A000301 -id:A000301 - cp's OEIS frontend

A061084 Fibonacci-type sequence based on subtraction: a(0) = 1, a(1) = 2 and a(n) = a(n-2) - a(n-1).

1, 2, -1, 3, -4, 7, -11, 18, -29, 47, -76, 123, -199, 322, -521, 843, -1364, 2207, -3571, 5778, -9349, 15127, -24476, 39603, -64079, 103682, -167761, 271443, -439204, 710647, -1149851, 1860498, -3010349, 4870847, -7881196, 12752043, -20633239, 33385282, -54018521

Offset: 0

Ulrich Schimke (ulrschimke(AT)aol.com)

If we drop 1 and start with 2 this is the Lucas sequence V(-1,-1). G.f.: (2+x)/(1+x-x^2). In this case a(n) is also the trace of A^(-n), where A is the Fibomatrix ((1,1), (1,0)). - Mario Catalani (mario.catalani(AT)unito.it), Aug 17 2002
The positive sequence with g.f. (1+x-2*x^2)/(1-x-x^2) gives the diagonal sums of the Riordan array (1+2*x, x/(1-x)). - Paul Barry, Jul 18 2005
Pisano period lengths:  1, 3, 8, 6, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, .... (is this A106291?). - R. J. Mathar, Aug 10 2012

			a(6) = a(4)-a(5) = -4 - 7 = -11.

Indranil Ghosh, Table of n, a(n) for n = 0..4771 (terms 0..500 from T. D. Noe)
Tanya Khovanova, Recursive Sequences
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Q. Wang, On generalized Lucas sequences, Contemp. Math. 531 (2010) 127-141, Table 2 (k=2).
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (-1,1).
Index entries for Lucas sequences

Cf. A061083 for division, A000301 for multiplication and A000045 for addition - the common Fibonacci numbers.

a061084 n = a061084_list !! n
a061084_list = 1 : 2 : zipWith (-) a061084_list (tail a061084_list)
-- Reinhard Zumkeller, Feb 01 2014

A061084:= func< n | Lucas(1-n) >; // G. C. Greubel, Jun 14 2025

LinearRecurrence[{-1,1},{1,2},40] (* Harvey P. Dale, Nov 22 2011 *)
LucasL[1-Range[0, 40]] (* G. C. Greubel, Jun 14 2025 *)

a(n)=([0,1; 1,-1]^n*[1;2])[1,1] \\ Charles R Greathouse IV, Feb 09 2017

def A061084(n): return lucas_number2(1-n,1,-1) # G. C. Greubel, Jun 14 2025

a(n) = (-1)^(n-1) * A000204(n-1).
O.g.f.: (1+3*x)/(1+x-x^2). - Len Smiley, Dec 02 2001
a(n) = A039834(n+1) + 3*A039834(n). - R. J. Mathar, Oct 30 2015
From G. C. Greubel, Jun 14 2025: (Start)
a(n) = A000032(1-n).
E.g.f.: exp(-x/2)*( cosh(p*x) + sqrt(5)*sinh(p*x) ), where p = sqrt(5)/2. (End)

Corrected by T. D. Noe, Oct 25 2006

A014565 Decimal expansion of rabbit constant.

Original entry on oeis.org

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

Offset: 0

Eric W. Weisstein, Dec 11 1999

Davison shows that the continued fraction is (essentially) A000301 and proves that this constant is transcendental. - Charles R Greathouse IV, Jul 22 2013
Using Davison's result we can find an alternating series representation for the rabbit constant r as r = 1 - sum {n >= 1} (-1)^(n+1)*(1 + 2^Fibonacci(3*n+1))/( (2^(Fibonacci(3*n - 1)) - 1)*(2^(Fibonacci(3*n + 2)) - 1) ). The series converges rapidly: for example, the first 10 terms of the series give a value for r accurate to more than 1.7 million decimal places. See A005614. - Peter Bala, Nov 11 2013
The rabbit constant is the number having the infinite Fibonacci word A005614 as binary expansion; its continued fraction expansion is A000301 = 2^A000045 (after a leading zero, depending on convention). - M. F. Hasler, Nov 10 2018

			0.709803442861291314641787399444575597012502205767...

