A000301 -id:A000301 - cp's OEIS frontend

A215271 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=8.

1, 8, 8, 64, 512, 32768, 16777216, 549755813888, 9223372036854775808, 5070602400912917605986812821504, 46768052394588893382517914646921056628989841375232, 237142198758023568227473377297792835283496928595231875152809132048206089502588928

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 := 7*sum {n = 1..inf} 1/8^floor(n*phi) (= 49*sum {n = 1..inf} floor(n/phi)/8^n) = 0.89040 80325 60827 28336 ... = 1/(1 + 1/(8 + 1/(8 + 1/(64 + 1/(512 + 1/(32768 + 1/(16777216 + ...))))))). 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} 1/8^(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, A215270, A215272, A014565.

Column k=8 of A244003.

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

a:= n-> 8^(<<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] == 8, a[n] == a[n - 1] a[n - 2]}, a[n], {n, 0, 15}]

a(n) = 8^Fibonacci(n).

A215272 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=9.

Original entry on oeis.org

1, 9, 9, 81, 729, 59049, 43046721, 2541865828329, 109418989131512359209, 278128389443693511257285776231761, 30432527221704537086371993251530170531786747066637049

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 := 8*sum {n = 1..inf} 1/9^floor(n*phi) (= 64*sum {n = 1..inf} floor(n/phi)/9^n) = 0.90109 74122 99938 29901 ... = 1/(1 + 1/(9 + 1/(9 + 1/(81 + 1/(729 + 1/(59049 + 1/(43046721 + ...))))))). 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} 1/9^(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, A215270, A215271, A014565.

Column k=9 of A244003.

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

a:= n-> 9^(<<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] == 9, a[n] == a[n - 1] a[n - 2]}, a[n], {n, 0, 15}]

a(n) = 9^fibonacci(n); \\ Jinyuan Wang, Apr 06 2019

a(n) = 9^Fibonacci(n).

A000304 a(n) = a(n-1)*a(n-2).

Original entry on oeis.org

2, 3, 6, 18, 108, 1944, 209952, 408146688, 85691213438976, 34974584955819144511488, 2997014624388697307377363936018956288, 104819342594514896999066634490728502944926883876041385836544

Offset: 2

N. J. A. Sloane

nonn

A038500(a(n)) = A010098(n-2); for n > 2: A006519(a(n)) = A000301(n-3); A001222(a(n)) = A000045(n-1). - Reinhard Zumkeller, Jul 06 2014

T. D. Noe, Table of n, a(n) for n = 2..18
Peter G. Anderson, Notes & Extensions for a Remarkable Continued Fraction, Fibonacci Quart. 55 (2017), no. 5, 9-14. Mentions this sequence.
Sergio Falcon, Fibonacci's multiplicative sequence, Int. J. Math. Edu. Sci. Technol. 34-2 (2003), 310-315. [_Sergio Falcon_, Nov 23 2009]

Cf. A000045, A000301, A001222, A006519, A010098, A038500, A249406.

a000304 n = a000304_list !! (n-2)
a000304_list = 2 : 3 : zipWith (*) a000304_list (tail a000304_list)
-- Reinhard Zumkeller, Jul 06 2014

A000304 := proc(n) option remember; if n <=3 then n else A000304(n-1)*A000304(n-2); fi; end;

nxt[{a_,b_}]:={b,a*b}; Transpose[NestList[nxt,{2,3},12]][[1]] (* Harvey P. Dale, Nov 16 2014 *)

For n>=4, a(n) = 2^A000045(n-3)*3^A000045(n-2). - Benoit Cloitre, Sep 26 2003

For n > 2: a(n) = A000301(n-3) * A010098(n-2). - Reinhard Zumkeller, Jul 06 2014

More terms from Vladimir Joseph Stephan Orlovsky, Feb 17 2010

A010099 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 4, 16, 64, 1024, 65536, 67108864, 4398046511104, 295147905179352825856, 1298074214633706907132624082305024, 383123885216472214589586756787577295904684780545900544

