490, A134491, A134492, A134493, A134494, A134495, A103134, A134497, A134498, A134499, A134500, A134501, A134502, A134503, A134504.
Cf. A103135.

[Fibonacci(6*n +4): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[6n+4], {n, 0, 30}]
LinearRecurrence[{18,-1},{3,55},20] (* Harvey P. Dale, Mar 29 2023 *)
Table[ChebyshevU[3*n+1, 3/2], {n, 0, 20}] (* Vaclav Kotesovec, May 27 2023 *)

a(n)=fibonacci(6*n+4) \\ Charles R Greathouse IV, Feb 05 2013

G.f.: (x+3)/(x^2-18*x+1).
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=3, a(1)=55. - Philippe Deléham, Nov 17 2008
a(n) = A007805(n) + A075796(n), as follows from comment above. - Richard R. Forberg, Sep 05 2013
a(n) = ((15-7*sqrt(5)+(9+4*sqrt(5))^(2*n)*(15+7*sqrt(5))))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
a(n) = S(3*n+1, 3) = 3*S(n,18) + S(n-1,18), with the Chebyshev S polynomials (A049310), S(-1, x) = 0, and S(n, 18) = A049660(n+1). - Wolfdieter Lang, May 08 2023

Edited by N. J. A. Sloane, Aug 10 2010

A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

Original entry on oeis.org

0, 2, 15, 104, 714, 4895, 33552, 229970, 1576239, 10803704, 74049690, 507544127, 3478759200, 23843770274, 163427632719, 1120149658760, 7677619978602, 52623190191455, 360684711361584, 2472169789339634, 16944503814015855, 116139356908771352, 796030994547383610

Offset: 0

R. K. Guy, Mar 01 2003

Another interpretation of this sequence is: nonnegative integers k such that (k + 1)^2 + (2k)^2 is a perfect square. So apart from a(0) = 0, a(n) + 1 and 2a(n) form the legs of a Pythagorean triple. - Nick Hobson, Jan 13 2007
Also solution y of Diophantine equation x^2 + 4*y^2 = k^2 for which x=y+1. - Carmine Suriano, Jun 23 2010
Also the index of the first of two consecutive heptagonal numbers whose sum is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 20 2014
Nonnegative integers k such that G(x) = k for some rational number x where G(x) = x/(1-x-x^2) is the generating function of the Fibonacci numbers. - Tom Edgar, Aug 24 2015
The integer solutions of the equation a(b+1) = (a-b)(a-b-1) or, equivalently, binomial(a, b) = binomial(a-1, b+1) are given by (a, b) = (a(n+1), A003482(n)=Fibonacci(2*n) * Fibonacci(2*n+3)) (Lind and Singmaster). - Tomohiro Yamada, May 30 2018

			G.f. = 2*x + 15*x^2 + 104*x^3 + 714*x^4 + 4895*x^5 + 33552*x^6 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 28.
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Colin Barker, Table of n, a(n) for n = 0..1000
Pridon Davlianidze, Problem B-1279, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; An Unusual Generalization, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
Dae S. Hong, When is the Generating Function of the Fibonacci Numbers an Integer?, The College Mathematics Journal, Vol. 46, No. 2 (March 2015), pp. 110-112
D. A. Lind, The quadratic field Q(sqrt(5)) and a certain diophantine equation, Fibonacci Quart., Vol. 6, No. 3 (1968), pp. 86-93.
David Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quart., Vol. 13, No. 4 (1973), pp. 295-298.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081017.
Partial sums of A033891. Bisection of A001654 and A059840.
Equals A089508 + 1.
Cf. A001622, A081007, A081016, A178898.

