A023039 -id:A023039 - cp's OEIS frontend

A075796 Numbers k such that 5*k^2 + 5 is a square.

2, 38, 682, 12238, 219602, 3940598, 70711162, 1268860318, 22768774562, 408569081798, 7331474697802, 131557975478638, 2360712083917682, 42361259535039638, 760141959546795802, 13640194012307284798, 244763350261984330562, 4392100110703410665318, 78813038642399407645162

Offset: 1

Gregory V. Richardson, Oct 13 2002

Bisection of A001077; a(n) = A001077(2*n-1). - Greg Dresden, Jun 08 2021
From Peter Bala, Aug 25 2022: (Start)
The aerated sequence (b(n))n>=1 = [2, 0, 38, 0, 682, 0, 1238, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). The sequence (1/2)*(b(n))n>=1 is the case P1 = 0, P2 = -16, Q = -1  of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. (End)

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A000290, A306380, A001077.

I:=[2,38]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 30 2011

[Lucas(6*n-3)/2: n in [1..20]]; // G. C. Greubel, Feb 13 2019

with(combinat); A075796:=n->fibonacci(6*n+3)+fibonacci(6*n)/2; seq(A075796(n), n=1..50); # Wesley Ivan Hurt, Nov 29 2013

LinearRecurrence[{18, -1}, {2, 38}, 50] (* Sture Sjöstedt, Nov 29 2011; typo fixed by Vincenzo Librandi, Nov 30 2011 *)
LucasL[6*Range[20]-3]/2 (* G. C. Greubel, Feb 13 2019 *)
CoefficientList[Series[2*(1+x)/( 1-18*x+x^2 ), {x,0,20}],x] (* Stefano Spezia, Mar 02 2019 *)

vector(20, n, (fibonacci(6*n-2) + fibonacci(6*n-4))/2) \\ G. C. Greubel, Feb 13 2019

[(fibonacci(6*n-2) + fibonacci(6*n-4))/2 for n in (1..20)] # G. C. Greubel, Feb 13 2019

a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n) + ((9 + 4*sqrt(5))^(n-1) - (9 - 4*sqrt(5))^(n-1)))/(4*sqrt(5)).
a(n) = 18*a(n-1) - a(n-2).
a(n) = 2*A049629(n-1).
Limit_{n->oo} a(n)/a(n-1) = 8*phi + 1 = 9 + 4*sqrt(5).
a(n+1) = 9*a(n) + 4*sqrt(5)*sqrt((a(n)^2+1)). - Richard Choulet, Aug 30 2007
G.f.: 2*x*(1 + x)/(1 - 18*x + x^2). - Richard Choulet, Oct 09 2007
From Johannes W. Meijer, Jul 01 2010: (Start)
a(n) = A000045(6*n+3) + A000045(6*n)/2.
a(n) = 2*A167808(6*n+4) - A167808(6*n+6).
Limit_{k->oo} a(n+k)/a(k) = A023039(n)*A060645(n)*sqrt(5).
(End)
5*A007805(n)^2 - 1 = a(n+1)^2. - Sture Sjöstedt, Nov 29 2011
From Peter Bala, Nov 29 2013: (Start)
a(n) = Lucas(6*n - 3)/2.
Sum_{n >= 1} 1/(a(n) + 5/a(n)) = 1/4. Compare with A002878, A005248, A023039. (End)
Limit_{n->oo} a(n)/A007805(n-1) = sqrt(5). - A.H.M. Smeets, May 29 2017
E.g.f.: (exp((9 - 4*sqrt(5))*x)*(- 5 + 2*sqrt(5) + (5 + 2*sqrt(5))*exp(8*sqrt(5)*x)))/(2*sqrt(5)). - Stefano Spezia, Feb 13 2019
Sum_{n > 0} 1/a(n) = (1/log(9 - 4*sqrt(5)))*(- 17 - 38/sqrt(5))*sqrt(5*(9 - 4*sqrt(5)))*(- 9 + 4*sqrt(5))*(psi_{9 - 4*sqrt(5)}(1/2) - psi_{9 - 4*sqrt(5)}(1/2 - (I*Pi)/log(9 - 4*sqrt(5)))) approximately equal to 0.527868600269500798938265500122302016..., where psi_q(x) is the q-digamma function. - Stefano Spezia, Feb 25 2019
a(n) = sinh((6*n - 3)*arccsch(2)). - Peter Luschny, May 25 2022

A060645 a(0) = 0, a(1) = 4, then a(n) = 18*a(n-1) - a(n-2).