Steven R. Finch, Mathematical Constants, Cambridge, 2003, p. 439.
M. Schroeder, Fractals, Chaos, Power Laws: Minutes from an Infinite Paradise, New York: W. H. Freeman, 1991.

Sean A. Irvine and Joerg Arndt, Table of n, a(n) for n = 0..2000
W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
P. G. Anderson, T. C. Brown, and P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
Joerg Arndt, Matters Computational (The Fxtbook), p. 754.
J. L. Davison, A series and its associated continued fraction, Proc. Amer. Math. Soc. 63 (1977), pp. 29-32.
Martin Griffiths, 96.12 The sum of a series: rational or irrational?, The Mathematical Gazette, Vol. 96, No. 535 (2012), pp. 121-124.
Clark Kimberling and K. B. Stolarsky, Slow Beatty sequences, devious convergence, and partitional divergence, Amer. Math. Monthly, 123 (No. 2, 2016), 267-273.
Eric Weisstein's World of Mathematics, Rabbit Constant.
Index entries for transcendental numbers

Cf. A005614, A073115, A119809, A119812.

Take[ RealDigits[ Sum[N[1/2^Floor[k*GoldenRatio], 120], {k, 0, 300}]-1][[1]], 103] (* Jean-François Alcover, Jul 28 2011, after Benoit Cloitre *)
RealDigits[ FromDigits[{Nest[Flatten[# /. {0 -> {1}, 1 -> {1, 0}}] &, {1}, 12], 0}, 2], 10, 111][[1]] (* Robert G. Wilson v, Mar 13 2014 *)
digits = 103; dm = 10; Clear[xi]; xi[b_, m_] := xi[b, m] = RealDigits[ ContinuedFractionK[1, b^Fibonacci[k], {k, 0, m}], 10, digits] // First; xi[2, dm]; xi[2, m = 2 dm]; While[xi[2, m] != xi[2, m - dm], m = m + dm]; xi[2, m] (* Jean-François Alcover, Mar 04 2015, update for versions 7 and up, after advice from Oleg Marichev *)

/* fast divisionless routine from fxtbook */
fa(y, N=17)=
{ my(t, yl, yr, L, R, Lp, Rp);
/* as powerseries correct up to order fib(N+2)-1 */
  L=0; R=1; yl=1; yr=y;
  for(k=1, N, t=yr; yr*=yl; yl=t; Lp=R; Rp=R+yr*L; L=Lp; R=Rp; );
  return( R )
}
a=0.5*fa(0.5) /* Joerg Arndt, Apr 15 2010 */

my(r=1,p=(3-sqrt(5))/2,n=1);while(r>r-=1.>>(n\p),n++);A014565=r \\ M. F. Hasler, Nov 10 2018

my(f(n)=1.<A098317 (=> 298, 1259, 5331, ... digits). - M. F. Hasler, Nov 10 2018

Equals Sum_{n>=1} 1/2^b(n) where b(n) = floor(n*phi) = A000201(n).
Equals -1 + A073115.
From Peter Bala, Nov 04 2013: (Start)
The results of Adams and Davison 1977 can be used to find a variety of alternative series representations for the rabbit constant r. Here are several examples (phi denotes the golden ratio (1/2)*(1 + sqrt(5))).
r = Sum_{n >= 2} ( floor((n+1)*phi) - floor(n*phi) )/2^n = (1/2)*Sum_{n >= 1} A014675(n)/2^n.
r = Sum_{n >= 1} floor(n/phi)/2^n = Sum_{n >= 1} A005206(n-1)/2^n.
r = ( Sum_{n >= 1} 1/2^floor(n/phi) ) - 2 and r = ( Sum_{n >= 1} floor(n*phi)/2^n ) - 2 = ( Sum_{n >= 1} A000201(n)/2^n ) - 2.
More generally, for integer N >= -1, r = ( Sum_{n >= 1} 1/2^floor(n/(phi + N)) ) - (2*N + 2) and for all integer N, r = ( Sum_{n >= 1} floor(n*(phi + N))/2^n ) - (2*N + 2).
Also r = 1 - Sum_{n >= 1} 1/2^floor(n*phi^2) = 1 - Sum_{n >= 1} 1/2^A001950(n) and r = 1 - Sum_{n >= 1} floor(n*(2 - phi))/2^n = 1 - Sum_{n >= 1} A060144(n)/2^n. (End)

