A188378 -id:A188378 - cp's OEIS frontend

A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.

1, 19, 341, 6119, 109801, 1970299, 35355581, 634430159, 11384387281, 204284540899, 3665737348901, 65778987739319, 1180356041958841, 21180629767519819, 380070979773397901, 6820097006153642399, 122381675130992165281, 2196050055351705332659, 39406519321199703822581

Offset: 0

Clark Kimberling

x(n) := 2*a(n) and y(n) := A007805(n), n >= 0, give all the positive solutions of the Pell equation x^2 - 5*y^2 = -1.
The Gregory V. Richardson formula follows from this. - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
2*a(n) = 2 o 2 o ... o 2 (2*n+1 terms). For example, 2 o 2 = 4*sqrt(5) and 2 o 2 o 2 = 2 o 4*sqrt(5) = 38 = 2*a(1). Cf. A084068.
a(n) = U(2*n+1) where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = sqrt(20)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/4)*( (sqrt(5) + 2)^n - (sqrt(5) - 2)^n ). (End)

			Pell, n=1: (2*19)^2 - 5*17^2 = -1.

G. C. Greubel, Table of n, a(n) for n = 0..795
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Tanya Khovanova, Recursive Sequences.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Eric Weisstein's World of Mathematics, Lehmer Number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials..
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Bisection of A001077 divided by 2.
Cf. A000045, A007805, A049310, A049660, A100047, A188378.
Cf. A005013, A033890, A049685, A084608.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.

[(Fibonacci(6*n+5) - Fibonacci(6*n+1))/4: n in [0..30]]; // G. C. Greubel, Dec 15 2017

with(numtheory): with(combinat):
seq((fibonacci(6*n+5)-fibonacci(6*n+1))/4,n=0..20); # Muniru A Asiru, Mar 25 2018

a[n_] := Simplify[(2 + Sqrt@5)^(2 n - 1) + (2 - Sqrt@5)^(2 n - 1)]/4; Array[a, 16] (* Robert G. Wilson v, Oct 28 2010 *)

my(x='x+O('x^30)); Vec((1+x)/(1 - 18*x + x^2)) \\ G. C. Greubel, Dec 15 2017

a(n) ~ (1/4)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all members x of the sequence, 20*x^2 + 5 is a square. Lim_{n -> oo} a(n)/a(n-1) = 9 + 2*sqrt(20) = 9 + 4*sqrt(5). The 20 can be seen to derive from the statement "20*x^2 + 5 is a square". - Gregory V. Richardson, Oct 12 2002
a(n) = (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)). - Gregory V. Richardson, Oct 12 2002
From R. J. Mathar, Nov 04 2008: (Start)
G.f.: (1+x)/(1 - 18x + x^2).
a(n) = A049660(n) + A049660(n+1). (End)
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=1, a(1)=19. - Philippe Deléham, Nov 17 2008
a(n) = S(n,18) + S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(6*n + 6 - 2*k) + Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) - Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer, k != 1.
The aerated sequence (b(n))n>=1 = [1, 0, 19, 0, 341, 0, 6119, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -16, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = (A188378(n)^3 + (A188378(n)-2)^3) / 8. - Altug Alkan, Jan 24 2016
a(n) = sqrt(5 * Fibonacci(3 + 6*n)^2 - 4)/4. - Gerry Martens, Jul 25 2016
a(n) = Lucas(6*n + 3)/4. - Ehren Metcalfe, Feb 18 2017
From Peter Bala, Mar 23 2018: (Start)
a(n) = 1/4*( (sqrt(5) + 2)^(2*n+1) - (sqrt(5) - 2)^(2*n+1) ).
a(n) = 9*a(n-1) + 2*sqrt(5 + 20*a(n-1)^2).
a(n) = (1/2)*sinh((2*n + 1)*arcsinh(2)). (End)
From Peter Bala, May 09 2025: (Start)
a(n)^2 - 18*a(n)*a(n+1) + a(n+1)^2 = 20.
More generally, for real x, a(n+x)^2 - 18*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 20 with a(n) := (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)) as given above.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/20 (telescoping series).
Product_{n >= 1} ((a(n) + 1)/(a(n) - 1)) = sqrt(5)/2 (telescoping product). (End)

A024851 Least m such that if r and s in {-F(2h) + tauF(2*h-1): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k, where F = A000045 (Fibonacci numbers) and tau = (1+sqrt(5))/2 (golden ratio).

