A058635 -id:A058635 - cp's OEIS frontend

A192222 a(n) = Fibonacci(2^n + 1).

1, 2, 5, 34, 1597, 3524578, 17167680177565, 407305795904080553832073954, 229265413057075367692743352179590077832064383222590237

Offset: 0

Jonathan Sondow, Jun 26 2011

a(n) is the numerator of the n-th iterate when Newton's method is applied to the function x^2 - x - 1 with initial guess x = 1. The n-th iterate is a(n)/A058635(n). - Jason Zimba, Jan 20 2023

John Gill and Matthew Miller, Newton's Method and Ratios of Fibonacci Numbers, Fibonacci Quarterly, 19(1):1-3, February 1981.
Jonathan Sondow, Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, Vol. 1385, No. 1 (2011), pp. 97-100, arXiv preprint, arXiv:1106.4246 [math.NT], 2011.
Yohei Tachiya, Transcendence of certain infinite products, J. Number Theory, Vol. 125, No. 1 (2007), pp. 182-200.

Cf. A000045 (Fibonacci numbers F(n)), A001622, A134973 (decimal expansion of 3/phi), A192223 (Lucas(2^n + 1)), A338305.

Table[Fibonacci[2^n + 1], {n, 0, 10}] (* T. D. Noe, Jan 11 2012 *)

a(n) = A000045(2^n + 1).
Product_{n>0} (1 + 1/a(n)) = 3/phi = A134973, where phi = (1+sqrt(5))/2 is the golden mean.
Sum_{n>=0} 1/a(n) = A338305. - Amiram Eldar, Oct 22 2020

A192223 a(n) = Lucas(2^n + 1).

Original entry on oeis.org

3, 4, 11, 76, 3571, 7881196, 38388099893011, 910763447271179530132922476, 512653048485188394162163283930413917147479973138989971

Offset: 0

Jonathan Sondow, Jun 26 2011

Product_{n>0} (1 + 1/a(n)) = 3 - phi = A094874, where phi = (1+sqrt(5))/2 is the golden mean.
From Peter Bala, Oct 28 2013: (Start)
Compare with A230600(n) = Lucas(2^n - 1).
Let x and b be positive real numbers. We define a Pierce expansion of x to the base b to be a (possibly infinite) increasing sequence of positive integers [a(1), a(2), a(3), ...] such that we have the alternating series representation x = b/a(1) - b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3)) - .... This definition generalizes the ordinary Pierce expansion of a real number 0 < x < 1, where the base b has the value 1. Depending on the values of x and b such a generalized Pierce expansion to the base b may not exist, and if it does exist it may not be unique.
Let Phi := 1/2*(sqrt(5) - 1) denote the reciprocal of the golden ratio. This sequence, apart from the initial term, provides a Pierce expansion of Phi^4 to the base Phi. That is we have the identity Phi^4 = Phi/4 - Phi^2/(4*11) + Phi^3/(4*11*76) - Phi^4/(4*11*76*3571) + ....
This result can be extended in two ways. Firstly, for k odd, the sequence {Lucas(k*(2^n + 1))}n>=1 gives a Pierce expansion of Phi^(4*k) to the base Phi^k. Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] gives a Pierce expansion of Phi^(2^n + 2) to the base Phi. See below for some examples. (End)

			Pierce series expansion of Phi^(2^n + 2) to the base Phi for n = 1 to 4:
n = 1:
Phi^4 = Phi/4 - Phi^2/(4*11) + Phi^3/(4*11*76) - Phi^4/(4*11*76*3571) + ...
n = 2:
Phi^6 = Phi/11 - Phi^2/(11*76) + Phi^3/(11*76*3571) - ...
n = 3:
Phi^10 = Phi/76 - Phi^2/(76*3571) + Phi^3/(76*3571*7881196) - ...
n = 4:
Phi^18 = Phi/3571 - Phi^2/(3571*7881196) + ...

J. Sondow, Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, vol. 1385, pp. 97-100.
Y. Tachiya, Transcendence of certain infinite products, J. Number Theory 125 (2007), 182-200.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A000032 (Lucas numbers L(n)), A094874 (decimal expansion of 3 - phi), A192222 (Fibonacci(2^n + 1)). A001622, A058635, A230600, A230601, A230602.

