A001566 -id:A001566 - 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.

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)

A072191 a(n) = a(n-1)^2 + 2.

Original entry on oeis.org

0, 2, 6, 38, 1446, 2090918, 4371938082726, 19113842599189892819591078, 365338978906606237729724396156395693696687137202086, 133472569508521677503139972517335009022889462418844369330479463819154657319297609174034202576402751398

Offset: 0

Miklos Kristof, Jul 02 2002

This shows that in the Mandelbrot set (with z^2 + c), the point c = 2 escapes to infinity. - Alonso del Arte, Apr 08 2016

			0^2 + 2 = 2, 2^2 + 2 = 6, 6^2 + 2 = 38 ...

Mordechai Ben-Ari, Mathematical Logic for Computer Science, Third edition, 173-203

Samuel R. Buss, Herbrand's Theorem, University of California, San Diego La Jolla, California 92093-0112, U.S.A.
Neil J. Calkin, Eunice Y. S. Chan, and Robert M. Corless, Some Facts and Conjectures about Mandelbrot Polynomials, Maple Trans., Vol. 1, No. 1, Article 1 (July 2021).
Alessandro Farinelli, Herbrand Universe
Eric Weisstein's World of Mathematics, Weakly Binary Tree
Wikipedia, Herbrand Structure
Damiano Zanardini, Computational Logic, UPM European Master in Computational Logic (EMCL) School of Computer Science Technical University of Madrid, 2009-2010.
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A001566 (a(n-1)^2-2), A003095 (a(n-1)^2+1).

[n le 1 select 0 else Self(n-1)^2+2: n in [1..10]]; // Vincenzo Librandi, Oct 05 2015

NestList[#^2 + 2 &, 0, 10]  (* Harvey P. Dale, Jan 23 2011 *)

a(n)=if(n<1, 0, 2+a(n-1)^2) /* Michael Somos, Mar 25 2006 */

a(n) ~ c^(2^n), where c = 1.57583423499194129500626808486999436507... - Vaclav Kotesovec, Sep 20 2013

a(n) mod 2 = 0. - Altug Alkan, Oct 04 2015

Edited by Robert G. Wilson v, Jul 03 2002

A144836 a(n) = round(phi^(4^n)) where phi is the golden ratio (A001622).

Original entry on oeis.org

2, 7, 2207, 23725150497407, 316837008400094222150776738483768236006420971486980607

Offset: 0

Artur Jasinski, Sep 22 2008

Amiram Eldar, Table of n, a(n) for n = 0..6

Cf. A000032, A001566, A001622, A006267, A144837, A144838, A144839.

Cf. A374883 (phi^4).

a := proc(n) option remember; if n = 0 then 2 elif n = 1 then 7 else a(n-1)^4 - 4*a(n-1)^2 + 2 end if; end proc: seq(a(n), n = 0..4); # Peter Bala, Nov 28 2022

Table[Round[GoldenRatio^(4^n)], {n, 0, 5}]
c = (1 + Sqrt[5])/2; Join[{2}, Table[Expand[c^(4^n) + (1 - c)^(4^n)], {n, 1, 5}]] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[2*Cosh[4^n*ArcCosh[Sqrt[5]/2]]], {n, 0, 5}] (* Artur Jasinski, Oct 09 2008 *)
a[n_] := LucasL[4^n]; a[0] = 2; Array[a, 5, 0] (* Amiram Eldar, Jul 12 2025 *)

a(n)=round(((1+sqrt(5))/2)^4^n) \\ Charles R Greathouse IV, Jul 29 2011

a(n) = Lucas(4^n) = A000032(4^n), n>0.
a(n) = phi^(4^n) + (1 - phi)^(4^n) = phi^(4^n) + (-phi)^(-4^n), where phi is golden ratio = (1 + sqrt(5))/2 = 1.6180339887..., n>0. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(4^n*arccosh(sqrt(5)/2)), n>0. - Artur Jasinski, Oct 09 2008
a(n+1) = a(n)^4 - 4*a(n-1)^2 + 2 with a(1) = 7. - Peter Bala, Nov 28 2022

Offset corrected by Charles R Greathouse IV, May 15 2013
Offset changed to 0 by Georg Fischer, Sep 02 2022
New name from Peter Bala, Nov 18 2022
Revised by editors, Jul 12 2025

A144838 a(n) = Lucas(6^n).