List([0..30], n-> (Lucas(1,-1, 4*n+1)[2] -1)/5); # G. C. Greubel, Jul 14 2019

[(Lucas(4*n+1)-1)/5: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+1)-1)/5) od: # James Sellers, Mar 03 2003

(LucasL[4*Range[0,30]+1]-1)/5 (* or *) LinearRecurrence[{8,-8,1}, {0,2,15}, 30] (* G. C. Greubel, Aug 24 2015, modified Jul 14 2019 *)

concat(0, Vec(x*(2-x)/((1-x)*(1-7*x+x^2)) + O(x^30))) \\ Colin Barker, Dec 20 2014

[(lucas_number2(4*n+1,1,-1) -1)/5 for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(3) + Fibonacci(7) + Fibonacci(11) + ... + Fibonacci(4n+3).
G.f.: x*(2-x)/((1-x)*(1-7*x+x^2)). - Colin Barker, Mar 30 2012
E.g.f.: (1/5)^(3/2)*((1+phi^2)*exp(phi^4*x) - (1 + (1/phi^2))*exp(x/phi^4) - sqrt(5)*exp(x)), where 2*phi = 1 + sqrt(5). - G. C. Greubel, Aug 24 2015
From - Michael Somos, Aug 27 2015: (Start)
a(n) = -A081016(-1-n) for all n in Z.
0 = a(n) - 7*a(n+1) + a(n+2) - 1 for all n in Z.
0 = a(n)*a(n+2) - a(n+1)^2 + a(n+1) + 2 for all n in Z.
0 = a(n)*(a(n) -7*a(n+1) -1) + a(n+1)*(a(n+1) - 1) - 2 for all n in Z. (End)
a(n) = (k(n) + sqrt(k(n)*(4 + 5*k(n))))/2, where k(n) = A049684(n). - Stefano Spezia, Mar 11 2021
Product_{n>=1} (1 + 1/a(n)) = phi (A001622) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021

More terms from James Sellers, Mar 03 2003

A134504 a(n) = Fibonacci(7n + 6).

Original entry on oeis.org

8, 233, 6765, 196418, 5702887, 165580141, 4807526976, 139583862445, 4052739537881, 117669030460994, 3416454622906707, 99194853094755497, 2880067194370816120, 83621143489848422977, 2427893228399975082453

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (29,1).

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888, A033889, A033890, A033891, A102312, A099100, A134490, A134491, A134492, A134493, A134494, A134495, A103134, A134497, A134498, A134499, A134500, A134501, A134502, A134503, A134504.

[Fibonacci(7*n +6): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[7n+6], {n, 0, 30}]
LinearRecurrence[{29,1},{8,233},20] (* Harvey P. Dale, Jul 21 2021 *)

a(n)=fibonacci(7*n+6) \\ Charles R Greathouse IV, Jun 11 2015

G.f.: (-8-x) / (-1 + 29*x + x^2). - R. J. Mathar, Jul 04 2011
a(n) = A000045(A017053(n)). - Michel Marcus, Nov 08 2013
a(n) = 29*a(n-1) + a(n-2). - Wesley Ivan Hurt, Mar 15 2023

Offset changed from 1 to 0 by Vincenzo Librandi, Apr 17 2011

A134494 a(n) = Fibonacci(6n+2).

Original entry on oeis.org

1, 21, 377, 6765, 121393, 2178309, 39088169, 701408733, 12586269025, 225851433717, 4052739537881, 72723460248141, 1304969544928657, 23416728348467685, 420196140727489673, 7540113804746346429, 135301852344706746049, 2427893228399975082453

Offset: 0

Artur Jasinski, Oct 28 2007

Colin Barker, Table of n, a(n) for n = 0..750
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001906, A001519, A015448, A014445, A033887-A033891, A049310, A049660, A099100, A102312, A103134, A134490 - A134504.

