A134492 -id:A134492 - cp's OEIS frontend

A049660 a(n) = Fibonacci(6*n)/8.

0, 1, 18, 323, 5796, 104005, 1866294, 33489287, 600940872, 10783446409, 193501094490, 3472236254411, 62306751484908, 1118049290473933, 20062580477045886, 360008399296352015, 6460088606857290384, 115921586524134874897, 2080128468827570457762

Offset: 0

Clark Kimberling

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 18's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,17}. - Milan Janjic, Jan 25 2015
10*a(n)^2 = Tri(4)*S(n-1, 18)^2 is the triangular number Tri((T(n, 9) - 1)/2), with Tri, S and T given in A000217, A049310 and A053120. This is instance k = 4 of the k-family of identities given in a comment on A001109. - Wolfdieter Lang, Feb 01 2016
Possible solutions for y in Pell equation x^2 - 80*y^2 = 1. The values for x are given in A023039. - Herbert Kociemba, Jun 05 2022

			a(3) = F(6 * 3) / 8 = F(18) / 8 = 2584 / 8 = 323. - _Indranil Ghosh_, Feb 06 2017

Indranil Ghosh, Table of n, a(n) for n = 0..796 (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.
R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Column m=6 of array A028412.
Partial sums of A007805.
Cf. A000045, A001076, A001077, A014445, A014448, A015448, A099843, A134492, A155179.

[Fibonacci(6*n)/8: n in [0..30]]; // G. C. Greubel, Dec 02 2017

with (combinat):seq(fibonacci(2*n,4)/4, n=0..16); # Zerinvary Lajos, Apr 20 2008

Fibonacci[6*Range[0,20]]/8 (* Harvey P. Dale, Nov 23 2011 *)
LinearRecurrence[{18,-1}, {0,1}, 30] (* G. C. Greubel, Dec 02 2017 *)
Table[ChebyshevU[-1 + n, 9], {n, 0, 18}] (* Herbert Kociemba, Jun 05 2022 *)

numlib::fibonacci(6*n)/8 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(6*n)/8  \\ Charles R Greathouse IV, Apr 17 2012

[lucas_number1(n,18,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008

[fibonacci(6*n)/8 for n in range(0, 17)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1 - 18*x + x^2).
a(n) = A134492(n)/8.
a(n) ~ (1/40)*sqrt(5)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all terms k of the sequence, 80*k^2 + 1 is a square. Limit_{n->oo} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 14 2002
a(n) = S(n-1, 18) with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n))/(8*sqrt(5)).
a(n) = sqrt((A023039(n)^2 - 1)/80) (cf. Richardson comment).
a(n) = 18*a(n-1) - a(n-2). - Gregory V. Richardson, Oct 14 2002
a(n) = A001076(2n)/4.
a(n) = 17*(a(n-1) + a(n-2)) - a(n-3) = 19*(a(n-1) - a(n-2)) + a(n-3). - Mohamed Bouhamida, May 26 2007
a(n+1) = Sum_{k=0..n} A101950(n,k)*17^k. - Philippe Deléham, Feb 10 2012
Product_{n>=1} (1 + 1/a(n)) = (1/2)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = (2/9)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
a(n) = (1/32)*(F(6*n + 3) - F(6*n - 3)).
Sum_{n>=1} 1/(4*a(n) + 1/(4*a(n))) = 1/4. Compare with A001906 and A049670. - Peter Bala, Nov 29 2013
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -G(x)*G(-x), where G(x) = Sum_{n >= 1} A001076(n)*x^n.
1 + 4*Sum_{n >= 1} a(n)*x^(2*n) = (1 + F(x))*(1 + F(-x)) = (1 + 2*x*G(x))*(1 - 2*x*G(-x)), where F(x) = Sum_{n >= 1} Fibonacci(3*n + 3)*x^n.
1 + 7*Sum_{n >= 1} a(n)*x^(2*n) = (1 + G(x))*(1 + G(-x)) = (1 + 7*G(x))*(1 + 7*G(-x)).
1 + 12*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*G(x))*(1 + 2*G(-x)) = (1 + 6*G(x))*(1 + 6*G(-x)) = (1 + A(x))*(1 + A(-x)), where A(x) = Sum_{n >= 1} Fibonacci(3*n)*x^n is the o.g.f for A014445.
1 + 15*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 5*G(x))*(1 + 5*G(-x)) = (1 + 3*G(x))*(1 + 3*G(-x)) = H(x)*H(-x), where H(x) = Sum_{n >= 0} A155179(n)*x^n.
1 + 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 4*G(x))*(1 + 4*G(-x)) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*(-x)^n) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*(-x)^n).
1 + 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} Lucas(3*n)*x^n)*(1 + Sum_{n >= 1} Lucas(3*n)*(-x)^n).
1 - 5*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} A001077(n+1)*x^n)*(1 + Sum_{n >= 1} A001077(n+1)*(-x)^n).
1 - 9*Sum_{n >= 1} a(n)*x^(2*n) = (1 - G(x))*(1 - G(-x)) = (1 + 9*G(x))*(1 + 9*G(-x)).
1 - 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*Sum_{n >= 1} A099843(n)*x^n)*(1 + 2*Sum_{n >= 1} A099843(n)*(-x)^n).
1 - 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 - 2*G(x))*(1 - 2*G(-x)) = (1 + 10*G(x))*(1 + 10*G(-x)).
(End)