Original entry on oeis.org

18, 33385282, 1384619022984618483717737087933569992335566082

Offset: 1

Artur Jasinski, Sep 22 2008

nonn

Previous name was: a(n) = round(phi^(6^n)) where phi = 1.6180339887498948482... = (sqrt(5)+1)/2.

General (hyperbolic) trigonometric formula for a(n) = round(phi^((2*k)^n)) = 2*cosh((2*k)^n*arccosh(sqrt(5)/2)) where phi = 1.6180339887498948482... = (sqrt(5)+1)/2. - Artur Jasinski, Oct 09 2008

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

Cf. A001566, A006267, A144836, A144837, A144838, A144839.

a := proc(n) option remember; if n = 1 then 18 else a(n-1)^6 - 6*a(n-1)^4 + 9*a(n-1)^2 - 2 end if; end;
seq(a(n), n = 1..5); # Peter Bala, Nov 28 2022

Table[Round[GoldenRatio^(6^n)], {n, 1, 5}]
c = (1 + Sqrt[5])/2; Table[Expand[c^(6^n) + (1 - c)^(6^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[2*Cosh[6^n*ArcCosh[Sqrt[5]/2]]], {n, 1, 4}] (* Artur Jasinski, Oct 09 2008 *)
Table[LucasL[6^n], {n, 1, 4}] (* Amiram Eldar, Jul 13 2025 *)

a(n) = G^(6^n) + (1 - G)^(6^n) = G^(6^n) + (-G)^(-6^n) where G is the golden ratio A001622. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(6^n*arccosh(sqrt(5)/2)). - Artur Jasinski, Oct 09 2008
From Peter Bala, Nov 28 2022: (Start)
a(n) = Lucas(6^n).
a(n+1) = a(n)^6 - 6*a(n)^4 + 9*a(n)^2 - 2 with a(1) = 18. (End)

New name from Peter Bala, Nov 28 2022

A144839 a(n) = Lucas(7^n).

Original entry on oeis.org

29, 17393796001, 481682208844384447843365760878364816732549453120338354329505085763436029

Offset: 1

Artur Jasinski, Sep 22 2008

nonn

Previous name was: a(n) = round(phi^(7^n)) where phi = 1.6180339887498948482... = (sqrt(5)+1)/2.

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

Cf. A000032, A001622, A001566, A006267, A144836, A144837, A144838, A144839.

a := proc(n) option remember; if n = 0 then 1 else a(n-1)^7 + 7*a(n-1)^5 + 14*a(n-1)^3 +7*a(n-1) end if; end;
seq(a(n), n = 1..5); # Peter Bala, Nov 28 2022

Table[Round[GoldenRatio^(7^n)], {n, 1, 5}]
c = (1 + Sqrt[5])/2; Table[Expand[c^(7^n) + (1 - c)^(7^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
Table[LucasL[7^n], {n, 1, 4}] (* Amiram Eldar, Jul 13 2025 *)

a(n) = G^(7^n) + (1 - G)^(7^n) = G^(7^n) + (-G)^(-7^n) where G is the golden ratio A001622. [Artur Jasinski, Oct 05 2008]
From Peter Bala, Nov 28 2022: (Start)
a(n) = Lucas(7^n).
a(n+1) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n) with a(0) = 1.
a(n) == 1 (mod 7).
a(n+1) == a(n) (mod 7^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
In the ring of 7-adic integers, the limit_{n -> oo} a(n) exists and is a root of the quartic equation x^4 + 4*x^2 + 2 = 0. (End)

New name from Peter Bala, Nov 28 2022

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

A094057 Number of decimal digits of Lucas(2^n).