Original entry on oeis.org

2, 5, 12, 30, 77, 200, 522, 1365, 3572, 9350, 24477, 64080, 167762, 439205, 1149852, 3010350, 7881197, 20633240

Offset: 2

Clark Kimberling

Possibly a duplicate of A188378. For a guide to related sequences, see A001000. - Clark Kimberling, Aug 09 2012

			Referring to the terminology introduced at A001000, m=5 is the (1st) separator of the set S = {f(1),f(2),f(3)}, where f(h) = - F(2*h) + tau*F(2*h-1).  That is, a(3) = 5, since 1/5 < f(3) < 2/5 < f(2) < 3/5 < f(1), whereas fractions k/m for m<5 do not separate the elements of S in this manner.

Cf. A001000.

f[n_] := f[n] = -Fibonacci[2 n] + GoldenRatio*Fibonacci[2 n - 1]
leastSeparator[seq_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
2 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Table[N[f[h], 40], {h, 1, 18}] (* A024851 *)
t1 = leastSeparator[t]
(* Peter J. C. Moses, Aug 01 2012 *)

Extended, corrected, and edited by Clark Kimberling, Aug 09 2012

a(19) from Sean A. Irvine, Jul 26 2019

A267797 Lucas numbers of the form (x^3 + y^3) / 2 where x and y are distinct positive integers.

Original entry on oeis.org

76, 1364, 24476, 439204, 7881196, 141422324, 2537720636, 45537549124, 817138163596, 14662949395604, 263115950957276, 4721424167835364, 84722519070079276, 1520283919093591604, 27280388024614569596, 489526700523968661124, 8784200221406821330636

Offset: 1

Altug Alkan, Jan 24 2016

Lucas numbers that are the averages of 2 distinct positive cubes.
Inspired by relation between sequence A024851 and A188378.
Corresponding indices are 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, ...
6*n + 3 is the corresponding form of indices.
Corresponding y values are listed by A188378, for n > 0. Note that corresponding x values are A188378(n) - 2, for n > 0.

			Lucas number 76 is a term because 76 = (3^3 + 5^3) / 2.
Lucas number 1364 is a term because 1364 = (10^3 + 12^3) / 2.
Lucas number 24476 is a term because 24476 = (28^3 + 30^3) / 2.
Lucas number 439204 is a term because 439204 = (75^3 + 77^3) / 2.
Lucas number 7881196 is a term because 7881196 = (198^3 + 200^3) / 2.
Lucas number 141422324 is a term because 141422324 = (520^3 + 522^3) / 2.

Colin Barker, Table of n, a(n) for n = 1..750
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A000032, A016945, A024670, A024851, A049629, A188378.

[Fibonacci(6*n+4)+Fibonacci(6*n+2): n in [1..20]]; // Vincenzo Librandi, Jan 24 2016

Table[Fibonacci[6 n + 4] + Fibonacci[6 n + 2], {n, 1, 20}] (* Vincenzo Librandi, Jan 24 2016 *)
LinearRecurrence[{18,-1},{76,1364},20] (* Harvey P. Dale, Jul 23 2024 *)

l(n) = fibonacci(n+1) + fibonacci(n-1);
is(n) = for(i=ceil(sqrtn(n\2+1, 3)), sqrtn(n-.5, 3), ispower(n-i^3, 3) && return(1));
for(n=1, 120, if(is(2*l(n)), print1(l(n), ", ")));

a(n) = ((5*fibonacci(n)*fibonacci(n+1) + 1 + (-1)^n)^3 + (5*fibonacci(n)*fibonacci(n+1) - 1 + (-1)^n)^3) / 2;

a(n) = (fibonacci(6*n+4) + fibonacci(6*n+2));

Vec(4*x*(19-x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016

a(n) = A000032(A016945(n)), for n > 0.
a(n) = A188378(n)^3 - 3*A188378(n)^2 + 6*A188378(n) - 4, for n > 0.
From Colin Barker, Jan 24 2016: (Start)
a(n) = (9+4*sqrt(5))^(-n)*(2-sqrt(5)+(2+sqrt(5))*(9+4*sqrt(5))^(2*n)).
a(n) = 18*a(n-1)-a(n-2) for n>2.
G.f.: 4*x*(19-x) / (1-18*x+x^2).
(End)

cp's OEIS Frontend

A049629 a(n) = (F(6*n+5) - F(6*n+1))/4 = (F(6*n+4) + F(6*n+2))/4, where F = A000045.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A024851 Least m such that if r and s in {-F(2*h) + tau*F(2*h-1): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k, where F = A000045 (Fibonacci numbers) and tau = (1+sqrt(5))/2 (golden ratio).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions

A267797 Lucas numbers of the form (x^3 + y^3) / 2 where x and y are distinct positive integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

PARI

PARI

Formula

A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.

A024851 Least m such that if r and s in {-F(2h) + tauF(2*h-1): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k, where F = A000045 (Fibonacci numbers) and tau = (1+sqrt(5))/2 (golden ratio).