Offset: 0

N. J. A. Sloane

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 := 3*sum {n = 1..inf} 1/4^floor(n*phi) (= 9*sum {n = 1..inf} floor(n/phi)/4^n) = 0.80938 42984 64421 90504 ... = 1/(1 + 1/(4 + 1/(4 + 1/(16 + 1/(64 + 1/(1024 + 1/(65536 + ...))))))). 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} 1/4^(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)

Indranil Ghosh, 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, 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, A014565, A214706, A214887, A215270, A215271, A215272.

Column k=4 of A244003.

a[ -1]:=1:a[0]:=4: a[1]:=4: 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

a(n) = 4^Fibonacci(n).

A124091 Decimal expansion of Fibonacci binary constant: Sum{i>=0} (1/2)^Fibonacci(i).

Original entry on oeis.org

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

Offset: 1

R. J. Mathar, Nov 25 2006

This constant is transcendental, see A084119. - Charles R Greathouse IV, Nov 12 2014

			2.4102787972078658917940430244710631444834239245952787725932...

Harry J. Smith, Table of n, a(n) for n = 1..20000
D. H. Bailey, J. M. Borwein, R. E. Crandall and C. Pomerance, On the binary expansions of algebraic numbers, Journal de Théorie des Nombres de Bordeaux 16 (2004), 487-518.
Index entries for transcendental numbers

Cf. A007404 (Kempner-Mahler number), A125600 (continued fraction), A084119 (essentially the same).

Cf. A000301.

RealDigits[ N[ Sum[(1/2)^Fibonacci[i], {i, 0, Infinity}], 111]][[1]] (* Robert G. Wilson v, Nov 26 2006 *)

a=0 ; for(n=0,30, a += .5^fibonacci(n) ; print(a) ; )

default(realprecision, 20080); x=suminf(k=0, 1/2^fibonacci(k)); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b124091.txt", n, " ", d)) \\ Harry J. Smith, May 04 2009

Equals Sum_{i>=0} 1/2^A000045(i).

Equals A084119 + 1.

More terms from Robert G. Wilson v, Nov 26 2006

A166470 a(n) = 2^F(n+1)*3^F(n), where F(n) is the n-th Fibonacci number, A000045(n).

Original entry on oeis.org

2, 6, 12, 72, 864, 62208, 53747712, 3343537668096, 179707499645975396352, 600858794305667322270155425185792, 107978831564966913814384922944738457859243070439030784

Offset: 0

Matthew Vandermast, Nov 05 2009

G. C. Greubel, Table of n, a(n) for n = 0..16

Subsequence of A025610 and hence of A003586 and A025487.

Cf. A000045, A000301, A002110, A010098, A115033, A166469, A174666, A230900.

[2^Fibonacci(n+1)*3^Fibonacci(n): n in [0..14]]; // G. C. Greubel, Jul 29 2024

3^First[#] 2^Last[#]&/@Partition[Fibonacci[Range[0,12]],2,1] (* Harvey P. Dale, Aug 20 2012 *)

a(n)=2^fibonacci(n+1)*3^fibonacci(n) \\ Charles R Greathouse IV, Sep 19 2022

[2^fibonacci(n+1)*3^fibonacci(n) for n in range(15)] # G. C. Greubel, Jul 29 2024

a(n) = A000301(n+1)*A010098(n).
For n > 1, a(n) = a(n-1)*a(n-2).
For m > 1, n > 1, A166469(A002110(m)*(a(n)^k)/12) = k*Fibonacci(m+n).
A166469(a(n)) = Fibonacci(n+2) + 1 = A001611(n+2).
a(n) = 2 * A174666(n+1). - Alois P. Heinz, Sep 16 2022
a(n) = 2^(Fibonacci(n+1) + c*Fibonacci(n)), with c=log_2(3). Cf. A000301 (c=1) & A010098 (c=2). - Andrea Pinos, Sep 29 2022
a(n) = A115033(2*n+1). - David Radcliffe, May 31 2025

Typo corrected by Matthew Vandermast, Nov 07 2009

A249406 Start with a(1) = 1, and extend by the rule that the next term is the product of the two most recent non-terms of the sequence.

