A045529 -id:A045529 - cp's OEIS frontend

A002814 For n > 1: a(n) = a(n-1)^3 + 3a(n-1)^2 - 3; a(0) = 1, a(1) = 2.

1, 2, 17, 5777, 192900153617, 7177905237579946589743592924684177

Offset: 0

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Next terms have 102 and 305 digits. - Harvey P. Dale, Jun 06 2011
From Peter Bala, Nov 15 2012: (Start)
The present sequence is the case x = 3 of the following general remarks. The recurrence equation a(n+1) = a(n)^3 + 3*a(n)^2 - 3 with the initial condition a(1) = x - 1 > 1 has the explicit solution a(n+1) = alpha^(3^n) + (1/alpha)^(3^n) - 1 for n >= 0, where alpha := {x + sqrt(x^2 - 4)}/2.
Two other recurrences satisfied by the sequence are a(n+1) = (a(1) + 3)*(Product_{k = 1..n} a(k)^2) - 3 and a(n+1) = 1 + (a(1) - 1)*Product_{k = 1..n} (a(k) + 2)^2, both with a(1) = x - 1.
The associated sequence b(n) := a(n) + 1 satisfies the recurrence equation b(n+1) = b(n)^3 - 3*b(n) with the initial condition b(1) = x. See A001999 for the case x = 3. The sequence c(n) := a(n) + 2 satisfies the recurrence equation c(n+1) = c(n)^3 - 3*c(n)^2 + 3 with the initial condition c(1) = x + 1.
The sequences a(n) and b(n) have been considered by Fine and Escott in connection with a product expansion for quadratic irrationals. We have the following identity, valid for x > 2: sqrt((x + 2)/(x - 2)) = (1 + 2/(x-1))*sqrt((y + 2)/(y - 2)), where y = x^3 - 3*x or, equivalently, y - 1 = (x - 1)^3 + 3*(x - 1)^2 - 3. Iterating the identity produces the product expansion sqrt((x+2)/(x-2)) = Product_{n >= 1} (1 + 2/a(n)), with a(1) = x - 1 and a(n+1) = a(n)^3 + 3*a(n)^2 - 3.
For similar results to the above see A145502. See also A219162. (End)
Conjecture: The sequence {a(n) - 2: n >= 1} is a strong divisibility sequence, that is, gcd(a(n) - 2, a(m) - 2) = a(gcd(n, m)) - 2 for n, m >= 1. - Peter Bala, Dec 08 2022

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 397.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..8
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
J. Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

Cf. A000045, A001566.
Cf. A045529. A001999, A145502, A219162.
Cf. A000244.

a002814 n = a002814_list !! n
a002814_list = 1 : zipWith div (tail xs) xs
   where xs = map a000045 a000244_list
-- Reinhard Zumkeller, Nov 24 2012

Join[{1}, NestList[#^3+3#^2-3&,2,5]] (* Harvey P. Dale, Apr 01 2011 *)

a[0]:1$ a[1]:2$ a[n]:=a[n-1]^3 + 3*a[n-1]^2-3$ A002814(n):=a[n]$
makelist(A002814(n),n,0,6); /* Martin Ettl, Nov 12 2012 */

a(n)=if(n<2,max(0,n+1),a(n-1)^3+3*a(n-1)^2-3)