Original entry on oeis.org

1, 1, 1, 2, 4, 7, 14, 27, 54, 108, 215, 429, 857, 1713, 3425, 6849, 13697, 27393, 54785, 109570, 219140, 438279, 876558, 1753116, 3506231, 7012462, 14024924, 28049847, 56099693, 112199386, 224398771, 448797541, 897595081, 1795190161, 3590380321, 7180760642, 14361521283

Offset: 0

Matthijs Coster, Apr 29 2004

a(n+1) is the number of decimal digits of A001566(n).
From Hans J. H. Tuenter, Jul 24 2025: (Start)
This sequence can be constructed by taking the first n digits of the binary expansion of
alpha = log_10(phi) = 0.00110 10110 00000 00011 ...
For example, expressing a(n)-1 in binary notation, gives
 a(0)-1 = 0,
 a(1)-1 = 0,
 a(2)-1 = 0,
 a(3)-1 = 1,
 a(4)-1 = 11,
 a(5)-1 = 110,
 a(6)-1 = 1101,
 a(7)-1 = 11010.
Another way of deriving the sequence is by the recurrence a(n+1)=2a(n)-1+d(n+1), with initial value a(0)=1, and d(n) the n-th digit in the binary expansion of alpha.
 a(0) = 1,
 a(1) = 2*1-1+0=1,
 a(2) = 2*1-1+0=1,
 a(3) = 2*1-1+1=2,
 a(4) = 2*2-1+1=4,
 a(5) = 2*4-1+0=7,
 a(6) = 2*7-1+1=14,
 a(7) = 2*14-1+0=27.
Alternatively, a(n) provides a encoding of the digits in the binary expansion of alpha,
 d(n) = a(n)+1-2a(n-1). (End)

			a(5)=7, as L(2^5)=L(32)=4870847 and has seven digits.

Hans J. H. Tuenter, Table of n, a(n) for n = 0..100

Cf. A000032, A001566, A001622, A055642, A114469.

a(n) = length(Str(fibonacci(2^(n+1))/fibonacci(2^n))); \\ adapted to new name by Michel Marcus, Jul 24 2025

a(n) = 1+floor(2^n*log_10(phi)), where phi=(1+sqrt(5))/2, the golden ratio. - Hans J. H. Tuenter, Jul 23 2025

a(n) = 1 + Sum_{i=0..n} d(i)*2^(n-i), where d(i) is the i-th digit in the binary expansion of log_10(phi). - Hans J. H. Tuenter, Jul 24 2025

More terms from Jason Earls, Apr 30 2004
a(23)-a(36) from Arkadiusz Wesolowski, Jul 20 2012
Name edited and a(0)=1 inserted by Hans J. H. Tuenter, Jul 23 2025

A186750 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 3.

Original entry on oeis.org

3, 6, 33, 1086, 1179393, 1390967848446, 1934791555410494424614913, 3743418362887760317407541271559358491868341997566

Offset: 0

Jonathan Vos Post, Feb 26 2011

This is to A001566 as 3 is to 2 (subtrahend). Unlike A001566, which begins with 4 consecutive primes, this sequence can never be prime after a(0) = 3, because the first two terms are both multiples of 3, hence all later terms are. This is the k = 3 row of the array A(k, 0) = 3, A(k, n) = A(k, n-1)^2 - k; and A001566 is the k = 2 row. A003096(n+1) is the k = 1 row.

Vincenzo Librandi, Table of n, a(n) for n = 0..11
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A001566, A003096.

RecurrenceTable[{a[0] == 3, a[n] == a[n-1]^2 - 3}, a, {n, 0, 10}] (* Vaclav Kotesovec, Dec 18 2014 *)
Drop[Abs[NestList[#^2 - 3 &, 0, 9]], 1] (* Alonso del Arte, Apr 08 2016 *)

a(n) ~ c^(2^n), where c = 2.3959550115176494685408322564302422183669584045032057908382914927198090627... - Vaclav Kotesovec, Dec 18 2014

cp's OEIS Frontend

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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A072191 a(n) = a(n-1)^2 + 2.

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A144836 a(n) = round(phi^(4^n)) where phi is the golden ratio (A001622).

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A144838 a(n) = Lucas(6^n).

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A144839 a(n) = Lucas(7^n).

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

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