Chebyshev and other comments from Wolfdieter Lang, Nov 08 2002

A053606 a(n) = (Fibonacci(6*n+3) - 2)/4.

Original entry on oeis.org

0, 8, 152, 2736, 49104, 881144, 15811496, 283725792, 5091252768, 91358824040, 1639367579960, 29417257615248, 527871269494512, 9472265593285976, 169972909409653064, 3050040103780469184, 54730748958638792256, 982103441151717791432

Offset: 0

N. J. A. Sloane, James Sellers, Jan 20 2000

Define a(1)=0, a(2)=8 with 5*(a(1)^2) + 5*a(1) + 1 = j(1)^2 = 1^2 and 5*(a(2)^2) + 5*a(2) + 1 = j(2)^2 = 19^2. Then a(n) = a(n-2) + 8*sqrt(5*(a(n-1)^2) + 5*a(n-1)+1). Another definition: a(n) such that 5*(a(n)^2) + 5*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005
It appears this sequence gives all nonnegative m such that 5*m^2 + 5*m + 1 is a square. - Gerald McGarvey, Apr 03 2005
sqrt(5*a(n)^2+5*a(n)+1) = A049629(n). - Gerald McGarvey, Apr 19 2005
a(n) is such that 5*a(n)^2 + 5*a(n) + 1 = j^2 = the square of A049629(n). Also A049629(n)/a(n) tends to sqrt(5) as n increases. - Pierre CAMI, Apr 21 2005
From Russell Jay Hendel, Apr 25 2015: (Start)
We prove the two McGarvey-CAMI conjectures mentioned at the beginning of the Comments section. Let, as usual, F(n) = A000045(n), the Fibonacci numbers. In the sequel we indicate equations with upper case letters ((A), (B), (C), (D)) for ease of reference.
Then we must prove (A), 5*((F(6*n+3)-2)/4)^2 + 5*((F(6*n+3)-2)/4) + 1 = ((F(6*n+5)-F(6*n+1))/4)^2. Let m = 3*n+1 so that 6*n+1, 6*n+3, and 6*n+5 are 2*m-1, 2*m+1, and 2*m+3 respectively. Define G(m) = F(6*n+3) = F(2*m+1) = A001519(m+1), the bisected Fibonacci numbers. We can now simplify equation (A) by i) multiplying the LHS and RHS by 16, ii) expanding squares, and iii) gathering like terms. This shows proof of (A) equivalent to proving (B), 5*G(m)^2-4 = (G(m+1)-G(m-1))^2.
By Jarden's theorem (D. Jarden, Recurring sequences, 2nd ed. Jerusalem, Riveon Lematematika, (1966)), if {H(n)}{n >=1} is any recursive sequence satisfying (C), H(n)=3H(n-1)-H(n-2), then {H(n)}^2{n >=1} is also a recursive sequence satisfying (D), H(n)^2=8H(n-1)^2-8H(n-2)^2+H(n-3)^2. As noted in the Formula section of A001519, {G(m)}_{m >= 1} satisfies (C).
Proof of (B) is now straightforward. Since {G(m)}{m >=1} satisfies (C), it follows that {G(m)^2}{m >=1} satisfies (D), and therefore, {5G(m)^2-4}_{m >=1} also satisfies (D).
Similarly, since {G(m)}{m >=1} satisfies (C), it follows that both {G(m+1)}{m >=1}, {G(m-1)}{m >=1} and their difference {G(m+1)-G(m-1)}{m >=1} satisfy (C), and therefore {G(m+1)-G(m-1)}^2_{m >=1} satisfies (D).
But then the LHS and RHS of (B) are equal for m=1,2,3 and satisfy the same recursion, (D). Hence the LHS and RHS of (B) are equal for all m. This completes the proof. (End)

G. C. Greubel, Table of n, a(n) for n = 0..790
F. Ellermann, Illustration of binomial transforms
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A049629.

