A167816 -id:A167816 - cp's OEIS frontend

A004187 a(n) = 7*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

0, 1, 7, 48, 329, 2255, 15456, 105937, 726103, 4976784, 34111385, 233802911, 1602508992, 10983760033, 75283811239, 516002918640, 3536736619241, 24241153416047, 166151337293088, 1138818207635569, 7805576116155895, 53500214605455696, 366695926122033977

Offset: 0

N. J. A. Sloane, R. K. Guy

Define the sequence T(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0 . A004187 (with initial 0 omitted) is T(1,7).
This is a divisibility sequence.
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 7'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
a(n) and b(n) := A056854(n) are the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 5*(3*a(n))^2 = +4. see the cross-reference to A056854 below. - Wolfdieter Lang, Jun 26 2013
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4,5,6}. - Milan Janjic, Jan 25 2015
The digital root is A253298, which shares its digital root with A253368. - Peter M. Chema, Jul 04 2016
Lim_{n->oo} a(n+1)/a(n) = 2 + 3*phi = 1+ A090550 = 6.854101...  - Wolfdieter Lang, Nov 16 2023

			a(2) = 7*a(1) - a(0) = 7*7 - 1 = 48. - _Michael B. Porter_, Jul 04 2016

Vincenzo Librandi, 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.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3 , example 12
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993
Zvonko Cerin, Some alternating sums of Lucas numbers, Centr. Eur. J. Math. vol 3 no 1 (2005) 1-13.
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).
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=7, q=-1.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
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), lhs, m=9.
Index entries for sequences related to Chebyshev polynomials.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).
Index entries for Pisot sequences

Cf. A000027, A001906, A001353, A004254, A001109, A049685, A033888. a(n)=sqrt((A056854(n)^2 - 4)/45).

Second column of array A028412.

[Fibonacci(4*n)/3 : n in [0..30]]; // Vincenzo Librandi, Jun 07 2011

/* By definition: */ [n le 2 select n-1 else 7*Self(n-1)-Self(n-2): n in [1..23]]; // Bruno Berselli, Dec 24 2012

seq(combinat:-fibonacci(4*n)/3, n = 0 .. 30); # Robert Israel, Jan 26 2015

LinearRecurrence[{7,-1},{0,1},30] (* Harvey P. Dale, Jul 13 2011 *)
CoefficientList[Series[x/(1 - 7*x + x^2), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 23 2012 *)

a[0]:0$ a[1]:1$ a[n]:=7*a[n-1] - a[n-2]$ A004187(n):=a[n]$ makelist(A004187(n),n,0,30); /* Martin Ettl, Nov 11 2012 */

numlib::fibonacci(4*n)/3 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(4*n)/3 \\ Charles R Greathouse IV, Mar 09 2012

