A145234 -id:A145234 - cp's OEIS frontend

A145277 a(n) = A145234(n+1)/A145234(n).

598364773, 27692759465311176949233529747775189817301578781117871380248013

Offset: 1

Artur Jasinski, Oct 06 2008

A member of the family of sequences of type:
(G^(k^(n + 1)) - (1 - G)^(k^(n + 1)))/(G^(k^n) - (1 - G)^(k^n)) where G = (1 + sqrt(5))/2
For k=2 see A001566.
For k=3 see A002814(n+2).
For k=4 see A145274.
For k=5 see A145275.
For k=6 see A145276.
For k=7 see this sequence.

Amiram Eldar, Table of n, a(n) for n = 1..3

Cf. A001566, A001622, A002814, A145234, A145274, A145275, A145276.

G = (1 + Sqrt[5])/2; Table[Expand[(G^(7^(n + 1)) - (1 - G)^(7^(n + 1)))/Sqrt[5]]/Expand[(G^(7^n) - (1 - G)^(7^n))/Sqrt[5]], {n, 1, 5}]

a(n) = (G^(7^(n + 1)) - (1 - G)^(7^(n + 1)))/(G^(7^n) - (1 - G)^(7^n)) where G = (1 + sqrt(5))/2.

A045529 a(n+1) = 5a(n)^3 - 3a(n), a(0) = 1.

Original entry on oeis.org

1, 2, 34, 196418, 37889062373143906, 271964099255182923543922814194423915162591622175362

Offset: 0

Jose Eduardo Blazek, Dec 11 1999

The next term, a(6), has 153 digits. - Harvey P. Dale, Oct 24 2011

Seiichi Manyama, Table of n, a(n) for n = 0..7
Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.
Zalman Usiskin, Problem B-265, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 333; Fibonacci Numbers for Powers of 3, Solution to Problem B-265 by Ralph Garfield and David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), this sequence (k=3), A145231 (k=4), A145232 (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

Cf. A000032, A000045, A001622, A001999, A002814, A006267, A094214.

a := proc(n) option remember; if n = 0 then 1 else 5*a(n-1)^3 - 3*a(n-1) end if; end:
seq(a(n), n = 0..5); # Peter Bala, Nov 24 2022

G = (1 + Sqrt[5])/2; Table[Expand[(G^(3^n) - (1 - G)^(3^n))/Sqrt[5]], {n, 1, 7}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[(4/5)^(1/2)*Cosh[3^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 4}] (* Artur Jasinski, Oct 05 2008 *)
RecurrenceTable[{a[0]==1,a[n]==5a[n-1]^3-3a[n-1]},a[n],{n,6}] (* Harvey P. Dale, Oct 24 2011 *)
NestList[5#^3-3#&,1,5] (* Harvey P. Dale, Dec 21 2014 *)

A045529(n):=fib(3^n)$
makelist(A045529(n),n,0,10); /* Martin Ettl, Nov 12 2012 */

The first example I know in which a(n) can be expressed as (4/5)^(1/2)*cosh(3^n*arccosh((5/4)^(1/2))).
a(n) = Fibonacci(3^n). - Leroy Quet, Mar 17 2002
a(n+1) = a(n)*A002814(n+1). - Lekraj Beedassy, Jun 16 2003
a(n) = (phi^(3^n) - (1 - phi)^(3^n))/sqrt(5), where phi is the golden ratio (A001622). - Artur Jasinski, Oct 05 2008
a(n) = Product_{k=0..n-1} (Lucas(2*3^k) - 1) (Usiskin, 1973). - Amiram Eldar, Jan 29 2022
From Peter Bala, Nov 24 2022: (Start)
a(2*n+2) == a(2*n) (mod 3^(2*n+1)); a(2*n+3) == a(2*n+1) (mod 3^(2*n+2));
a(2*n+1) + a(2*n) == 0 (mod 3^(2*n+1)).
a(2*n) == 1 (mod 3) and a(2*n+1) == 2 (mod 3).
5*a(n)^2 == 2 (mod 3^(n+1)).
In the ring of 3-adic integers, the sequences {a(2*n)} and {a(2*n+1)} are both Cauchy sequences and converge to the pair of 3-adic roots of the quadratic equation 5*x^2 - 2 = 0. (End)
From Amiram Eldar, Jan 07 2023: (Start)
Product_{n>=1} (1 + 2/(sqrt(5)*a(n)-1)) = phi (A001622).
Product_{n>=1} (1 - 2/(sqrt(5)*a(n)+1)) = 1/phi (A094214).
Both formulas are from Duverney and Kurosawa (2022). (End)