More terms from Simon Plouffe, Dec 11 1999

A063896 a(n) = 2^Fibonacci(n) - 1.

Original entry on oeis.org

0, 1, 1, 3, 7, 31, 255, 8191, 2097151, 17179869183, 36028797018963967, 618970019642690137449562111, 22300745198530623141535718272648361505980415

Offset: 0

Robert G. Wilson v, Aug 29 2001

The recurrence can also be written a(n)+1 = (a(n-1)+1)*(a(n-2)+1) or log_p(a(n)+1) = log_p(a(n-1)+1) + log_p(a(n-2)+1), respectively. Setting a(1)=p-1 for any natural p>1, it follows that log_p(a(n)+1)=Fibonacci(n). Hence any other sequence p^Fibonacci(n)-1 could also serve as a valid solution to that recurrence, depending only on the value of the term a(1). - Hieronymus Fischer, Jun 27 2007
Written in binary, a(n) contains Fibonacci(n) 1's. Thus the sequence converted to base-2 is A007088(a(n)) = 0, 1, 1, 11, 111, 11111, 11111111, ... . - Hieronymus Fischer, Jun 27 2007
In general, if b(n) is defined recursively by b(0) = p, b(1) = q, b(n) = b(n-1)*b(n-2) + b(n-1) + b(n-2) for n >= 2 then b(n) = p^Fibonacci(n-1) * q^Fibonacci(n) - 1. - Rahul Goswami, Apr 15 2020
a(n) is also the numerator of the continued fraction [2^F(0), 2^F(1), 2^F(2), 2^F(3), ..., 2^F(n-2)] for n>0. For the denominator, see A005203. - Chinmay Dandekar and Greg Dresden, Sep 19 2020

Alois P. Heinz, Table of n, a(n) for n = 0..18
M. Tamba and Y. S. Valaulikar, A nonlinear extension of Fibonacci sequence, Turkish Journal of Analysis and Number Theory, Vol. 4, No. 4, 2016.

Cf. A000045 (Fibonacci), A000225, A000301, A005203, A061107.

See A131293 for a base-10 analog with Fib(n) 1's.

a:= n-> 2^(<<0|1>, <1|1>>^n)[1,2]-1:
seq(a(n), n=0..15);  # Alois P. Heinz, Aug 12 2017

2^Fibonacci[Range[0,15]]-1 (* Harvey P. Dale, May 20 2014 *)
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == (a[n - 1] + 1)*(a[n - 2] + 1) - 1}, a[n], {n, 0, 12}] (* Ray Chandler, Jul 30 2015 *)

a(n) = 2^fibonacci(n) - 1 \\ Charles R Greathouse IV, Oct 03 2016

The solution to the recurrence a(0) = 0; a(1) = 1; a(n) = a(n-1)*a(n-2) + a(n-1) + a(n-2).
a(n) = A000301(n) - 1. - R. J. Mathar, Apr 26 2007
a(n) = a(n-2)*2^ceiling(log_2(a(n-1))) + a(n-1) for n>1. - Hieronymus Fischer, Jun 27 2007
a(n) = A000225(A000045(n)). - Alois P. Heinz, Mar 19 2020

A080674 a(n) = (4/3)*(4^n - 1).