Related to sum of Fibonacci(kn) over n.. A000071, A027941, A099919, A058038, A138134.

List([0..30], n-> (Fibonacci(6*n+3)-2)/4); # G. C. Greubel, May 16 2019

[(Fibonacci(6*n+3)-2)/4: n in [0..30]]; // Vincenzo Librandi, Apr 20 2011

A053606 := proc(n) add(combinat[fibonacci](6*k),k=0..n) ;end proc:
seq(A053606(n),n=0..30) ;

Table[(Fibonacci[6n+3] -2)/4, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)

a(n)=fibonacci(6*n+3)\4 \\ Charles R Greathouse IV, Jul 02 2013

[(fibonacci(6*n+3)-2)/4 for n in (0..30)] # G. C. Greubel, May 16 2019

a(n) = 8*A049664(n).
a(n+1) = 9*a(n) + 2*sqrt(5*(2*a(n)+1)^2-1) + 4. - Richard Choulet, Aug 30 2007
G.f.: 8*x/((1-x)*(1-18*x+x^2)). - Richard Choulet, Oct 09 2007
a(n) = 18*a(n-1) - a(n-2) + 8, n > 1. - Gary Detlefs, Dec 07 2010
a(n) = Sum_{k=0..n} A134492(k). - Gary Detlefs, Dec 07 2010
a(n) = (Fibonacci(6*n+6) - Fibonacci(6*n) - 8)/16. - Gary Detlefs, Dec 08 2010

A079343 Period 6: repeat [0, 1, 1, 2, 3, 1]; also F(n) mod 4, where F(n) = A000045(n).

Original entry on oeis.org

0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1

Offset: 0

Jon Perry, Jan 04 2003

This sequence shows that every sixth Fibonacci number (A134492) is divisible by 4. - Alonso del Arte, Jul 27 2013

			a(5) = F(5) mod 4 = 5 mod 4 = 1.
a(6) = F(6) mod 4 = 8 mod 4 = 0.
a(7) = F(7) mod 4 = 13 mod 4 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jon Maiga, Upper bound of Fibonacci entry points, (2019).
Index entries for linear recurrences with constant coefficients, signature (1,-1,1,-1,1).

Cf. A000045, A079344, A079345, A134492.

[Fibonacci(n) mod 4: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

A079343:=n->[0, 1, 1, 2, 3, 1][(n mod 6)+1]: seq(A079343(n), n=0..100); # Wesley Ivan Hurt, Jun 20 2016

PadLeft[{}, 108, {0, 1, 1, 2, 3, 1}] (* Harvey P. Dale, Aug 10 2011 *)
Table[Mod[Fibonacci[n], 4], {n, 0, 127}] (* Alonso del Arte, Jul 27 2013 *)
LinearRecurrence[{1, -1, 1, -1, 1},{0, 1, 1, 2, 3},105] (* Ray Chandler, Aug 27 2015 *)

for (n=0,100,print1(fibonacci(n)%4","))

a(n) = 2^(1 - P(3, n) + P(6, n+2))*3^P(6, n+3) - 1, where P(k, n) = floor(1/2*cos(2*n*Pi/k) + 1/2). [Gary Detlefs, May 16 2011]
a(n) = 4/3 - cos(Pi*n/3) - sin(Pi*n/3)/sqrt(3) - cos(2*Pi*n/3)/3 + sin(2*Pi*n/3)/sqrt(3). - R. J. Mathar, Oct 08 2011
G.f.: x*(1+2*x^2+x^3) / ( (1-x)*(1-x+x^2)*(1+x+x^2) ). - R. J. Mathar, Jul 14 2012
a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + a(n-5) for n>4. - Wesley Ivan Hurt, Jun 20 2016
E.g.f.: 2*(2*exp(x) - sqrt(3)*sin(sqrt(3)*x/2)*sinh(x/2) - cos(sqrt(3)*x/2)*(sinh(x/2) + 2*cosh(x/2)))/3. - Ilya Gutkovskiy, Jun 20 2016

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

Original entry on oeis.org

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

A134497 a(n) = Fibonacci(6n+5).

Original entry on oeis.org

5, 89, 1597, 28657, 514229, 9227465, 165580141, 2971215073, 53316291173, 956722026041, 17167680177565, 308061521170129, 5527939700884757, 99194853094755497, 1779979416004714189, 31940434634990099905, 573147844013817084101, 10284720757613717413913

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, A049660, A101858, A103134, A134492, A134493, A134494, A134495.