Original entry on oeis.org

1, 6, 20, 56, 90, 132, 182, 240, 306, 399, 506, 600, 702, 812, 930, 1056, 1190, 1332, 1482, 1640, 1806, 1980, 2162, 2352, 2550, 2756, 2970, 3306, 3540, 3782, 4032, 4290, 4556, 4830, 5112, 5402, 5700, 6006, 6320, 6642, 6972, 7310, 7656, 8099, 8556, 8930, 9312

Offset: 1

Reinhard Zumkeller, Oct 31 2014

nonn

Compare to A075326, where not products, but sums of the two most recent non-terms are considered;
a(195) = 159200 is the smallest even term not of the form m*(m+1); see also A249408, the set of all non-oblong terms of this sequence.
a(10) = 399 is the smallest odd term.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A000301, A000304, A075326, A249055, A249407 (complement), subsequence of A002808.

import Data.List ((\\))
a249406 n = a249406_list !! (n-1)
a249406_list = 1 : f [2..] where
   f ws@(u:v:_) = y : f (ws \\ [u, v, y]) where y = u * v

A073115 Decimal expansion of sum(k>=0, 1/2^floor(k*phi) ) where phi = (1+sqrt(5))/2.

Original entry on oeis.org

1, 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, 3

Offset: 1

Benoit Cloitre, Aug 19 2002

Number whose digits are obtained from the substitution system (1->(1,0),0->(1)).
The n-th term of the continued fraction is 2^Fibonacci(n-2) (cf. A000301).
This number is known to be transcendental.

			1.70980344286129131464178739944457559701250220576786...

S. Wolfram, "A new kind of science", p. 913

J. L. Davison, A series and its associated continued fraction, Proc. Amer. Math. Soc. 63 (1977), pp. 29-32.
Index entries for transcendental numbers

Take[ RealDigits[ Sum[N[1/2^Floor[k*GoldenRatio], 120], {k, 0, 300}]][[1]], 105] (* Jean-François Alcover, Jul 28 2011 *)

phi=(1+sqrt(5))/2; suminf(n=0,2.^-(n*phi\1)) \\ Charles R Greathouse IV, Jul 22 2013

phi=(1+sqrt(5))/2; suminf(n=1, (phi*n\1)/2^n) - 1 /* Michael Somos, May 22 2021 */

Equals 1 + A014565.

A000336 a(n) = a(n-1)a(n-2)a(n-3)*a(n-4); for n < 5, a(n) = n.

Original entry on oeis.org

1, 2, 3, 4, 24, 576, 165888, 9172942848, 21035720123168587776, 18437563379178327736384102280592359424, 590180110002114158896983994712576414865667267958188575935810179040280576

Offset: 1

N. J. A. Sloane

nonn

The next term has 139 digits. - Harvey P. Dale, Jan 21 2019

T. D. Noe, Table of n, a(n) for n = 1..15
B. K. Agarwala and F. C. Auluck, Statistical mechanics and partitions into non-integral powers of integers, Proc. Camb. Phil. Soc., 47 (1951), 207-216. [Annotated scanned copy]

Cf. A000301, A000308, A001631, A003586, A251656.