concat(0, Vec(x/(1-7*x+x^2) + O(x^99))) \\ Altug Alkan, Jul 03 2016

[lucas_number1(n,7,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

[fibonacci(4*n)/3 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1-7*x+x^2).
a(n) = F(4*n)/3 = A033888(n)/3, where F=A000045 (the Fibonacci sequence).
a(n) = S(2*n-1, sqrt(9))/sqrt(9) = S(n-1, 7); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = Sum_{i = 0..n-1} C(2*n-1-i, i)*5^(n-i-1). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
[A049685(n-1), a(n)] = [1,5; 1,6]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167816(4*n). - Reinhard Zumkeller, Nov 13 2009
a(n) = (((7+sqrt(45))/2)^n-((7-sqrt(45))/2)^n)/sqrt(45). - Noureddine Chair, Aug 31 2011
a(n+1) = Sum_{k = 0..n} A101950(n,k)*6^k. - Philippe Deléham, Feb 10 2012
a(n) = (A081072(n)/3)-1. - Martin Ettl, Nov 11 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = (1/5)*(5 + 3*sqrt(5)).
Product {n >= 2} (1 - 1/a(n)) = (1/14)*(5 + 3*sqrt(5)). (End)
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -A(x)*A(-x), where A(x) = Sum_{n >= 1} Fibonacci(2*n)* x^n.
1 + 5*Sum_{n >= 1} a(n)*x^(2*n) = F(x)*F(-x) = G(x)*G(-x), where F(x) = 1 + A(x) and G(x) = 1 + 5*A(x).
1 + Sum_{n >= 1} a(n)*x^(2*n) = H(x)*H(-x) = I(x)*I(-x), where H(x) = 1 + Sum_{n >= 1} Fibonacci(2*n + 3)*x^n and I(x) = 1 + x + x*Sum_{n >= 1} Fibonacci(2*n - 1)*x^n. (End)
E.g.f.: 2*exp(7*x/2)*sinh(3*sqrt(5)*x/2)/(3*sqrt(5)). - Ilya Gutkovskiy, Jul 03 2016
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*9^k*binomial(n+k, 2*k+1). - Peter Bala, Jul 17 2023
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*7^(n-2*k)*binomial(n-k, k). - Greg Dresden, Aug 03 2024
From Peter Bala, Jul 22 2025: (Start)
The following products telescope:
Product {n >= 2} (1 + (-1)^n/a(n)) = (3/14)*(3 + sqrt(5)).
Product {n >= 1} (1 - (-1)^n/a(n)) = (1/3)*(3 + sqrt(5)).
Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = (3/5)*sqrt(5). (End)

Entry improved by comments from Michael Somos and Wolfdieter Lang, Aug 02 2000

A033890 a(n) = Fibonacci(4*n + 2).

Original entry on oeis.org

1, 8, 55, 377, 2584, 17711, 121393, 832040, 5702887, 39088169, 267914296, 1836311903, 12586269025, 86267571272, 591286729879, 4052739537881, 27777890035288, 190392490709135, 1304969544928657, 8944394323791464, 61305790721611591, 420196140727489673

Offset: 0

N. J. A. Sloane

(x,y) = (a(n), a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033888. - Floor van Lamoen, Dec 10 2001
This sequence consists of the odd-indexed terms of A001906 (whose terms are the values of x such that 5*x^2 + 4 is a square). The even-indexed terms of A001906 are in A033888. Limit_{n->infinity} a(n)/a(n-1) = phi^4 = (7 + 3*sqrt(5))/2. - Gregory V. Richardson, Oct 13 2002
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
Indices of square numbers which are also 12-gonal. - Sture Sjöstedt, Jun 01 2009
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with 3'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
If we let b(0) = 0 and, for n >= 1, b(n) = A033890(n-1), then the sequence b(n) will be F(4n-2) and the first difference is L(4n) or A056854. F(4n-2) is also the ratio of golden spiral length (rounded to the nearest integer) after n rotations. L(4n) is also the pitch length ratio. See illustration in links. - Kival Ngaokrajang, Nov 03 2013
The aerated sequence (b(n))n>=1 = [1, 0, 8, 0, 55, 0, 377, 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 = -5, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
Solutions y of Pell equation x^2 - 5*y^2 = 4; corresponding x values are in A342710 (see A342709). - Bernard Schott, Mar 19 2021

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.
Nathan D. Cahill and Darren A. Narayan, Fibonacci and Lucas Numbers as Tridiagonal Matrix Determinants, Fib. Quart. 42, no. 3, Aug. 2004, pp. 216-221. See p. 219.
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
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
Kival Ngaokrajang, Illustration of golden spiral length and pitch ratio
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 (7,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001906, A100047.

Cf. A342709, A342710.