Original entry on oeis.org

0, 4, 20, 84, 340, 1364, 5460, 21844, 87380, 349524, 1398100, 5592404, 22369620, 89478484, 357913940, 1431655764, 5726623060, 22906492244, 91625968980, 366503875924, 1466015503700, 5864062014804, 23456248059220, 93824992236884, 375299968947540, 1501199875790164

Offset: 0

N. J. A. Sloane, Mar 02 2003

a(n) is the number of steps which are made when generating all n-step random walks that begin in a given point P on a two-dimensional square lattice. To make one step means to move along one edge on the lattice. - Pawel P. Mazur (Pawel.Mazur(AT)pwr.wroc.pl), Mar 10 2005
Conjectured to be the number of integers from 0 to (10^n)-1 that lack 0, 1, 2, 3, 4 and 5 as a digit. - Alexandre Wajnberg, Apr 25 2005
Gives the values of m such that binomial(4*m + 4,m) is odd. Cf. A002450, A020988 and A263132. - Peter Bala, Oct 11 2015
Also the partial sums of 4^n for n>0, cf. A000302. - Robert G. Wilson v, Sep 18 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..170
Peter Bala, A characterization of A002450, A020988 and A080674.
Mattia Fregola, Cellular Automata RULE13 generating OEIS sequence A080674
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Row n = 4 of A228275.

Cf. A000301, A000302, A002450, A020988, A024036, A080674, A263132.

[(4/3)*(4^n-1): n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

Table[4*(4^n-1)/3,{n,0,100}]  (* Vladimir Joseph Stephan Orlovsky, Jan 30 2012 *)
LinearRecurrence[{5,-4},{0,4},40] (* Harvey P. Dale, May 05 2018 *)

vector(100, n, n--; (4/3)*(4^n-1)) \\ Altug Alkan, Oct 11 2015

Vec(4*x/((x-1)*(4*x-1)) + O(x^40)) \\ Colin Barker, Oct 12 2015

a(n) = 2*A020988(n) = A002450(n+1) - 1 = 4*A002450(n).
a(n) = Sum_{i = 1..n} 4^i. - Adam McDougall (mcdougal(AT)stolaf.edu), Sep 29 2004
a(n) = 4*a(n-1) + 4. - Alexandre Wajnberg, Apr 25 2005
a(n) = 4^n + a(n-1) (with a(0) = 0). - Vincenzo Librandi, Aug 08 2010
From Colin Barker, Oct 12 2015: (Start)
a(n) = 5*a(n-1) - 4*a(n-2).
G.f.: 4*x / ((x-1)*(4*x-1)). (End)
E.g.f.: 4*exp(x)*(exp(3*x) - 1)/3. - Elmo R. Oliveira, Dec 17 2023

A010098 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 3, 9, 27, 243, 6561, 1594323, 10460353203, 16677181699666569, 174449211009120179071170507, 2909321189362570808630465826492242446680483, 507528786056415600719754159741696356908742250191663887263627442114881

Offset: 0

N. J. A. Sloane

nonn

From Peter Bala, Nov 01 2013: (Start)
Let phi = (1/2)*(1 + sqrt(5)) denote the golden ratio A001622. This sequence gives the simple continued fraction expansion of the constant c := 2*Sum_{n>=1} 1/3^floor(n*phi) (= 4*Sum_{n>=1} floor(n/phi)/3^n) = 0.768597560593155198508 ... = 1/(1 + 1/(3 + 1/(3 + 1/(9 + 1/(27 + 1/(243 + 1/(6561 + ...))))))). The constant c is known to be transcendental (see Adams and Davison 1977). Cf. A014565.
Furthermore, for k = 0,1,2,... if we put X(k) = sum {n >= 1} 1/3^(n*Fibonacci(k) + Fibonacci(k+1)*floor(n*phi)) then the real number X(k+1)/X(k) has the simple continued fraction expansion [0; a(k+1), a(k+2), a(k+3), ...] (apply Bowman 1988, Corollary 1). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..17
W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
P. G. Anderson, T. C. Brown, and P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
D. Bowman, A new generalization of Davison's theorem, Fib. Quart. Volume 26 (1988), 40-45

Cf. A000045, A000301, A000304, A010099, A010100, A014565, A214706, A214887, A215270, A215271, A215272.