[Fibonacci(6*n +5): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[6n+5], {n, 0, 30}]
Take[Fibonacci[Range[100]],{5,-1,6}] (* Harvey P. Dale, Jun 18 2013 *)

a(n)=fibonacci(6*n+5) \\ Charles R Greathouse IV, Jun 11 2015

Vec((5-x)/(1-18*x+x^2) + O(x^100)) \\ Altug Alkan, Jan 24 2016

G.f.: ( 5-x ) / ( 1-18*x+x^2 ). a(n) = 5*A049660(n+1)-A049660(n). - R. J. Mathar, Apr 17 2011
a(n) = A000045(A016969(n)). - Michel Marcus, Nov 08 2013
a(n) = ((25-11*sqrt(5)+(9+4*sqrt(5))^(2*n)*(25+11*sqrt(5))))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
a(n) = 5*S(n, 18) - S(n-1, 18), n >= 0, with the Chebyshev S-polynomials S(n-1, 18) = A049660(n). (See the g.f.) - Wolfdieter Lang, Jul 10 2018
From Peter Bala, Aug 11 2022: (Start)
Let n ** m =  n*m + floor(phi*n)*floor(phi*m), where phi = (1 + sqrt(5))/2, denote the Porta-Stolarsky star product of the integers n and m (see A101858). Then a(n) = 5 ** 5 ** ... ** 5 (n+1 factors).
a(2*n+1) = a(n) ** a(n) = Fibonacci(12*n+11); a(3*n+2) = a(n) ** a(n) ** a(n) = Fibonacci(18*n+17) and so on. (End)

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

A134493 a(n) = Fibonacci(6*n+1).

Original entry on oeis.org

1, 13, 233, 4181, 75025, 1346269, 24157817, 433494437, 7778742049, 139583862445, 2504730781961, 44945570212853, 806515533049393, 14472334024676221, 259695496911122585, 4660046610375530309, 83621143489848422977, 1500520536206896083277, 26925748508234281076009

Offset: 0

Artur Jasinski, Oct 28 2007

For positive n, a(n) equals (-1)^n times the permanent of the (6n)X(6n) tridiagonal matrix with ((-1)^(1/6))'s along the three central diagonals. - John M. Campbell, Jul 12 2011

a(n) = x + y where those two values 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
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000032, A000045, A049660, A134492, A134494, A134495, A103134, A134497.

[Fibonacci(6*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Mathematica
```
Table[Fibonacci[6n+1], {n, 0, 30}]
```

a(n)=fibonacci(6*n+1) \\ Charles R Greathouse IV, Jul 15 2011

Vec((1-5*x)/(1-18*x+x^2) + O(x^100)) \\ Altug Alkan, Jan 24 2016

From R. J. Mathar, Apr 17 2011: (Start)
G.f.: ( 1-5*x ) / ( 1-18*x+x^2 ).
a(n) = A049660(n+1) - 5*A049660(n). (End)
a(n) = Fibonacci(3*n+1)^2 + Fibonacci(3*n)^2. - Gary Detlefs, Oct 12 2011
a(n) = 18*a(n-1) - a(n-2). - Richard R. Forberg, Sep 05 2013
a(n) = A060645(n) + A023039(n), as derives from comment above. - Richard R. Forberg, Sep 05 2013
a(n) = ((5-sqrt(5)+(5+sqrt(5))*(9+4*sqrt(5))^(2*n)))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
2*a(n) = Fibonacci(6*n) + Lucas(6*n). - Bruno Berselli, Oct 13 2017
a(n) = S(n, 18) - 5*S(n-1, 18), n >= 0, with the Chebyshev S-polynomials S(n-1, 18) = A049660(n). (See the g.f.) - Wolfdieter Lang, Jul 10 2018

Offset changed to 0 by Vincenzo Librandi, Apr 16 2011

A134495 a(n) = Fibonacci(6n + 3).

Original entry on oeis.org

2, 34, 610, 10946, 196418, 3524578, 63245986, 1134903170, 20365011074, 365435296162, 6557470319842, 117669030460994, 2111485077978050, 37889062373143906, 679891637638612258, 12200160415121876738

Offset: 0

Artur Jasinski, Oct 28 2007

From Tanya Khovanova, Jan 06 2023: (Start)
Fibonacci(6n+3) are divisible by 2 but not by 4.
These numbers are not divisible by 3. (End)

Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, 2*A007805, A049660, A134492, A134493, A134494, A103134.

Subsequence of A091999.