[Fibonacci(4*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

A033890 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,8]);
    else
        7*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

Table[Fibonacci[4n + 2], {n, 0, 14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
LinearRecurrence[{7, -1}, {1, 8}, 50] (* G. C. Greubel, Jul 13 2017 *)
a[n_] := (GoldenRatio^(2 (1 + 2 n)) - GoldenRatio^(-2 (1 + 2 n)))/Sqrt[5]
Table[a[n] // FullSimplify, {n, 0, 21}] (* Gerry Martens, Aug 20 2025 *)

PARI
```
a(n)=fibonacci(4*n+2);
```

G.f.: (1+x)/(1-7*x+x^2).
a(n) = 7*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=8.
a(n) = S(n,7) + S(n-1,7) = S(2*n,sqrt(9) = 3), where S(n,x) = U(n,x/2) are Chebyshev's polynomials of the 2nd kind. Cf. A049310. S(n,7) = A004187(n+1), S(n,3) = A001906(n+1).
a(n) = ((7+3*sqrt(5))^n - (7-3*sqrt(5))^n + 2*((7+3*sqrt(5))^(n-1) - ((7-3*sqrt(5))^(n-1)))) / (3*(2^n)*sqrt(5)). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = (-1)^n*q(n, -9). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-7)*(-1)^n, where L is defined as in A108299; see also A049685 for L(n,+7). - Reinhard Zumkeller, Jun 01 2005
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),7/2) + f(a(n-2),7/2). - Marcos Carreira, Dec 27 2006
a(n+1) = 8*a(n) - 8*a(n-1) + a(n-2); a(1)=1, a(2)=8, a(3)=55. - Sture Sjöstedt, May 27 2009
a(n) = A167816(4*n+2). - Reinhard Zumkeller, Nov 13 2009
a(n)=b such that (-1)^n*Integral_{0..Pi/2} (cos((2*n+1)*x))/(3/2-sin(x)) dx = c + b*log(3). - Francesco Daddi, Aug 01 2011
a(n) = A000045(A016825(n)). - Michel Marcus, Mar 22 2015
a(n) = A001906(2*n+1). - R. J. Mathar, Apr 30 2017
E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + 3*sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Apr 14 2025
From Peter Bala, Jun 08 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/9 [telescoping series: 3/(a(n) - 1/a(n)) = 1/Fibonacci(4*n+4) + 1/Fibonacci(4*n)].
Product_{n >= 1} (a(n) + 3)/(a(n) - 3) = 5/2 [telescoping product:
(a(n) + 3)/(a(n) - 3) = b(n)/b(n-1), where b(n) = (Lucas(4*n+4) - 3)/(Lucas(4*n+4) + 3)].
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(9/5) [telescoping product:
(a(n) + 1)/(a(n) - 1) = c(n)/c(n-1) for n >= 1, where c(n) = Fibonacci(2*n+2)/Lucas(2*n+2)]. (End)
From Gerry Martens, Aug 20 2025: (Start)
a(n) = ((3 + sqrt(5))^(1 + 2*n) - (3 - sqrt(5))^(1 + 2*n)) / (2^(1 + 2*n)*sqrt(5)).
a(n) = Sum_{k=0..2*n} binomial(2*n + k + 1, 2*k + 1). (End)

A033891 a(n) = Fibonacci(4*n+3).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

G. C. Greubel, 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. A000045, A004187, A049685, A101858, A167816.

List([0..30], n-> Fibonacci(4*n+3)); # G. C. Greubel, Jul 14 2019

[Fibonacci(4*n+3): n in [0..30]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[4*n+3],{n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
LinearRecurrence[{7,-1},{2,13},31] (* or *) CoefficientList[Series[ (2-x)/(1-7x+x^2),{x,0,30}],x] (* Harvey P. Dale, May 03 2011 *)

a(n)=fibonacci(4*n+3) \\ Charles R Greathouse IV, Sep 24 2015

[fibonacci(4*n+3) for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = 7*a(n-1) - a(n-2). - Floor van Lamoen, Dec 10 2001
G.f.: (2-x)/(1-7*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = A167816(4*n+3). - Reinhard Zumkeller, Nov 13 2009
a(n) = Fibonacci(2*n+2)^2 + Fibonacci(2*n+1)^2. - Gary Detlefs, Oct 12 2011
a(n) = 2*A004187(n+1) - A004187(n). - R. J. Mathar, Nov 26 2011
a(n) = A004187(n+1) + A049685(n). - Yuriy Sibirmovsky, Sep 15 2016
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) = 2 ** 2 ** ... ** 2 (n+1 factors).
a(2*n+1) = a(n) ** a(n) = Fibonacci(8*n+7); a(3*n+2) = a(n) ** a(n) ** a(n) = Fibonacci(12*n+11) and so on. (End)

A033889 a(n) = Fibonacci(4*n + 1).