[Fibonacci(6*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

seq( combinat[fibonacci](6*n+2),n=0..10) ; # R. J. Mathar, Apr 17 2011

Table[Fibonacci[6n+2], {n, 0, 30}]
Table[ChebyshevU[3*n, 3/2], {n, 0, 20}] (* Vaclav Kotesovec, May 27 2023 *)

a(n)=fibonacci(6*n+2) \\ Charles R Greathouse IV, Jun 11 2015

Vec((1+3*x)/(1-18*x+x^2) + O(x^100)) \\ Altug Alkan, Jan 24 2016

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: ( 1+3*x ) / ( 1-18*x+x^2 ).
a(n) = 3*A049660(n)+A049660(n+1). (End)
a(n) = A000045(A016933(n)). - Michel Marcus, Nov 07 2013
a(n) = ((5-3*sqrt(5)+(5+3*sqrt(5))*(9+4*sqrt(5))^(2*n)))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
a(n) = S(3*n, 3) = S(n,18) + 3*S(n-1,18), with the Chebyshev S polynomials (A049310), S(-1, x) = 0, and S(n, 18) = A049660(n+1). - Wolfdieter Lang, May 08 2023

Index in definition corrected by T. D. Noe, Joerg Arndt, Apr 17 2011

A167816 Numerator of x(n) = x(n-1) + x(n-2), x(0)=0, x(1)=1/3; denominator=A167817.

Original entry on oeis.org

0, 1, 1, 2, 1, 5, 8, 13, 7, 34, 55, 89, 48, 233, 377, 610, 329, 1597, 2584, 4181, 2255, 10946, 17711, 28657, 15456, 75025, 121393, 196418, 105937, 514229, 832040, 1346269, 726103, 3524578, 5702887, 9227465, 4976784, 24157817, 39088169, 63245986, 34111385

Offset: 0

Reinhard Zumkeller, Nov 13 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Fibonacci number
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 7, 0, 0, 0, -1).

Cf. A000045, A167808.

[0,1,1] cat [Numerator(Fibonacci(n)/Fibonacci(2*n-4)): n in [3..40]]; // Vincenzo Librandi, Jun 28 2016

Numerator[LinearRecurrence[{1,1},{0,1/3},40]] (* Harvey P. Dale, Dec 07 2014 *)
LinearRecurrence[{0, 0, 0, 7, 0, 0, 0, -1},{0, 1, 1, 2, 1, 5, 8, 13},39] (* Ray Chandler, Aug 03 2015 *)

a(n) = (a(n-1)*A093148(n+2) + a(n-2)*A093148(n+1))/A093148(n-1) for n>1.
a(4*n) = A004187(n) = (a(4*n-1) + a(4*n-2))/3;
a(4*n+1) = A033889(n) = 3*a(4*n-1) + a(4*n-2);
a(4*n+2) = A033890(n) = a(4*n-1) + 3*a(4*n-2);
a(4*n+3) = A033891(n) = a(4*n-1) + a(4*n-2).
Numerator of Fibonacci(n) / Fibonacci(2n-4) for n>=3. - Gary Detlefs, Dec 20 2010

Definition corrected by D. S. McNeil, May 09 2010

A134501 a(n) = Fibonacci(7n + 3).

Original entry on oeis.org

2, 55, 1597, 46368, 1346269, 39088169, 1134903170, 32951280099, 956722026041, 27777890035288, 806515533049393, 23416728348467685, 679891637638612258, 19740274219868223167, 573147844013817084101, 16641027750620563662096

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (29,1).

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888, A033889, A033890, A033891, A102312, A099100, A134490-A134495, A103134, A134497 - A134504.

[Fibonacci(7*n+3): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Mathematica
```
Table[Fibonacci[7n+3], {n, 0, 30}]
```

a(n)=fibonacci(7*n+3) \\ Charles R Greathouse IV, Jun 11 2015

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: (-2+3*x) / (-1 + 29*x + x^2).
a(n) = 2*A049667(n+1) - 3*A049667(n). (End)
a(n) = A000045(A017017(n)). - Michel Marcus, Nov 07 2013

Offset changed to 0 by Vincenzo Librandi, Apr 16 2011

A134502 a(n) = Fibonacci(7n + 4).

Original entry on oeis.org

3, 89, 2584, 75025, 2178309, 63245986, 1836311903, 53316291173, 1548008755920, 44945570212853, 1304969544928657, 37889062373143906, 1100087778366101931, 31940434634990099905, 927372692193078999176, 26925748508234281076009

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (29,1)

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888, A033889, A033890, A033891, A102312, A099100, A134490-A134495, A103134, A134497-A134504.