Column k=3 of A244003.

a010098 n = a010098_list !! n
a010098_list = 1 : 3 : zipWith (*) a010098_list (tail a010098_list)
-- Reinhard Zumkeller, Jul 06 2014

[3^Fibonacci(n): n in [0..12]]; // G. C. Greubel, Jul 29 2024

a[-1]:=1: a[0]:=3: a[1]:=3: for n from 2 to 13 do a[n]:=a[n-1]*a[n-2] od: seq(a[n], n=-1..10); # Zerinvary Lajos, Mar 19 2009

3^Fibonacci[Range[0,13]] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
RecurrenceTable[{a[0]==1,a[1]==3,a[n]==a[n-1]a[n-2]},a,{n,15}] (* Harvey P. Dale, Jan 21 2021 *)

[3^fibonacci(n) for n in range(13)] # G. C. Greubel, Jul 29 2024

a(n) = 3^Fibonacci(n).

a(n+1) = A000304(n+3) / A000301(n). - Reinhard Zumkeller, Jul 06 2014

A244003 A(n,k) = k^Fibonacci(n); square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 2, 1, 0, 1, 4, 3, 4, 1, 0, 1, 5, 4, 9, 8, 1, 0, 1, 6, 5, 16, 27, 32, 1, 0, 1, 7, 6, 25, 64, 243, 256, 1, 0, 1, 8, 7, 36, 125, 1024, 6561, 8192, 1, 0, 1, 9, 8, 49, 216, 3125, 65536, 1594323, 2097152, 1, 0

Offset: 0

Alois P. Heinz, Jun 17 2014

			Square array A(n,k) begins:
  1, 1,   1,    1,     1,      1,       1, ...
  0, 1,   2,    3,     4,      5,       6, ...
  0, 1,   2,    3,     4,      5,       6, ...
  0, 1,   4,    9,    16,     25,      36, ...
  0, 1,   8,   27,    64,    125,     216, ...
  0, 1,  32,  243,  1024,   3125,    7776, ...
  0, 1, 256, 6561, 65536, 390625, 1679616, ...

Alois P. Heinz, Antidiagonals n = 0..20, flattened

Columns k=0-10 give: A000007, A000012, A000301, A010098, A010099, A214706, A215270, A214887, A215271, A215272, A010100.
Rows n=0, 1+2, 3-8 give: A000012, A001477, A000290, A000578, A000584, A001016, A010801, A010809.
Main diagonal gives: A152915.
Cf. A000045, A103323.

A:= (n, k)-> k^(<<1|1>, <1|0>>^n)[1, 2]:
seq(seq(A(n, d-n), n=0..d), d=0..12);

A[0, 0] = 1; A[n_, k_] := k^Fibonacci[n]; Table[A[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Nov 11 2015 *)

A(n,k) = k^A000045(n).

A(0,k) = 1, A(1,k) = k, A(n,k) = A(n-1,k) * A(n-2,k) for n>=2.

A010100 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=10.

Original entry on oeis.org

1, 10, 10, 100, 1000, 100000, 100000000, 10000000000000, 1000000000000000000000, 10000000000000000000000000000000000, 10000000000000000000000000000000000000000000000000000000

Offset: 0

N. J. A. Sloane

nonn

From Peter Bala, Nov 11 2013: (Start)
Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio A001622. This sequence is the simple continued fraction expansion of the constant c := 9*sum {n = 1..inf} 1/10^floor(n*phi) (= 81*sum {n = 1..inf} floor(n/phi)/10^n) = 0.90990 90990 99090 99090 ... = 1/(1 + 1/(10 + 1/(10 + 1/(100 + 1/(1000 + 1/(100000 + 1/(100000000 + ...))))))). The constant c is known to be transcendental (see Adams and Davison 1977). Cf. A014565 and A005614.
Furthermore, for k = 0,1,2,... if we define the real number X(k) = sum {n >= 1} 1/10^(n*Fibonacci(k) + Fibonacci(k+1)*floor(n*phi)) then the real number X(k+1)/X(k) has the simple continued fraction expansion [0; a(k+1), a(k+2), a(k+3), ...] (apply Bowman 1988, Corollary 1). (End)