Original entry on oeis.org

1, 5, 34, 233, 1597, 10946, 75025, 514229, 3524578, 24157817, 165580141, 1134903170, 7778742049, 53316291173, 365435296162, 2504730781961, 17167680177565, 117669030460994, 806515533049393, 5527939700884757, 37889062373143906, 259695496911122585, 1779979416004714189

Offset: 0

N. J. A. Sloane

For positive n, a(n) equals (-1)^n times the permanent of the (4n) X (4n) tridiagonal matrix with sqrt(i)'s along the three central diagonals, where i is the imaginary unit. - John M. Campbell, Jul 12 2011

a(n) = 5^n*a(n; 3/5) = (16/5)^n*a(2*n; 3/4), and F(4*n) = 5^n*b(n; 3/5) = (16/5)^n*b(2*n; 3/4), where a(n; d) and b(n; d), n=0, 1, ..., d in C, denote the delta-Fibonacci numbers defined in comments to A014445. Two of these identities from the following relations follows: F(k+1)^n*a(n; F(k)/F(k+1)) = F(k*n+1) and F(k+1)^n*b(n; F(k)/F(k+1)) = F(k*n) (see also Witula's et al. papers). - Roman Witula, Jul 24 2012

Bruno Berselli, Table of n, a(n) for n = 0..500
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
Tanya Khovanova, Recursive Sequences.
Kai Wan, Problem H-90, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 3 (2022), p. 282; Solution to Problem H-90, by Ángel Plaza, ibid., Vol. 62, No. 1 (2024), pp. 95-96.
Roman Wituła, Binomials Transformation Formulae of Scaled Lucas Numbers, Demonstratio Mathematica, Vol. XLVI, No. 1 (2013), pp. 15-27.
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009), 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A014445, A016813, A122070, A167816, A094567.

Cf. A003881, A049684, A081018, A081016, A081068, A172968.

[Fibonacci(4*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Table[Fibonacci[4*n+1], {n,0,14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)

a(n)=fibonacci(4*n+1) \\ Charles R Greathouse IV, Jul 15 2011

Vec((1-2*x)/(1-7*x+x^2) + O(x^100)) \\ Altug Alkan, Dec 10 2015

a(n) = 7*a(n-1) - a(n-2) for n >= 2. - Floor van Lamoen, Dec 10 2001
From R. J. Mathar, Jan 17 2008: (Start)
O.g.f.: (1 - 2*x)/(1 - 7*x + x^2).
a(n) = A004187(n+1) - 2*A004187(n). (End); corrected by Klaus Purath, Jul 29 2020
a(n) = A167816(4*n+1). - Reinhard Zumkeller, Nov 13 2009
a(n) = sqrt(1 + 2 * Fibonacci(2*n) * Fibonacci(2*n + 1) + 5 * (Fibonacci(2*n) * Fibonacci(2*n + 1))^2). - Artur Jasinski, Feb 06 2010
a(n) = Sum_{k=0..n} A122070(n,k)*2^k. - Philippe Deléham, Mar 13 2012
a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n)*Fibonacci(2*n+2) + 1. - Gary Detlefs, Apr 18 2012
a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n+1)^2. - Bruno Berselli, Apr 19 2012
a(n) = Sum_{k = 0..n} A238731(n,k)*4^k. - Philippe Deléham, Mar 05 2014
a(n) = A000045(A016813(n)). - Michel Marcus, Mar 05 2014
2*a(n) = Fibonacci(4*n) + Lucas(4*n). - Bruno Berselli, Oct 13 2017
a(n) = A094567(n-1) + A094567(n), assuming A094567(-1) = 0. - Klaus Purath, Jul 29 2020
Sum_{n>=0} (-1)^n * arctan(3/a(n)) = Pi/4 (A003881) (Wan, 2022). - Amiram Eldar, Mar 01 2024
E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