[Fibonacci(7*n +4): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Mathematica
```
Table[Fibonacci[7n+4], {n, 0, 30}]
```

a(n)=fibonacci(7*n+4) \\ Charles R Greathouse IV, Jun 11 2015

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: (-3-2*x) / (-1 + 29*x + x^2).
a(n) = 2*A049667(n) + 3*A049667(n+1). (End)
a(n) = A000045(A017029(n)). - Michel Marcus, Nov 07 2013

Offset changed from 1 to 0 by Vincenzo Librandi, Apr 17 2011

A168229 Decimal expansion of arctan(sqrt(7)).

Original entry on oeis.org

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

Offset: 1

Jonathan Vos Post, Nov 20 2009

This constant is the least x > 0 satisfying cos(4*x) = (cos x)^2. - Clark Kimberling, Oct 15 2011

An identity resembling Machin's Pi/4 = arctan(1/1) = 4*arctan(1/5) - arctan(1/239) is arctan(sqrt(7)/1) = 5*arctan(sqrt(7)/11) + 2*arctan(sqrt(7)/181), which can also be expressed as arcsin(sqrt(7/2^3)) = 5*arcsin(sqrt(7/2^7)) + 2*arcsin(sqrt(7/2^15)) (cf. A038198). - Joerg Arndt, Nov 09 2012

			arctan(sqrt(7)) = 1.209429202888189... .

G. C. Greubel, Table of n, a(n) for n = 1..5000
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
Djurdje Cvijovic, A dilogarithmic integral arising in quantum field theory, arXiv:0911.3773 [math.CA], 2009.
Index entries for transcendental numbers

Cf. A000032, A000045, A033891, A038198, A195699.

[Arctan(Sqrt(7))]; // G. C. Greubel, Nov 18 2017

RealDigits[ArcTan[Sqrt[7]], 10, 50][[1]] (* G. C. Greubel, Nov 18 2017 *)

atan(sqrt(7)) \\ Michel Marcus, Mar 11 2013

Smallest positive solution of cos(x) + sqrt(1 + cos^2(x)) = sqrt(2). - Geoffrey Caveney, Apr 24 2014
Equals Sum_{k >= 1} atan(5*sqrt(7)*F(4k-1)/L(2*(4k-1))) where L=A000032 and F=A000045. See also A033891. - Michel Marcus, Mar 29 2016
Equals arccos(1/(2*sqrt(2))). - Amiram Eldar, May 28 2021

More digits from R. J. Mathar, Dec 06 2009

A288913 a(n) = Lucas(4*n + 3).

Original entry on oeis.org

4, 29, 199, 1364, 9349, 64079, 439204, 3010349, 20633239, 141422324, 969323029, 6643838879, 45537549124, 312119004989, 2139295485799, 14662949395604, 100501350283429, 688846502588399, 4721424167835364, 32361122672259149, 221806434537978679, 1520283919093591604

Offset: 0

Bruno Berselli, Jun 19 2017

a(n) mod 4 gives A101000.

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A101000.
Cf. A033891: fourth quadrisection of A000045.
Partial sums are in A081007 (after 0).
Positive terms of A098149, and subsequence of A001350, A002878, A016897, A093960, A068397.
Quadrisection of A000032: A056854 (first), A056914 (second), A246453 (third, without 11), this sequence (fourth).