Alois P. Heinz, Table of n, a(n) for n = 0..16
W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
P. G. Anderson, T. C. Brown, P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
D. Bowman, A new generalization of Davison's theorem, Fib. Quart. Volume 26 (1988), 40-45

Cf. A000045, A000301, A010098, A010099. A005614, A014565, A214706, A214887, A215270, A215271, A215272.

Column k=10 of A244003.

a:= n-> 10^(<<1|1>, <1|0>>^n)[1, 2]:
seq(a(n), n=0..12);  # Alois P. Heinz, Jun 17 2014

10^Fibonacci[Range[0,10]] (* Harvey P. Dale, Feb 12 2023 *)

a(n) = 10^fibonacci(n); \\ Michel Marcus, Oct 25 2017

a(n) = 10^Fibonacci(n).

A214706 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=5.

Original entry on oeis.org

1, 5, 5, 25, 125, 3125, 390625, 1220703125, 476837158203125, 582076609134674072265625, 277555756156289135105907917022705078125, 161558713389263217748322010169914619837072677910327911376953125

Offset: 0

Vincenzo Librandi, Aug 01 2012

a(17) has 1117 digits.
From Peter Bala, Nov 01 2013: (Start)
Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio A001622. This sequence is the simple continued fraction expansion of the constant c := 4*Sum_{n = 1..oo} 1/5^floor(n*phi) (= 16*Sum_{n = 1..oo} floor(n/phi)/5^n) = 0.83866 83869 91037 14262 ... = 1/(1 + 1/(5 + 1/(5 + 1/(25 + 1/(125 + 1/(3125 + 1/(390625 + ...))))))). The constant c is known to be transcendental (see Adams and Davison 1977). Cf. A014565.
Furthermore, for k = 0,1,2,... if we define the real number X(k) = sum {n >= 1} 1/5^(n*Fibonacci(k) + Fibonacci(k+1)*floor(n*phi)) then the real number X(k+1)/X(k) has the simple continued fraction expansion [0; a(k+1), a(k+2), a(k+3), ...] (apply Bowman 1988, Corollary 1). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..16
W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
P. G. Anderson, T. C. Brown, and P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
D. Bowman, A new generalization of Davison's theorem, Fib. Quart. Volume 26 (1988), 40-45.

Cf. A000045, A000301, A001622, A010098, A010099, A010100, A014565, A214706, A214887, A215270, A215271, A215272.

Column k=5 of A244003.

Magma
```
[5^Fibonacci(n): n in [0..13]];
```

a:= n-> 5^(<<1|1>, <1|0>>^n)[1, 2]:
seq(a(n), n=0..12);  # Alois P. Heinz, Jun 17 2014

5^Fibonacci[Range[0,11]]
nxt[{a_,b_}]:={b,a*b}; NestList[nxt,{1,5},12][[All,1]] (* Harvey P. Dale, Oct 14 2018 *)

[5^fibonacci(n) for n in range(15)] # G. C. Greubel, Jan 07 2024

a(n) = 5^Fibonacci(n).

A214887 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=7.