A093148 a(n) = gcd(Fibonacci(n+5), Fibonacci(n+1)).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1

Offset: 0

Paul Barry, Apr 02 2004

From Klaus Brockhaus, May 30 2010: (Start)
Periodic sequence: Repeat [1, 1, 1, 3].
Continued fraction expansion of (9+sqrt(165))/14.
Decimal expansion of 371/3333. (End)
Final nonzero digit of n^n in base 4. - José María Grau Ribas, Jan 19 2012

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Cf. A000045, A061347, A167816, A177704, A178591.

[1+2*0^((n+1) mod 4) : n in [0..100]]; // Wesley Ivan Hurt, Oct 23 2014

A093148:=n->1+2*0^(n+1 mod 4): seq(A093148(n), n=0..100); # Wesley Ivan Hurt, Oct 23 2014

f[n_] := Switch[Mod[n, 4], 0, 1, 1, 1, 2, 1, 3, 3]; Array[f, 105, 0] (* Robert G. Wilson v, Jan 23 2012 *)
PadRight[{},120,{1,1,1,3}] (* Harvey P. Dale, Sep 03 2021 *)

a(n)=if(n%4==3,3,1) \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (1+x+x^2+3*x^3)/(1-x^4); a(n) = 3/2-sin(Pi*n/2)-cos(Pi*n)/2.
From Klaus Brockhaus, May 30 2010: (Start)
a(n) = a(n-4) for n > 3; a(0) = a(1) = a(2) = 1, a(3) = 3.
a(n) = (3-(-1)^n+(1-(-1)^n)*i*i^n)/2 where i = sqrt(-1). (End)
a(n) = 1 + 2*0^mod(n+1, 4). - Wesley Ivan Hurt, Oct 23 2014

A167817 Period 4: repeat [1, 3, 3, 3].

Original entry on oeis.org

1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1

Offset: 0

Reinhard Zumkeller, Nov 13 2009

Denominator of x(n) = x(n-1) + x(n-2), x(0)=0, x(1)=1/3; numerator = A167816(n).

Continued fraction expansion of (33 + sqrt(2805))/66. - Klaus Brockhaus, May 06 2010

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Cf. A130196.

Cf. A177344 (decimal expansion of (33+sqrt(2805))/66). - Klaus Brockhaus, May 06 2010

&cat[[1, 3, 3, 3]: n in [0..50]]; // Vincenzo Librandi, Dec 28 2010

seq(op([1, 3, 3, 3]), n=0..50); # Wesley Ivan Hurt, Jul 09 2016

Denominator[LinearRecurrence[{1,1},{0,1/3},110]] (* or *) PadRight[{},110,{1,3,3,3}] (* Harvey P. Dale, Dec 07 2014 *)
LinearRecurrence[{0, 0, 0, 1},{1, 3, 3, 3},105] (* Ray Chandler, Aug 03 2015 *)

a(n) = 3 - 2 * 0^(n mod 4).
G.f.: (1 + 3*x + 3*x^2 + 3*x^3)/(1-x^4). - Klaus Brockhaus, May 06 2010
a(n) = 5/2 - cos(Pi*n/2) - (-1)^n/2. - R. J. Mathar, Oct 08 2011
E.g.f.: -cos(x) + 3*sinh(x) + 2*cosh(x). - Ilya Gutkovskiy, Jun 27 2016
a(n) = a(n-4) for n>3. - Wesley Ivan Hurt, Jul 09 2016

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

cp's OEIS Frontend

A004187 a(n) = 7*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

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A033890 a(n) = Fibonacci(4*n + 2).

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A033891 a(n) = Fibonacci(4*n+3).

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A033889 a(n) = Fibonacci(4*n + 1).

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A093148 a(n) = gcd(Fibonacci(n+5), Fibonacci(n+1)).

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A167817 Period 4: repeat [1, 3, 3, 3].

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