Jose Eduardo Blazek - cp's OEIS frontend

A238290 a(n+1) = a(n) + 6 + 2(n - 2floor(n/2)) for n > 0, a(0) = 0.

0, 8, 14, 22, 28, 36, 42, 50, 56, 64, 70, 78, 84, 92, 98, 106, 112, 120, 126, 134, 140, 148, 154, 162, 168, 176, 182, 190, 196, 204, 210, 218, 224, 232, 238, 246, 252, 260, 266, 274, 280, 288, 294, 302, 308, 316, 322, 330, 336, 344, 350, 358, 364, 372, 378

Offset: 0

Jose Eduardo Blazek, Feb 22 2014

			G.f.: 8*x + 14*x^2 + 22*x^3 + 28*x^4 + 36*x^5 + 42*x^6 + 50*x^7 + 56*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1, 1, -1).

Cf. A047345.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(2*x*(4+3*x)/((1-x)*(1-x^2)))); // G. C. Greubel, Aug 07 2018

CoefficientList[Series[2 x (4 + 3 x)/((1 - x) (1 - x^2)), {x, 0, 80}], x] (* Vincenzo Librandi, Feb 26 2014 *)
Table[(14 n - (-1)^n + 1)/2, {n, 0, 60}] (* Bruno Berselli, Feb 26 2014 *)

x='x+O('x^50); concat([0], Vec(2*x*(4+3*x)/((1-x)*(1-x^2)))) \\ G. C. Greubel, Aug 07 2018

a(n) = floor((2*n+1)/4) + 2*floor((2*n+1)/2) + 3*floor(3*(2*n+1)/4).
a(n) = 8*floor((n+1)/2) + 6*floor(n/2).
G.f.: 2 * x * (4 + 3*x) / ((1 - x) * (1 - x^2)). - Michael Somos, Feb 24 2014
a(n) = 2*A047345(n+1) = (14*n - (-1)^n + 1)/2. - Bruno Berselli, Feb 26 2014
E.g.f.: 7*exp(x)*x + sinh(x). - Stefano Spezia, May 15 2021

A213680 a(n) = 2*a(n-1)^2/3-3 with a(0)=6.

Original entry on oeis.org

6, 21, 291, 56451, 2124476931, 3008934820234119171, 6035792501611554034238453484153151491, 24287194081673507672666338605180770497437885188248737771493963111463682051

Offset: 0

Jose Eduardo Blazek, Mar 04 2013

nonn

			a(1) = 2*a(0)^2/3-3 = 2*6^2/3-3 = 21,
a(2) = 2*a(1)^2/3-3 = 2*21^2/3-3 = 291,
a(3) = 2*a(2)^2/3-3 = 2*291^2/3-3 = 2*84681/3-3 = 56451.
Or, by the first formula:
a(3) = 3*cosh(2^3*arccosh(2)) = 56451,
a(4) = 3*cosh(2^4*arccosh(2)) = 2124476931.

Cf. A002812.

var('x')
p=2*x^2/3-3
s=[6]
for i in [0..8]:
    s=s+[p(s[i])]
show(s)

a(n) = 3*cosh(2^n*arccosh(2)).

a(n) = 3*A002812(n). [Giovanni Resta, Mar 04 2013]

A213681 a(n) = a(n-1)^2/2 - 4 with a(0) = 6.

Original entry on oeis.org

6, 14, 94, 4414, 9741694, 47450300994814, 1125765532249223239027447294, 633674016800188444301553476967536472012841942973961214

Offset: 0

Jose Eduardo Blazek, Mar 04 2013

nonn

The next term has 108 digits. - Harvey P. Dale, Apr 21 2018

			a(2) = a(1)^2/2-4 = 14^2/2-4 = 94.

Cf. A005248.

NestList[#^2/2-4&,6,8] (* Harvey P. Dale, Apr 21 2018 *)

var('x')
p=x^2/2-4
s=[6]
for i in [0..8]:
    s=s+[p(s[i])]
show(s)

a(n) = 4*cosh(2^n*arccosh(3/2)).

a(n) = 2*A005248(2^n). [Bruno Berselli, Mar 04 2013]

A094680 a(n+1) = 4a(n)^3 - 3a(n), with a(0) = 2.

Original entry on oeis.org

2, 26, 70226, 1385331749802026, 10634604778476758291777057017318241822792488226

Offset: 0

Jose Eduardo Blazek, Jun 07 2004

Smallest positive integer x satisfying the Pell equation x^2 - 3^(2*n-3) * y^2 = 1. - A.H.M. Smeets, Sep 29 2017

Term a(5) has 139 decimal digits and a(6) has 417 decimal digits. - Andrew Howroyd, Feb 25 2018

Andrew Howroyd, Table of n, a(n) for n = 0..6

Cf. A001075, A002812.

NestList[4 #^3 - 3 # &, 2, 5] (* Michael De Vlieger, Oct 02 2017 *)

a(n) = if (n==0, 2, 4*a(n-1)^3 - 3*a(n-1)); \\ Michel Marcus, Oct 03 2017

a(n) = polchebyshev(3^n, 1, 2); \\ Michel Marcus, Oct 03 2017

a(n) = cosh(3^n*arccosh(2)).
a(n) = ChebyshevT(3^n, 2). - Vladeta Jovovic, Jun 11 2004
From A.H.M. Smeets, Oct 02 2017: (Start)
a(n) = A001075(3^(n-2))
a(n) = A002350(3^(2n-3)). (End)

More terms from Vladeta Jovovic, Jun 11 2004

Offset corrected by Michel Marcus, Oct 03 2017

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)

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User: Jose Eduardo Blazek

A238290 a(n+1) = a(n) + 6 + 2*(n - 2*floor(n/2)) for n > 0, a(0) = 0.

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A213680 a(n) = 2*a(n-1)^2/3-3 with a(0)=6.

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A213681 a(n) = a(n-1)^2/2 - 4 with a(0) = 6.

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A094680 a(n+1) = 4*a(n)^3 - 3*a(n), with a(0) = 2.

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

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A238290 a(n+1) = a(n) + 6 + 2(n - 2floor(n/2)) for n > 0, a(0) = 0.

A094680 a(n+1) = 4a(n)^3 - 3a(n), with a(0) = 2.

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