Table[LucasL[2^n + 1], {n, 0, 10}] (* T. D. Noe, Jan 11 2012 *)

a(n) = A000032(2^n + 1).
From Peter Bala, Oct 28 2013: (Start)
a(n) = phi^(2^n + 1) - (1/phi)^(2^n + 1), where phi = 1/2*(1 + sqrt(5)) denotes the golden ratio A001622.
Recurrence equation: a(0) = 3, a(1) = 4 and a(n) = floor(1/phi*a(n-1)^2) + 2 for n >= 2. (End)

A343202 Decimal expansion of Sum_{k>=0} 1/(k! * Fibonacci(2^k)).

Original entry on oeis.org

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

Offset: 1

Amiram Eldar, Jul 07 2021

The transcendence of this constant was proved independently by Mignotte (1974) and Mahler (1975).

			2.17464539389651955644333792522982188971668174552838...

Maurice Mignotte, Quelques problèmes d'effectivité en théorie des nombres, Thesis, Univ. Paris XIII, Paris, 1974.

Kurt Mahler, On the transcendency of the solutions of a special class of functional equations, Bulletin of the Australian Mathematical Society, Vol. 13, No. 3 (1975), pp. 389-410.

Cf. A000045, A000142, A058635, A079585.

RealDigits[Sum[1/(n!*Fibonacci[2^n]), {n, 0, 20}], 10, 100][[1]]

suminf(k=0, 1/(k!*fibonacci(2^k))) \\ Michel Marcus, Jul 07 2021

A230600 a(n) = Lucas(2^n - 1).

Original entry on oeis.org

2, 1, 4, 29, 1364, 3010349, 14662949395604, 347880681146567910619198829, 195816040085094172011386545446645681141059001652009364, 62041768337314169100816125405238438263014895218124648720624536859920496229610874552659773465281966850403949

Offset: 0

Peter Bala, Oct 28 2013

Compare with A192223(n) = Lucas(2^n + 1).
Let x and b be positive real numbers. We define a Pierce expansion of x to the base b to be a (possibly infinite) increasing sequence of positive integers [a(1), a(2), a(3), ...] such that we have the alternating series representation x = b/a(1) - b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3)) - .... This definition generalizes the ordinary Pierce expansion of a real number 0 < x < 1, where the base b is taken equal to 1. Depending on the values of x and b such a generalized Pierce expansion to the base b may not exist, and if it does exist it may not be unique.
Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio A001622. This sequence, apart from the initial term, provides a Pierce expansion of 1 to the base phi. That is, we have the identity 1 = phi/1 - phi^2/(1*4) + phi^3/(1*4*29) - phi^4/(1*4*29*1364) + ....
This result can be extended in two ways. Firstly, for k odd, the sequence {Lucas(k*(2^n - 1))} n>=1 gives a Pierce expansion of 1 to the base phi^k. Secondly, for n = 1,2,3,..., the sequence [a(n),a(n+1),a(n+2),...] provides a Pierce expansion of the quadratic irrational 1/phi^(2^n - 2) = Fibonacci(2^n - 1) - Fibonacci(2^n - 2)*phi to the base phi. Some examples are given below.

			A Pierce expansion of 1/phi^(2^n - 2) = Fibonacci(2^n - 1) - Fibonacci(2^n - 2)*phi to the base phi for n = 1 to 4.
n = 1: 1        = phi/1 - phi^2/(1*4) + phi^3/(1*4*29) - phi^4/(1*4*29*1364) + ...
n = 2: 1/phi^2  = phi/4 - phi^2/(4*29) + phi^3/(4*29*1364) - phi^4/(4*29*1364*3010349) + ...
n = 3: 1/phi^6  = phi/29 - phi^2/(29*1364) + phi^3/(29*1364*3010349) - ...
n = 4: 1/phi^14 = phi/1364 - phi^2/(1364*3010349) + ...

Iain Fox, Table of n, a(n) for n = 0..12
Iain Fox, Table of n, a(n) for n = 0..24 (terms too large for b-file)
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A000032, A001622, A058635, A192223, A230601, A230602.

[Lucas(2^n -1): n in [0..10]]; // G. C. Greubel, Dec 22 2017

Mathematica
```
Table[LucasL[2^n - 1], {n, 0, 10}]
```

for(n=0,10, print1(fibonacci(2^n) + fibonacci(2^n -2), ", ")) \\ G. C. Greubel, Dec 22 2017

a(n) = Lucas(2^n - 1) = A000032(2^n-1) = phi^(2^n-1) + (-1/phi)^(2^n-1).
Recurrence equation: a(0) = 2, a(1) = 1 and a(n) = floor(phi*a(n-1)^2) + 4 for n >= 2.
Product {n = 2..k} (1 - 1/a(n)) = 1/2 + Fibonacci(2^k - 1)/(2*Lucas(2^k - 1)).
Product {n >= 2} (1 - 1/a(n)) = (5 + sqrt(5))/10.
a(n) = A000032(A000225(n)). - Omar E. Pol, Oct 28 2013

A079613 a(n) = F(3*2^n) where F(k) denotes the k-th Fibonacci number.