A000336 := proc(n) option remember; if n <=4 then n else A000336(n-1)*A000336(n-2)*A000336(n-3)*A000336(n-4); fi; end;

t = {1, 2, 3, 4}; Do[AppendTo[t, t[[-1]]*t[[-2]]*t[[-3]]*t[[-4]]], {n, 5, 15}] (* T. D. Noe, Jun 19 2012 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,a b c d}; NestList[nxt,{1,2,3,4},10][[All,1]] (* Harvey P. Dale, Jan 21 2019 *)

a(n,a=[24,1,2,3,4])={for(n=6,n,a[n%5+1]=a[(n-1)%5+1]^2\a[n%5+1]);a[n%5+1]} \\ M. F. Hasler, Apr 22 2018

first(n) = n = max(n, 5); my(res = vector(n)); for(i=1, 4, res[i] = i); res[5]=24; for(i = 6, n, res[i] = res[i-1]^2 / res[i - 5]); res \\ David A. Corneth, Apr 22 2018

a(n) = 2^A251656(n) * 3^A001631(n-1). - Vaclav Kotesovec, Feb 02 2016

a(n) = a(n-1)^2 / a(n-5), for n > 5. - M. F. Hasler, Apr 22 2018

A230900 a(n) = 2^Lucas(n).

Original entry on oeis.org

4, 2, 8, 16, 128, 2048, 262144, 536870912, 140737488355328, 75557863725914323419136, 10633823966279326983230456482242756608, 803469022129495137770981046170581301261101496891396417650688

Offset: 0

Peter Bala, Oct 31 2013

Compare with A000301(n) = 2^Fibonacci(n).
The sequence a(n) for n >= 1 gives the sequence of partial quotients (other than the first) in the continued fraction expansion of the transcendental real constant c := sum {n >= 1} 1/2^floor(n*(5 + sqrt(5))/2) = 0.13385 44229 67609 80592 ... = 1/(7 + 1/(2 + 1/(8 + 1/(16 + 1/(128 + 1/(2048 + ...)))))). See Adams Davison 1977. Cf. A014565.
The constant c has various series representations including
c = 1 - sum {n >= 1} 1/2^floor(n*(5 - sqrt(5))/2),
c = sum {n >= 1} floor(n*(5 - sqrt(5))/10)/2^n,
c = 3 - sum {n >= 1} 1/2^floor(n*(15 - sqrt(5))/22) and
c = sum {n >= 1} 1/2^floor(n*(15 + sqrt(5))/22) - 2.

William 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.
J. L. Davison, A series and its associated continued fraction, Proc. Amer. Math. Soc. 63 (1977), pp. 29-32.

Cf. A000032, A000301, A014565, A121821.

[2^(Lucas(n)): n in [0..10]]; // G. C. Greubel, Dec 22 2017

a := proc(n) option remember; if n = 0 then 4 elif n = 1 then 2 else a(n-1)*a(n-2); fi; end; seq(a(n), n = 0..10);

2^LucasL[Range[0,15]] (* Harvey P. Dale, Jul 21 2015 *)

for(n=0,10, print1(2^(fibonacci(n+1) + fibonacci(n-1)), ", ")) \\ G. C. Greubel, Dec 22 2017

a(n) = 2^Lucas(n) = 2^A000032(n).
Recurrence: a(n) = a(n-1)*a(n-2) with a(0) = 4, a(1) = 2.
Sum_{n>=1} 1/a(n) = A121821. - Amiram Eldar, Oct 27 2020

cp's OEIS Frontend

A215271 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=8.

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A215272 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=9.

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A000304 a(n) = a(n-1)*a(n-2).

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A010099 a(n) = a(n-1)*a(n-2) with a(0)=1, a(1)=4.

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A124091 Decimal expansion of Fibonacci binary constant: Sum{i>=0} (1/2)^Fibonacci(i).

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A166470 a(n) = 2^F(n+1)*3^F(n), where F(n) is the n-th Fibonacci number, A000045(n).

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A249406 Start with a(1) = 1, and extend by the rule that the next term is the product of the two most recent non-terms of the sequence.

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A000336 a(n) = a(n-1)a(n-2)a(n-3)*a(n-4); for n < 5, a(n) = n.