[Lucas(4*n + 3): n in [0..30]]; // G. C. Greubel, Dec 22 2017

LucasL[4 Range[0, 21] + 3]
LinearRecurrence[{7,-1}, {4,29}, 30] (* G. C. Greubel, Dec 22 2017 *)

Vec((4 + x)/(1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017

from sympy import lucas
def a(n):  return lucas(4*n + 3)
print([a(n) for n in range(22)]) # Michael S. Branicky, Apr 29 2021

def L():
    x, y = -1, 4
    while True:
        yield y
        x, y = y, 7*y - x
r = L(); [next(r) for  in (0..21)] # _Peter Luschny, Jun 20 2017

G.f.: (4 + x)/(1 - 7*x + x^2).
a(n) = 7*a(n-1) - a(n-2) for n>1, with a(0)=4, a(1)=29.
a(n) = ((sqrt(5) + 1)^(4*n + 3) - (sqrt(5) - 1)^(4*n + 3))/(8*16^n).
a(n) = Fibonacci(4*n+4) + Fibonacci(4*n+2).
a(n) = 4*A004187(n+1) + A004187(n).
a(n) = 5*A003482(n) + 4 = 5*A081016(n) - 1.
a(n) = A002878(2*n+1) = A093960(2*n+3) = A001350(4*n+3) = A068397(4*n+3).
a(n+1)*a(n+k) - a(n)*a(n+k+1) = 15*Fibonacci(4*k). Example: for k=6, a(n+1)*a(n+6) - a(n)*a(n+7) = 15*Fibonacci(24) = 695520.

A172968 a(n) = 7*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 13, 89, 610, 4181, 28657, 196418, 1346269, 9227465, 63245986, 433494437, 2971215073, 20365011074, 139583862445, 956722026041, 6557470319842, 44945570212853, 308061521170129, 2111485077978050, 14472334024676221

Offset: 0

Artur Jasinski, Feb 06 2010

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marta Na Chen, Wenchang Chu, Four classes of symmetric sums over cyclically binomial products, Symmetry 17 (2025) no 2
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A081016.

Essentially the same as A033891.

List([0..30], n-> Fibonacci(4*n-1)); # G. C. Greubel, Jul 15 2019

[n le 2 select n else 7*Self(n-1)-Self(n-2): n in [1..30]]; // Bruno Berselli, Mar 29 2016

with(combinat):F:= n-> fibonacci(n):L:=n-> 2*F(n+1)-F(n):
seq(1/2*(L(4*n)-F(4*n)), n=0..20); # Gary Detlefs, Nov 28 2010

Table[Sqrt[1-2m+5m^2]/.m ->Fibonacci[2n+1]Fibonacci[2n+2], {n, -1, 30}]
CoefficientList[Series[(1-5x)/(1-7x+x^2), {x, 0, 30}], x] (* Michael De Vlieger, Mar 29 2016 *)
Fibonacci[4*Range[0, 30] -1] (* G. C. Greubel, Jul 15 2019 *)

x='x+O('x^30); Vec((1-5*x)/(1-7*x+x^2)) \\ Altug Alkan, Mar 29 2016

[fibonacci(4*n-1) for n in (0..30)] # G. C. Greubel, Jul 15 2019

a(n) = (1/10)*((5+sqrt(5))*((7-3*sqrt(5))/2)^n + ((5-sqrt(5))*((7+3*sqrt(5))/2)^n)).
a(n) = sqrt(1 - 2*F(2n+1)*F(2n+2) + 5*(F(2n+1)*F(2n+2))^2), where F = A000045.
a(n) = sqrt(1 - 2*A081016(n) + 5*A081016(n)^2).
a(n) = A033891(n-1), n>0. - R. J. Mathar, Feb 08 2010
a(n) = (Lucas(4*n) - Fibonacci(4*n))/2, where Lucas = A000032. - Gary Detlefs, Nov 28 2010
G.f.: (1 - 5*x)/(1 - 7*x + x^2). - Bruno Berselli, Mar 29 2016
a(n) = Fibonacci(4*n-1). - G. C. Greubel, Jul 15 2019
a(n) = (a(n-1)^2 + 9)/a(n-2). - Klaus Purath, Aug 30 2020

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A103134 a(n) = Fibonacci(6n+4).

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A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

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A134504 a(n) = Fibonacci(7n + 6).

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A134494 a(n) = Fibonacci(6n+2).

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A167816 Numerator of x(n) = x(n-1) + x(n-2), x(0)=0, x(1)=1/3; denominator=A167817.

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A134501 a(n) = Fibonacci(7n + 3).

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