A250486 A(n,k) is the n^k-th Fibonacci number; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 3, 2, 1, 0, 1, 21, 34, 3, 1, 0, 1, 987, 196418, 987, 5, 1, 0, 1, 2178309, 37889062373143906, 10610209857723, 75025, 8, 1

Offset: 0

Alois P. Heinz, Nov 24 2014

			Square array A(n,k) begins:
  1,  0,  0,      0,       0,    0,  0,  0,   ...
  1,  1,  1,      1,       1,    1,  1,  1,   ...
  1,  1,  3,      21,      987,  2178309,     ...
  1,  2,  34,     196418,  37889062373143906, ...
  1,  3,  987,    10610209857723,             ...
  1,  5,  75025,  59425114757512643212875125, ...
  1,  8,  14930352,                           ...
  1,  13, 7778742049,                         ...

Alois P. Heinz, Antidiagonals n = 0..10, flattened
Wikipedia, Fibonacci number

Columns k=0-8 give: A000012, A000045, A054783, A182149, A250490, A250491, A250492, A250493, A250494.
Rows n=0-10 give: A000007, A000012, A058635, A045529, A145231, A145232, A145233, A145234, A250487, A250488, A250489.
Main diagonal gives A250495.
Cf. A000045.

A:= (n, k)-> (<<0|1>, <1|1>>^(n^k))[1, 2]:
seq(seq(A(n, d-n), n=0..d), d=0..8);

A[n_, k_] := MatrixPower[{{0, 1}, {1, 1}}, n^k][[1, 2]]; A[0, 0] = 1;
Table[A[n, d-n], {d, 0, 8}, {n, 0, d}] // Flatten (* Jean-François Alcover, May 28 2019, from Maple *)

A(n,k) = [0, 1; 1, 1]^(n^k)[1,2].

A145231 a(n) = Fibonacci(4^n).

Original entry on oeis.org

1, 3, 987, 10610209857723, 141693817714056513234709965875411919657707794958199867

Offset: 0

Artur Jasinski, Oct 05 2008

This sequence has the property that a(n+1) is divisible by a(n). Conjecture: each prime divisor can occur only once (i.e., all terms are squarefree). - Artur Jasinski, Oct 05 2008

Seiichi Manyama, Table of n, a(n) for n = 0..6

Cf. A000045, A341601, A341603.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), A045529 (k=3), this sequence (k=4), A145232 (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

a := proc(n) option remember; if n = 1 then 3 else a(n-1)*(5*a(n-1)^2 + 2)*sqrt(5*a(n-1)^2 + 4) end if; end: seq(a(n), n = 1..5); # Peter Bala, Nov 14 2022

G = (1 + Sqrt[5])/2; Table[Expand[(G^(4^n) - (1 - G)^(4^n))/Sqrt[5]], {n, 1, 6}]
Table[Round[(4/5)^(1/2)*Cosh[4^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 7}]
Fibonacci[4^Range[5]] (* Harvey P. Dale, Mar 28 2012 *)

a(n) = (G^(4^n) - (1-G)^(4^n) )/sqrt(5) where G = (1 + sqrt 5)/2 = A001622.
a(n) = round( sqrt(4/5) *cosh( 4^n*arccosh (sqrt(5/4)) )).
a(n)= A000045(A000302(n)). - Michel Marcus, Nov 07 2013
From Peter Bala, Nov 11 2022: (Start)
a(n+1) = a(n)*(5*a(n)^2 + 2)*sqrt(5*a(n)^2 + 4) for n >= 1.
a(n) == 3 (mod 4) for n >= 1.
a(n+1) == a(n) (mod 2^(2*n+1)).
A341601(n) == a(n) (mod 2^n) for n >= 2.
In the ring of 2-adic integers, the sequence {Fibonacci(4^n)} converges to the 2-adic integer A341603. (End)

A145232 a(n) = Fibonacci(5^n).

