A002814 -id:A002814 - cp's OEIS frontend

A174652 Partial sums of A002814.

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}.

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)

A001999 a(n) = a(n-1)*(a(n-1)^2 - 3).

Original entry on oeis.org

3, 18, 5778, 192900153618, 7177905237579946589743592924684178

Offset: 0

N. J. A. Sloane

The next terms in the sequence contain 102 and 305 digits. - Harvey P. Dale, Jun 09 2011
From Peter Bala, Nov 13 2012: (Start)
The present sequence is the case x = 3 of the following general remarks. For other cases see A219160 (x = 4), A219161 (x = 5) and A112845 (x = 6).
Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(3^n) + (1/alpha)^(3^n). Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^3 - 3*a(n) with the initial condition a(0) = x.
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. Iterating the identity produces the product expansion sqrt((x+2)/(x-2)) = Product_{n = 0..oo} (1 + 2/(a(n) - 1)), with a(0) = x and a(n+1) = a(n)^3 - 3*a(n). The rate of convergence is cubic (Fine).
For similar results to the above see A001566 and A219162. (End)
Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. - Peter Bala, Dec 08 2022

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).

G. C. Greubel, Table of n, a(n) for n = 0..7
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, Vol. 44, No. 10 (1937), pp. 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8 (Oct. 1977), pp. 629-630.
Walther Janous, Problem B-916, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 39, No. 2 (2001), p. 181; Subscript Is Power, Solution to Problem B-916 by H.-J. Seiffert, ibid., Vol. 40, No. 1 (2002), p. 86.
Hideyuki Ohtsu, Problem B-1316, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 4 (2022), p. 365.
Eric Weisstein's World of Mathematics, Pierce Expansion.

Cf. A000032, A006276, A001566, A002814, A045529, A112845, A219160, A219161, A219162.

NestList[#(#^2-3)&,3,6] (* Harvey P. Dale, Jun 09 2011 *)
RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 3}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)

a(n)=2*fibonacci(2*3^n+1)-fibonacci(2*3^n)

a(n) = 2*F(2*3^n+1) - F(2*3^n) = ceiling(tau^(2*3^n)) where F(k) = A000045(k) is the k-th Fibonacci number and tau is the golden ratio. - Benoit Cloitre, Nov 29 2002
From Peter Bala, Nov 13 2012: (Start)
a(n) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n).
Product_{n >= 0} (1 + 2/(a(n) - 1)) = sqrt(5).
a(n) = A002814(n+1) + 1. (End)
a(n) = 2*T(3^n,3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A219161. - Peter Bala, Feb 01 2017
From Amiram Eldar, Jan 12 2022: (Start)
a(n) = A000032(2*3^n).
a(n) = A006267(n)^2 + 2.
Product_{k=0..n} (a(k)-1) = Fibonacci(3^(n+1)) = A045529(n+1) (Janous, 2001). (End)
Sum_{n>=0} arctanh(1/a(n)) = log(5)/4 (Ohtsu, 2022). - Amiram Eldar, Dec 15 2022

A145275 a(n) = A145232(n+1)/A145232(n).

Original entry on oeis.org

15005, 792070839820228500005, 311759807762174781605301007201736860141952393239819056256447450170889021063181630442743411596527196875005

Offset: 1

Artur Jasinski, Oct 06 2008

nonn

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 this sequence.
For k=6 see A145276.
For k=7 see A145277.

Amiram Eldar, Table of n, a(n) for n = 1..4

Cf. A001566, A001622, A002814, A145232, A145274, A145276, A145277.

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

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

A002813 a(0) = 4; for n > 0, a(n) = a(n-1)^3 - 3*a(n-1)^2 + 3.

Original entry on oeis.org

4, 19, 5779, 192900153619, 7177905237579946589743592924684179, 369822356418414944143680173221426891716916679027557977938929258031490127514207143830378340325399155219

Offset: 0

N. J. A. Sloane

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
The next term, a(7), has 305 digits. - Harvey P. Dale, Jul 19 2011
From Peter Bala, Nov 22 2012: (Start)
The present sequence is the case x = 1 of the following general remarks about the recurrence a(n+1) = a(n)^3 - 3*a(n-1)^2 + 3. Cf. A002814.
Define a sequence of polynomials P(n,x) inductively by setting P(0,x) = x^2 + 3 and P(n+1,x) = P(n,x^3 + 3*x) for n >= 0. Then P(n,x) satisfies the cubic recurrence P(n+1,x) = P(n,x)^3 - 3*P(n-1,x)^2 + 3 with the initial condition P(0,x) = x^2 + 3.
An explicit formula is P(n,x) = Q(3^(n+1),x)/Q(3^n,x), where Q(n,x) = ((x + sqrt(x^2 + 4))/2)^n + ((x - sqrt(x^2 + 4))/2)^n.
Alternatively, P(n,x) = ((x^2 + 2 + sqrt(x^4 + 4*x^2))/2)^(3^n) + ((x^2 + 2 - sqrt(x^4 + 4*x^2))/2)^(3^n) + 1.
Iterating the algebraic identity x/sqrt(x^2 + 4) = (1 - 2/(x^2 + 3))*y/sqrt(y^2 + 4), where y = x^3 + 3*x, leads to the product expansion x/sqrt(x^2 + 4) = Product_{n = 0..oo} (1 - 2/P(n,x)). See Escott and also Fine.
The sequence A(n,x) := x*Product_{k = 0..n} P(k,x) satisfies the recurrence A(n+1,x) = A(n,x)^3 + 3*A(n,x). These sequences occur in the continued cotangent expansions of Lehmer. Cases currently in the database are A006267 (x = 1), A006266 (x = 2), A006268 (x = 3), A006269 (x = 5) and A145180 through A145189 (x = 6 through x = 15).
(End)

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).