Original entry on oeis.org

2, 8, 144, 46368, 4807526976, 51680708854858323072, 5972304273877744135569338397692020533504, 79757008057644623350300078764807923712509139103039448418553259155159833079730688

Offset: 0

Benoit Cloitre, Jan 29 2003

nonn

Let b = sqrt(5)/5. We have the alternating series identity (10 - 4*sqrt(5))/5 = b/2 - b^2/(2*8) + b^3/(2*8*144) - b^4/(2*8*144*46368) + ..., so this sequence is a generalized Pierce expansion of (10 - 4*sqrt(5))/5 to the base b as defined in A058635. - Peter Bala, Nov 04 2013

Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete mathematics, second edition, Addison Wesley, 1994, p. 557, ex. 6.61.

Amiram Eldar, Table of n, a(n) for n = 0..10
V. E. Hoggatt, Jr. and Marjorie Bicknell, A Reciprocal Series of Fibonacci Numbers with Subscripts 2^n k, The Fibonacci Quarterly, Vol. 14, No. 5 (1976), p. 453-455.

Cf. A000045, A001622 (phi), A007283, A058635, A081976, A083523.

[Fibonacci(3*2^n) : n in [0..7]]; // Wesley Ivan Hurt, Apr 05 2023

Table[Fibonacci[3*2^n], {n, 0, 7}] (* Amiram Eldar, Jan 29 2022 *)

Sum_{n>=0} 1/a(n) = 5/4 - 1/phi = 0.6319660112... since Sum_{k=0..n} 1/a(k) = 5/4 - F(3*2^n-1)/F(3*2^n).
a(n) = A081976(n+1)*A081976(n+2). - Amarnath Murthy, Apr 03 2003
a(n) = (1/sqrt(5))*( (2 + sqrt(5))^2^n - 1/(2 + sqrt(5))^2^n ) for n >= 1. - Peter Bala, Nov 04 2013
a(n) = A000045(A007283(n)). - Amiram Eldar, Jan 29 2022

A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)sqrt(21a(n-1)^2 + 4) for n >= 1.

Original entry on oeis.org

1, 5, 115, 60605, 16831644835, 1298263252133919638045, 7723873922612696850892381990249713732303715, 273388347343560518533856033712658350781293745092679040607342582493129736504927611387805

Offset: 0

Peter Bala, Oct 30 2013

For integer N, the recurrence equation a(n) = a(n-1)*sqrt((N^2 - 4)*a(n-1)^2 + 4) for n >= 1, with starting value a(0) = 1, produces an integer sequence. The present sequence is the case N = 5. Cf. A000079 (case N = 2), A058635 (case N = 3) and A071579 (case N = 4).
Sequence of numerators in the Engel series representation of 1/2*(7 - sqrt(21)) = 1 + 1/5 + 1 /115 + 1/60605 + .... The corresponding Engel expansion is A003487.
The sequence also has a description as a Pierce expansion of the quadratic irrational 1/2*(5 - sqrt(21)) to the base b := 1/sqrt(21) (see A058635 for a definition of this term).
The associated Pierce series representation of 1/2*(5 - sqrt(21)) to the base b begins 1/2*(5 - sqrt(21)) = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + ....
More generally, for n >= 0, the sequence [a(n), a(n+1), a(n+2), ...] gives a Pierce expansion of ( 1/2*(5 - sqrt(21)) )^(2^n) to the base b = 1/sqrt(21). Some examples are given below.

			Let b = 1/sqrt(21) and x = 1/2*(5 - sqrt(21)). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + b^5/(1*5*115*60605*16831644835) - ....
x^2 = b/5 - b^2/(5*115) + b^3/(5*115*60605) - b^4/(5*115*60605*16831644835) + ....
x^4 = b/115 - b^2/(115*60605) + b^3/(115*60605*16831644835) - ....
x^8 = b/60605 - b^2/(60605*16831644835) + ....