Original entry on oeis.org

1, 5, 75025, 59425114757512643212875125, 18526362353047317310282957646406309593963452838196423660508102562977229905562196608078556292556795045922591488273554788881298750625

Offset: 0

Artur Jasinski, Oct 05 2008

Seiichi Manyama, Table of n, a(n) for n = 0..5
Robert Frontczak, Problem B-1341, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 1 (2024), p. 84.
Thomas Koshy and Zhenguang Gao, Polynomial Extensions of a Diminnie Delight, Fibonacci Quart. 55 (2017), no. 1, 13-20.
Achilleas Sinefakopoulos, Solution to Problem 1909, Crux Mathematicorum, 20 (1994), 295-296.

Cf. A000045.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), A045529 (k=3), A145231 (k=4), this sequence (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

a := proc(n) option remember; if n = 0 then 1 else 25*a(n-1)^5 - 25*a(n-1)^3 + 5*a(n-1) end if; end:
seq(a(n), n = 0..5); # Peter Bala, Nov 24 2022

G = (1 + Sqrt[5])/2; Table[Expand[(G^(5^n) - (1 - G)^(5^n))/Sqrt[5]], {n, 1, 6}]
Table[Round[N[(2/Sqrt[5])*Cosh[5^n*ArcCosh[Sqrt[5]/2]], 1000]], {n, 1, 4}]
Fibonacci[5^Range[0,4]] (* Harvey P. Dale, Nov 29 2018 *)

a(n) = (G^(5^n) - (1 - G)^(5^n))/sqrt(5) where G = (1 + sqrt(5))/2.
a(n) = (2/sqrt(5))*cosh((2*k+1)^n*arccosh(sqrt(5)/2)).
a(n) = (2/sqrt(5))*cosh(5^n*arccosh(sqrt(5)/2)).
a(n) = (5^n)*A128935(n). - R. J. Mathar, Nov 04 2010
a(n) = A000045(A000351(n)). - Michel Marcus, Nov 07 2013
a(n+1) = 25*a(n)^5 - 25*a(n)^3 + 5*a(n) with a(0) = 1. - Peter Bala, Nov 24 2022
a(n) = 5^n * Product_{k=0..n-1} (5*a(k)^4 - 5*a(k)^2 + 1) (Frontczak, 2024). - Amiram Eldar, Feb 29 2024

A145233 a(n) = Fibonacci(6^n).

Original entry on oeis.org

1, 8, 14930352, 619220451666590135228675387863297874269396512

Offset: 0

Artur Jasinski, Oct 05 2008

nonn

The next term, a(4), has 271 digits. - Harvey P. Dale, Jul 18 2011

Seiichi Manyama, Table of n, a(n) for n = 0..4

Cf. A000045.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), A045529 (k=3), A145231 (k=4), A145232 (k=5), this sequence (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

[Fibonacci(6^n): n in [0..5]]; // Vincenzo Librandi, Apr 02 2011

A145233 := proc(n) combinat[fibonacci](6^n) ; end proc: # R. J. Mathar, Apr 01 2011

G = (1 + Sqrt[5])/2; Table[Expand[(G^(6^n) - (1 - G)^(6^n))/Sqrt[5]], {n, 1, 6}]
(* Second program: *)
Fibonacci[6^Range[4]] (* Harvey P. Dale, Jul 18 2011 *)

a(n) = (G^(6^n) - (1 - G)^(6^n))/sqrt(5) where G = (1 + sqrt(5))/2.

cp's OEIS Frontend

A145277 a(n) = A145234(n+1)/A145234(n).

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A045529 a(n+1) = 5*a(n)^3 - 3*a(n), a(0) = 1.

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A250486 A(n,k) is the n^k-th Fibonacci number; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A145231 a(n) = Fibonacci(4^n).

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A145232 a(n) = Fibonacci(5^n).

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A145233 a(n) = Fibonacci(6^n).

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A045529 a(n+1) = 5a(n)^3 - 3a(n), a(0) = 1.