a(n) = Fibonacci(3^n)/Fibonacci(3^(n-1)). - Henry Bottomley, Jul 10 2001
a(n+1) = 5*(f(n))^2 - 3, where f(n) = Fibonacci(3^n) = product of first n entries. - Lekraj Beedassy, Jun 16 2003
From Artur Jasinski, Oct 05 2008: (Start)
a(n+2) = (G^(3^(n + 1)) - (1 - G)^(3^(n + 1)))/((G^(3^n)) - (1 - G)^(3^n)) where G = (1 + sqrt(5))/2;
a(n+2) = A045529(n+1)/A045529(n). (End)
From Peter Bala, Nov 15 2012: (Start)
a(n+1) = (1/2*(3 + sqrt(5)))^(3^n) + (1/2*(3 - sqrt(5)))^(3^n) - 1.
The sequence b(n):= a(n) + 2 is a solution to the recurrence b(n+1) = b(n)^3 - 3*b(n)^2 + 3 with b(1) = 4.
Other recurrence equations:
a(n+1) = -3 + 5*(Product_{k = 1..n} a(k)^2) with a(1) = 2.
a(n+1) = 1 + Product_{k = 1..n} (a(k) + 2)^2 with a(1) = 2.
Thus Y := Product_{k = 1..n} a(k) and X := Product_{k = 1..n} (a(k) + 2) give a solution to the Diophantine equation X^2 - 5*Y^2 = -4.
sqrt(5) = Product_{n >= 1} (1 + 2/a(n)). The rate of convergence is cubic. Fine remarks that twelve factors of the product would give well over 300,000 correct decimal digits for sqrt(5).
5 - {Product_{n = 1..N} (1 + 2/a(n))}^2 = 20/(a(N+1) + 3). (End)
a(n) = 2*T(3^(n-1),3/2) - 1 for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 06 2022

Definition improved by Reinhard Zumkeller, Feb 29 2012

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.

A145234 a(n) = Fibonacci(7^n).

Original entry on oeis.org

1, 13, 7778742049, 215414832505658809004682396169711233230800418578767753330908886771798637

Offset: 0

Artur Jasinski, Oct 05 2008

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), A145233 (k=6), this sequence (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

[Fibonacci(7^n): n in [0..5]]; // Vincenzo Librandi, Apr 02 2011

A145234 := proc(n) combinat[fibonacci](7^n) ; end proc:
seq(A145234(n),n=1..3) ; # R. J. Mathar, Apr 01 2011

G = (1 + Sqrt[5])/2; Table[Expand[(G^(7^n) - (1 - G)^(7^n))/Sqrt[5]], {n, 1, 6}]
(* Second program: *)
Table[Round[N[(2/Sqrt[5])*Cosh[7^n*ArcCosh[Sqrt[5]/2]], 1000]], {n, 1, 4}]

a(n) = (G^(7^n) - (1 - G)^(7^n))/sqrt(5) where G = (1 + sqrt(5))/2.
a(n) = (2/sqrt(5))*cosh(7^n*arccosh(sqrt(5)/2)).
a(n+1) = 125*a(n)^7 - 175*a(n)^5 + 70*a(n)^3 - 7*a(n) with a(0) = 1. - Peter Bala, Nov 25 2022

A244847 Decimal expansion of rho_c = (5-sqrt(5))/10, the asymptotic critical density for the hard hexagon model.

Original entry on oeis.org

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

Offset: 0

Jean-François Alcover, Nov 12 2014

The vertical distance between the accumulation point and the outermost point of a golden spiral inscribed inside a golden rectangle with dimensions phi and 1 along the x and y axes, respectively (the horizontal distance is A176015). - Amiram Eldar, May 18 2021

The limiting frequency of the digit 1 in the base phi representation of real numbers in the range [0,1], where phi is the golden ratio (A001622) (Rényi, 1957). - Amiram Eldar, Mar 18 2025

			0.2763932022500210303590826331268723764559381640388474275729102754589479...

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 1.2 The Golden Mean, phi, p. 7.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 5.12.1 Phase transitions in Lattice Gas Models, p. 347.

Rodney J. Baxter Hard hexagons: exact solution, Journal of Physics A: Mathematical and General, Vol. 13, No. 3 (1980), pp. L61-L70, alternative link.
P. S. Bruckman and I. J. Good, A Generalization of a Series of De Morgan with Applications of Fibonacci Type, The Fibonacci Quarterly, Vol. 14, No. 3 (1976), pp. 193-196.
Alfréd Rényi, Representations for real numbers and their ergodic properties, Acta Math. Acad. Sci. Hungar., Vol.8, No. 3-4 (1957), pp. 477-493.
Eric Weisstein's MathWorld, Hard Hexagon Entropy Constant.
Wikipedia, Hard Hexagon Model.
Index entries for algebraic numbers, degree 2.

Cf. A242671, A244593.
Cf. A000032, A000045.
Essentially the same sequence of digits as A229760 and A187799.