Seiichi Manyama, Table of n, a(n) for n = 0..10

Cf. A003487, A004254, A058635, A071579.

a[n_] := a[n - 1]*Sqrt[21 a[n - 1]^2 + 4]; a[0] = 1; Array[a, 8, 0] (* Robert G. Wilson v, Mar 19 2014 *)
a[ n_] := If[ n < 0, 0, ChebyshevU[2^n - 1, 5/2]]; (* Michael Somos, Dec 06 2016 *)

a(n) = if( n<0, 0, imag( (5 + quadgen(84))^2^n) / 2^(2^n - 1)); /* Michael Somos, Dec 06 2016 */

a(n) = 1/sqrt(21)*( alpha^(2^n) - (1/alpha)^(2^n) ), where alpha = 1/2*(5 + sqrt(21)).
a(n) = product {k = 0..n-1} A003487(k).
Defining recurrence equation:
a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.
Other recurrence equations:
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = 21*a(n-2)^2 + 2 for n >= 2.
a(n) = A004254(2^n). - Michael Somos, Dec 06 2016

A083697 a(n) = 2^(2^n - 1) * Fibonacci(2^n).

Original entry on oeis.org

1, 2, 24, 2688, 32342016, 4677882957791232, 97861912906883207538212742365184, 42829440312913272520181533609472356498655100482256687829780267008

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), May 22 2003

A083696(n)/a(n) converges to sqrt(5).

Similar to A081460: a(n) is the denominator of the same mapping f(r)=(1/2)(r+5/r) but with initial value r=1.

G. C. Greubel, Table of n, a(n) for n = 0..10

Cf. A000045, A058635, A058891, A081460, A083696.

[2^(2^n -1)*Fibonacci(2^n): n in [0..8]]; // G. C. Greubel, Jan 14 2022

Table[Sum[Product[2^n -k, {k,0,2*r}]k^r/(2*r+1)!, {r,0,2^n -1}], {n,0,8}]
Table[2^(2^n -1)*Fibonacci[2^n], {n,0,8}] (* G. C. Greubel, Jan 14 2022 *)

[2^(2^n -1)*lucas_number1(2^n, 1, -1) for n in (0..8)] # G. C. Greubel, Jan 14 2022

a(n) = 2*a(n-1)*A083696(n-1).
a(n) = A058635(n) * A058891(n).
a(n) = 2^(2^n - 1) * A000045(2^n).
a(n) = Sum_{r=0..(2^n -1)} (5^r/(2*r+1)!)*Product_{k=0..2*r} (2^n - k).

The next term is too large to include.

Better description from Ralf Stephan, Aug 29 2004

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

A074715 Number of prime factors of F(2^n) where F(m) is the m-th Fibonacci number.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 10, 13, 17, 20

Offset: 1

Benoit Cloitre, Sep 04 2002

a(11) = 20: F(2^11)/F(2^10) = 10077 286735 077005 660982 008061 065073 068074 475300 466012 444629 388487 574769 652115 651763 500026 128367 679301 744790 365920 278775 601766 000217 455997 930809 875108 639504 578766 853603 625505 162682 177708 433023 235042 368022 152858 871807 = 85 386449 571091 580927 (Curve 32) x 2359 309429 082740 633601 (Curve 34) x 50023 002657 441668 505734 051281 798632 006164 815525 563108 785740 688607 215220 816049 385513 420452 786904 754250 297381 898342 562475 311659 851996 080980 941281 231807 930745 342088 837303 971841
From Robert Israel, Apr 15 2015: (Start)
a(n+1) > a(n) since gcd(F(2^(n+1)), F(2^n)) = F(2^n).
a(n+1) = a(n) + 1 iff F(2^(n+1))/F(2^n) = F(2^n-1) + F(2^n+1) is prime, which is true for n <= 4 but not from n = 5 to at least 17. (End)
Prime factors counted with multiplicity. - Harvey P. Dale, Jun 09 2024

			a(10) = 17: 4506 699633 677819 813104 383235 728886 049367 860596 218604 830803 023149 600030 645708 721396 248792 609141 030396 244873 266580 345011 219530 209367 425581 019871 067646 094200 262285 202346 655868 899711 089246 778413 354004 103631 553925 405243 = 3 x 7 x 47 x 127 x 1087 x 2207 x 4481 x 21503 x 34303 x 119809 x 73327 699969 (Curve 2) x 186812 208641 x 455666 699738 584063 (Curve 27) x 4698 167634 523379 875583 x 1 019850 606646 830767 915009 (Curve 318) x 125 960894 984050 328038 716298 487435 392001 x 10 045901 211945 615185 770822 340063 765796 721091 525356 133475 714047

Cf. A001221 (omega(n)), A058635 (Fibonacci(2^n)).

