A215271 -id:A215271 - cp's OEIS frontend

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

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).

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).

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).

cp's OEIS Frontend

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

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

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

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

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

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