Original entry on oeis.org

1, 7, 7, 49, 343, 16807, 5764801, 96889010407, 558545864083284007, 54116956037952111668959660849, 30226801971775055948247051683954096612865741943

Offset: 0

Vincenzo Librandi, Aug 01 2012

a(17) has 1350 digits.
From Peter Bala, Nov 01 2013: (Start)
Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio A001622. This sequence is the simple continued fraction expansion of the constant c := 6*sum {n = 1..inf} 1/7^floor(n*phi) (= 36*sum {n = 1..inf} floor(n/phi)/7^n) = 0.87718 67194 00499 51922 ... = 1/(1 + 1/(7 + 1/(7 + 1/(49 + 1/(343 + 1/(16807 + 1/(5764801 + ...))))))). The constant c is known to be transcendental (see Adams and Davison 1977). Cf. A014565.
Furthermore, for k = 0,1,2,... if we define the real number X(k) = sum {n >= 1} 1/7^(n*Fibonacci(k) + Fibonacci(k+1)*floor(n*phi)) then the real number X(k+1)/X(k) has the simple continued fraction expansion [0; a(k+1), a(k+2), a(k+3), ...] (apply Bowman 1988, Corollary 1). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..16
W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
P. G. Anderson, T. C. Brown, P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
D. Bowman, A new generalization of Davison's theorem, Fib. Quart. Volume 26 (1988), 40-45

Cf. A000045, A000301, A010098, A010099, A010100, A214706, A014565, A215270, A215271, A215272.

Column k=7 of A244003.

Magma
```
[7^Fibonacci(n): n in [0..10]];
```

a:= n-> 7^(<<1|1>, <1|0>>^n)[1, 2]:
seq(a(n), n=0..12);  # Alois P. Heinz, Jun 17 2014

7^Fibonacci[Range[0,10]]
nxt[{a_,b_}]:={b,a*b}; Transpose[NestList[nxt,{1,7},10]][[1]] (* Harvey P. Dale, Jun 10 2014 *)

a(n) = 7^Fibonacci(n).

A215270 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 6, 36, 216, 7776, 1679616, 13060694016, 21936950640377856, 286511799958070431838109696, 6285195213566005335561053533150026217291776, 1800782593726645086383198950649858141454002621435149880441896326660096

Offset: 0

Bruno Berselli, Aug 07 2012

From Peter Bala, Nov 01 2013: (Start)
Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio A001622. This sequence is the simple continued fraction expansion of the constant c := 5*sum {n = 1..inf} 1/6^floor(n*phi) (= 25*sum {n = 1..inf} floor(n/phi)/6^n) = 0.86045 01626 86090 61353 ... = 1/(1 + 1/(6 + 1/(6 + 1/(36 + 1/(216 + 1/(7776 + 1/(1679616 + ...))))))). The constant c is known to be transcendental (see Adams and Davison 1977). Cf. A014565.
Furthermore, for k = 0,1,2,... if we define the real number X(k) = sum {n >= 1} 1/6^(n*Fibonacci(k) + Fibonacci(k+1)*floor(n*phi)) then the real number X(k+1)/X(k) has the simple continued fraction expansion [0; a(k+1), a(k+2), a(k+3), ...] (apply Bowman 1988, Corollary 1). (End)

Bruno Berselli, Table of n, a(n) for n = 0..15
W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
P. G. Anderson, T. C. Brown, P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
D. Bowman, A new generalization of Davison's theorem, Fib. Quart. Volume 26 (1988), 40-45

Cf. A000045, A000301, A010098-A010100, A214706, A214887, A215271, A215272.
Cf. A166470 (same recurrence with initial values 2, 6). A014565.
Column k=6 of A244003.

Magma
```
[6^Fibonacci(n): n in [0..11]];
```

a:= n-> 6^(<<1|1>, <1|0>>^n)[1, 2]:
seq(a(n), n=0..12);  # Alois P. Heinz, Jun 17 2014

RecurrenceTable[{a[0] == 1, a[1] == 6, a[n] == a[n - 1] a[n - 2]}, a[n], {n, 0, 15}]

a(n) = 6^Fibonacci(n).

cp's OEIS Frontend

A061084 Fibonacci-type sequence based on subtraction: a(0) = 1, a(1) = 2 and a(n) = a(n-2) - a(n-1).

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A063896 a(n) = 2^Fibonacci(n) - 1.

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A010098 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=3.

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A244003 A(n,k) = k^Fibonacci(n); square array A(n,k), n>=0, k>=0, read by antidiagonals.

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