PrimeOmega[Fibonacci[2^Range[11]]] (* Harvey P. Dale, Jun 09 2024 *)

PARI
```
a(n)=omega(fibonacci(2^n))
```

a(n) = A001221(A058635(n)). - Michel Marcus, Apr 15 2015

a(10) and a(11) from Jorge Coveiro, Jan 28 2006
Entry revised by N. J. A. Sloane, Feb 17 2006
a(9) corrected by Kellen Shenton, May 20 2022

A232326 Pierce expansion of 1 to the base Pi.

Original entry on oeis.org

3, 69, 310, 1017, 36745, 214369, 966652, 11159821, 74039764, 550021544, 4481549430, 16543857917, 87205978613, 476981856953, 30989048525367, 203786458494160, 711639924282497, 3174772986229899, 29814569078896025, 100158574806804154

Offset: 0

Peter Bala, Nov 26 2013

Let r and b be positive real numbers. We define a Pierce expansion of r to the base b to be a (possibly infinite) increasing sequence of positive integers [a(0), a(1), a(2), ...] such that we have the alternating series representation r = b/a(0) - b^2/(a(0)*a(1)) + b^3/(a(0)*a(1)*a(2)) - .... Depending on the values of r and b such an expansion may not exist, and if it does exist it may not be unique. When b = 1 and 0 < r < 1 we recover the ordinary Pierce expansion of r.
See A058635, A192223 and A230600 for some predictable Pierce expansions to a base b other than 1.
In the particular case that the base b >= 1 and 0 < r < b then we can find a Pierce expansion of r to the base b as follows:
Define the map f(x) (which depends on the base b) by f(x) = x/b*ceiling(b/x) - 1 and let f^(n)(x) denote the n-th iterate of the map f(x), with the convention that f^(0)(x) = x.
For n = 0,1,2,... define a(n) = ceiling(b/f^(n)(-r)) until f^n(-r) = 0.
Then it can be shown that the sequence of positive integers |a(n)| is a Pierce expansion of r to the base b.
For the present sequence we apply this algorithm with r := 1 and with base b := Pi. See A232325 for an Engel expansion of 1 to the base Pi.

Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A014014, A006784, A058635, A061233, A192223, A230600, A232325, A232327, A232328.

# Define the n-th iterate of the map f(x) = x/b*ceiling(b/x) - 1
map_iterate := proc(n,b,x) option remember;
if n = 0 then
   x
else
   -1 + 1/b*thisproc(n-1,b,x)*ceil(b/thisproc(n-1,b,x))
end if
end proc:
# Define the (signed) terms of the expansion of x to the base b
a := n -> ceil(evalf(b/map_iterate(n,b,x))):
Digits:= 500:
# Choose values for x and b
x := -1: b:= Pi:
seq(abs(a(n)), n = 0..19);

a(n) = ceiling(Pi/f^(n)(-1)), where f^(n)(x) denotes the n-th iterate of the map f(x) = x/Pi*ceiling(Pi/x) - 1, with the convention that f^(0)(x) = x.
Pierce series expansion of 1 to the base Pi:
1 = Pi/3 - Pi^2/(3*69) + Pi^3/(3*69*310) - Pi^4/(3*69*310*1017) + ....
The associated power series F(z) := 1 - ( z/3 - z^2/(3*69) + z^3/(3*69*310) - z^4/(3*69*310*1017) + ...) has a zero at z = Pi. Truncating the series F(z) to n terms produces a polynomial F_n(z) with rational coefficients which has a real zero close to Pi.

cp's OEIS Frontend

A192222 a(n) = Fibonacci(2^n + 1).

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A192223 a(n) = Lucas(2^n + 1).

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A343202 Decimal expansion of Sum_{k>=0} 1/(k! * Fibonacci(2^k)).

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A230600 a(n) = Lucas(2^n - 1).

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A079613 a(n) = F(3*2^n) where F(k) denotes the k-th Fibonacci number.

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Magma

Mathematica

Formula

A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.

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Mathematica

PARI

Formula

A083697 a(n) = 2^(2^n - 1) * Fibonacci(2^n).

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Magma

A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)sqrt(21a(n-1)^2 + 4) for n >= 1.