[Fibonacci(6*n +3): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[6n+3], {n, 0, 30}]
LinearRecurrence[{18,-1},{2,34},20] (* Harvey P. Dale, Jul 28 2018 *)

a(n)=fibonacci(6*n+3) \\ Charles R Greathouse IV, Jun 11 2015

From R. J. Mathar, Apr 17 2011: (Start)
G.f.: (2-2*x) / (1 - 18*x + x^2).
a(n) = 2*A007805(n). (End)
a(n) = A000045(A016945(n)). - Michel Marcus, Nov 08 2013
a(n) = 2*(S(n, 18) - S(n-1, 18)), n >= 0, with the Chebyshev S-polynomials S(n-1, 18) = A049660(n). (See the g.f.) - Wolfdieter Lang, Jul 10 2018

Index in definition and offset corrected by R. J. Mathar, Apr 17 2011

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

A037917 Numbers n such that the Fibonacci number F(n) is divisible by a square.

Original entry on oeis.org

6, 12, 18, 24, 25, 30, 36, 42, 48, 50, 54, 56, 60, 66, 72, 75, 78, 84, 90, 91, 96, 100, 102, 108, 110, 112, 114, 120, 125, 126, 132, 138, 144, 150, 153, 156, 162, 168, 174, 175, 180, 182, 186, 192, 198, 200, 204, 210, 216, 220, 222, 224, 225, 228, 234, 240

Offset: 1

Marc LeBrun

nonn

Is a(n) asymptotic to C*n with 4 < C < 4.5 ? - Benoit Cloitre, Sep 04 2002
Numbers are a superset of the multiples of 6 (A008588), because 8 divides Fibonacci(6m) = A134492(m). Sequence apparently also contains the multiples of 25. Are all a(n) composite? Members not divisible by 6 or 25 are 56, 91, 110, 112, 153, 182, 220, 224, 273, 280, ... - Ralf Stephan, Jan 26 2014
These numbers are the positive multiples of A065069. - Charles R Greathouse IV, Feb 02 2014
To address Cloitre's question, if such C exists it must be less than 4.3 using the known terms of A065069. - Charles R Greathouse IV, Feb 04 2014

Amiram Eldar, Table of n, a(n) for n = 1..328 (terms 1..235 from T. D. Noe, based on Kelly's table)
Blair Kelly, Fibonacci Factorizations
Eric Weisstein's World of Mathematics, Fibonacci Number.

Cf. A000045, A037918, A001177, A030426.

Select[Range[100], MoebiusMu[Fibonacci[#]] == 0 &] (* Alonso del Arte, Jan 26 2014 *)

is(n)=!issquarefree(fibonacci(n)) \\ Charles R Greathouse IV, Feb 02 2014

More terms from Eric W. Weisstein

A280592 Numbers n such that phi(n) is a Fibonacci number.

Original entry on oeis.org

1, 2, 3, 4, 6, 15, 16, 20, 24, 30, 185, 219, 273, 285, 292, 296, 304, 315, 364, 370, 380, 432, 438, 444, 456, 468, 504, 540, 546, 570, 630, 3235, 5176, 6470, 7764, 46843, 47423, 47693, 48053, 50431, 52403, 56231, 57965, 59555, 62855, 67655, 67865, 70735, 72123, 72297, 73473

Offset: 1

Altug Alkan, Jan 06 2017

Note that the sequence of corresponding Fibonacci numbers is not the same as A134492. See also A280681.

			24 is a term because phi(24) = 8 is a Fibonacci number.

Robert Israel, Table of n, a(n) for n = 1..2075

Cf. A000010, A000045, A134492, A280681.

fibs:= {seq(combinat:-fibonacci(n),n=1..30)}:
N:= combinat:-fibonacci(31):
sort(convert(select(t -> t <= N, map(t -> op(numtheory:-invphi(t)), fibs)), list)); # Robert Israel, Nov 20 2019

Module[{fibs=Fibonacci[Range[30]]},Select[Range[80000],MemberQ[ fibs,EulerPhi[ #]]&]] (* Harvey P. Dale, Jul 04 2021 *)

isFibonacci(n)=my(k=n^2); issquare(k+=(k+1)<<2) || (n>0 && issquare(k-8));
is(n)=isFibonacci(eulerphi(n));

cp's OEIS Frontend

A049660 a(n) = Fibonacci(6*n)/8.

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A053606 a(n) = (Fibonacci(6*n+3) - 2)/4.

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A079343 Period 6: repeat [0, 1, 1, 2, 3, 1]; also F(n) mod 4, where F(n) = A000045(n).

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

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A134497 a(n) = Fibonacci(6n+5).

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A134493 a(n) = Fibonacci(6*n+1).

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