RealDigits[(5 - Sqrt[5])/10, 10, 102] // First

Equals 1/(sqrt(5)*phi), where phi = (1+sqrt(5))/2 = A001622. - Vaclav Kotesovec, Nov 13 2014
Equals lim_{n -> infinity} A000045(n)/A000032(n+1). - Bruno Berselli, Jan 22 2018
Equals Sum_{n>=1} A000045(3^(n-1))/A000032(3^n) = Sum_{n>=1} A045529(n-1)/A006267(n). - Amiram Eldar, Dec 20 2018
Equals 1 - A242671. - Amiram Eldar, Mar 18 2025

A250488 a(n) = Fibonacci(9^n).

Original entry on oeis.org

1, 34, 37889062373143906

Offset: 0

Alois P. Heinz, Nov 24 2014

Alois P. Heinz, Table of n, a(n) for n = 0..3

Row n = 9 of A250486. Bisection of A045529.

Cf. A000045.

a:= n-> (<<0|1>, <1|1>>^(9^n))[1, 2]:
seq(a(n), n=0..4);

a(n) = A000045(9^n).
From Peter Bala, Nov 25 2022: (Start)
a(n+1) = 625*a(n)^9 - 1125*a(n)^7 + 675*a(n)^5 - 150*a(n)^3 + 9*a(n) with a(0) = 1.
a(n) == 7 (mod 9) for n >= 1.
a(n+1) == a(n) mod (9^n).
5*a(n)^2 == 2 (mod 9^n).
In the ring of 9-adic integers, the sequence {a(n)} is a Cauchy sequence. It converges to a 9-adic root of the quadratic equation 5*x^2 - 2 = 0 (the 9-adic Cauchy sequence {Fibonacci(3*9^n)} converges to the other root). (End)

A128935 a(n) = Fibonacci(5^n) / 5^n.

Original entry on oeis.org

1, 1, 3001, 475400918060101145703001, 29642179764875707696452732234250095350341524541114277856812964100763567848899514572925690068090872073476146381237687662210078001

Offset: 0

Alexander Adamchuk, May 11 2007

Numbers k such that k divides Fibonacci(k) are listed in A023172.
All powers of 5 belong to A023172.
5^n divides Fibonacci(5^n).
a(n) == 1 (mod 1000).
{a(n+1)/a(n)} = {1, 3001, 158414167964045700001, 62351961552434956321060201440347372028390478647963811251289490034177804212636326088548682319305439375001, ...}.

Cf. A023172, A121169, A121170, A058635, A045529.

a := proc(n) option remember; if n = 0 then 1 else 5^(4*n-3)*a(n-1)^5 - 5^(2*n-1)*a(n-1)^3 + a(n-1) end if; end proc: seq(a(n), n = 0..5); # Peter Bala, Nov 24 2022

Table[ Fibonacci[ 5^n ] / 5^n, {n,0,4} ]

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

a(n+1) = 5^(4*n+1)*a(n)^5 - 5^(2*n+1)*a(n)^3 + a(n) with a(0) = 1. - Peter Bala, Nov 24 2022

A174652 Partial sums of A002814.

Original entry on oeis.org

1, 3, 20, 5797, 192900159414, 7177905237579946589743785824843591, 369822356418414944143680173221426891716916679027557977938929258031497305419444723776968084111223998808

Offset: 0

Jonathan Vos Post, Mar 25 2010

nonn

Partial sums of an infinite coprime sequence defined by recursion. 3 is the only prime through a(6). Is it computationally feasible to find another?

			a(3) = 1 + 2 + 17 + 5777 = 5797 = 11 * 17 * 31.

Cf. A002814, A000045, A001566, A045529.

a(n) = SUM[i=0..n] A002814(i) = SUM[i=0..n] {a(n) = a(n-1)^3 + 3a(n-1)^2 - 3}.

cp's OEIS Frontend

A002814 For n > 1: a(n) = a(n-1)^3 + 3a(n-1)^2 - 3; a(0) = 1, a(1) = 2.

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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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A145234 a(n) = Fibonacci(7^n).

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A244847 Decimal expansion of rho_c = (5-sqrt(5))/10, the asymptotic critical density for the hard hexagon model.

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