Vincenzo Librandi, 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)
J. Shallit Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.
Eric Weisstein's World of Mathematics, Lehmer Cotangent Expansion

Cf. A002814, A006267.

[Lucas(2*3^n)+1: n in [0..5]]; // Vincenzo Librandi, Jul 20 2011

NestList[#^3-3#^2+3&,4,6] (* Harvey P. Dale, Jul 19 2011 *)

a(n)=if(n<1,4*(n==0),a(n-1)^3-3*a(n-1)^2+3)

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

a(n) = L(2*3^n)+1 where L=Lucas numbers.
a(n) = L(3^(n+1))/L(3^n). - Benoit Cloitre, Sep 18 2005
a(n) = A001999(n)+1. - R. J. Mathar, Apr 22 2007
From Peter Bala, Nov 22 2012: (Start)
a(n) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n) + 1.
(1/5)*sqrt(5) = Product_{n = 0..oo} (1 - 2/a(n)).
A006267(n+1) = Product_{k = 0..n} a(k).
A002814(n+1) = a(n) - 2. (End)

A145274 a(n) = A145231(n+1)/A145231(n).

Original entry on oeis.org

329, 10749959329, 13354478338703157414450712411084788083329

Offset: 1

Artur Jasinski, Oct 06 2008

nonn

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 this sequence.
For k=5 see A145275.
For k=6 see A145276.
For k=7 see A145277.

Amiram Eldar, Table of n, a(n) for n = 1..5

Cf. A001566, A002814, A145231, A145274, A145275, A145276, A145277.

G = (1 + Sqrt[5])/2; Table[Expand[(G^(4^(n + 1)) - (1 - G)^(4^(n + 1)))/Sqrt[5]]/Expand[(G^(4^n) - (1 - G)^(4^n))/Sqrt[5]], {n, 1, 5}]

a(n) = (G^(4^(n + 1)) - (1 - G)^(4^(n + 1)))/(G^(4^n) - (1 - G)^(4^n)) where G = (1 + sqrt(5))/2.

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

Original entry on oeis.org

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.

A219162 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 3.

Original entry on oeis.org

3, 47, 4870847, 562882766124611619513723647

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A001566. Compare the following remarks with A001999.
The present sequence is the case x = 3 of the following general remarks. For other cases see A219163 (x = 4), A219164 (x = 5) and A219165 (x = 6).
Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(4^n) + (1/alpha)^(4^n). Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^4 + 4*a(n)^2 - 2 with the initial condition a(0) = x.
We have the product expansion sqrt((x + 2)/(x - 2)) = Product_{n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)).

Cf. A000032, A001566, A001999, A002814, A145502, A219163, A219164, A219165.

a(n)={if(n==0,3,a(n-1)^4-4*a(n-1)^2+2)} \\ Edward Jiang, Sep 11 2014

Let alpha = 1/2*(3 + sqrt(5)) then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A001566(2*n) = A000032(2*4^n).
Product {n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(5).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)

A145276 a(n) = A145233(n+1)/A145233(n).

Original entry on oeis.org

1866294, 41473935220454958813340461622291147206

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 this sequence.
For k=7 see A145277.

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

Cf. A001566, A001622, A002814, A145233, A145274, A145275, A145277.

G = (1 + Sqrt[5])/2; Table[Expand[(G^(6^(n + 1)) - (1 - G)^(6^(n + 1)))/Sqrt[5]]/Expand[(G^(6^n) - (1 - G)^(6^n))/Sqrt[5]], {n, 1, 5}]

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

A181419 Numbers of the form Fibonacci(p^{k+1})/Fibonacci(p^k) where p are primes, k>=1.

Original entry on oeis.org

3, 7, 17, 47, 2207, 5777, 15005, 4870847, 598364773, 192900153617, 23725150497407, 792070839820228500005, 97415813466381445596089, 562882766124611619513723647, 400009475456580321242184872389193

Offset: 1

Vladimir Shevelev, Oct 18 2010

nonn

The union of A001566 (p=2), A002814 except the first two terms (p=3), A145275 (p=5), A145277 (p=7), etc.

Robert Israel, Table of n, a(n) for n = 1..44

Cf. A001566, A002814, A145275, A145277.

N:= 10^50: # for terms <= N
S:= {}: p:= 1:
do
 p:= nextprime(p);
 v:= combinat:-fibonacci(p);
 for k from 2 do
   w:= v;
   v:= combinat:-fibonacci(p^k);
   r:= v/w;
   if r > N then break fi;
   S:= S union {r};
 od;
 if k = 2 then break fi;
od:
sort(convert(S,list)); # Robert Israel, Apr 09 2024

t = Sort@ Flatten[ Table[ {Prime[n]^(e + 1), Prime[n]^e}, {n, 8}, {e, 10}], 1]; u = Select[t, First@# < 350 &]; Sort[ Fibonacci[ #[[1]]]/Fibonacci[ #[[2]]] & /@ u] (* Robert G. Wilson v, Oct 21 2010 *)

a(11) onwards from Robert G. Wilson v, Oct 21 2010

cp's OEIS Frontend

A174652 Partial sums of A002814.

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

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A001999 a(n) = a(n-1)*(a(n-1)^2 - 3).

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A145275 a(n) = A145232(n+1)/A145232(n).

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A002813 a(0) = 4; for n > 0, a(n) = a(n-1)^3 - 3*a(n-1)^2 + 3.

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Magma

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A145274 a(n) = A145231(n+1)/A145231(n).

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

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A219162 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 3.

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