Original entry on oeis.org

0, 4, 72, 1292, 23184, 416020, 7465176, 133957148, 2403763488, 43133785636, 774004377960, 13888945017644, 249227005939632, 4472197161895732, 80250321908183544, 1440033597185408060, 25840354427429161536

Offset: 0

Lekraj Beedassy, Apr 17 2001

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 5*y^2 = 1, the third simplest case of the Pell-Fermat type. The corresponding x values are in A023039.

Numbers k such that 5*k^2 = floor(sqrt(5)*k*ceiling(sqrt(5)*k)). - Benoit Cloitre, May 10 2003

			Given a(1) = 4, a(2) = 72 we have, for instance, a(4) = 18*a(3) - a(2) = 18*{18*a(2) - a(1)} - a(2), i.e., a(4) = 323*a(2) - 18*a(1) = 323*72 - 18*4 = 23184.

Harry J. Smith, Table of n, a(n) for n = 0..200
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
John Robertson, Home page
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A023039.

A060645 := proc(n) option remember: if n=1 then RETURN(4) fi: if n=2 then RETURN(72) fi: 18*A060645(n -1)- A060645(n-2): end: for n from 1 to 30 do printf(`%d,`, A060645(n)) od:

CoefficientList[ Series[4x/(1 - 18x + x^2), {x, 0, 16}], x] (* Robert G. Wilson v *)
LinearRecurrence[{18, -1} {0, 4}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
Table[4 ChebyshevU[-1 + n, 9], {n, 0, 16}] (* Herbert Kociemba, Jun 05 2022 *)

g(n,k) = for(y=0,n,x=k*y^2+1;if(issquare(x),print1(y",")))

a(n)=fibonacci(6*n)/2 \\ Benoit Cloitre

for (i=1,10000,if(Mod(sigma(5*i^2+1),2)==1,print1(i,",")))

{ for (n=0, 200, write("b060645.txt", n, " ", fibonacci(6*n)/2); ) } \\ Harry J. Smith, Jul 09 2009

a(n) = 18*a(n-1) - a(n-2), with a(1) = denominator of continued fraction [2; 4] and a(2) = denominator of [2; 4, 4, 4].
G.f.: 4x/(1-18*x+x^2). - Cino Hilliard, Feb 02 2006
a(n) may be computed either as (i) the denominator of the (2n-1)-th convergent of the continued fraction [2; 4, 4, 4, ...] = sqrt(5), or (ii) as the coefficient of sqrt(5) in (9+sqrt(5))^n.
Numbers k such that sigma(5*k^2 + 1) mod 2 = 1. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004
a(n) = 4*A049660(n), a(n) = A000045(6*n)/2. - Benoit Cloitre, Feb 03 2006
a(n) = 17*(a(n-1) + a(n-2)) - a(n-3) = 19*(a(n-1) - a(n-2)) + a(n-3). - Mohamed Bouhamida, Sep 20 2006
From Johannes W. Meijer, Jul 01 2010: (Start)
Limit_{k->infinity} a(n+k)/a(k) = A023039(n) + A060645(n)*sqrt(5).
Limit_{n->infinity} A023039(n)/a(n) = sqrt(5). (End)
a(n) = Fibonacci(6*n)/2. - Gary Detlefs, Apr 02 2012
a(n) = 4*S(n-1, 18), with Chebyshev's S-polynomials. See A049310. S(-1, x)= 0. - Wolfdieter Lang, Aug 24 2014

More terms from James Sellers, Apr 19 2001

Entry revised by N. J. A. Sloane, Aug 13 2006

A134493 a(n) = Fibonacci(6*n+1).

Original entry on oeis.org

1, 13, 233, 4181, 75025, 1346269, 24157817, 433494437, 7778742049, 139583862445, 2504730781961, 44945570212853, 806515533049393, 14472334024676221, 259695496911122585, 4660046610375530309, 83621143489848422977, 1500520536206896083277, 26925748508234281076009

Offset: 0

Artur Jasinski, Oct 28 2007

For positive n, a(n) equals (-1)^n times the permanent of the (6n)X(6n) tridiagonal matrix with ((-1)^(1/6))'s along the three central diagonals. - John M. Campbell, Jul 12 2011

a(n) = x + y where those two values are solutions to: x^2 = 5*y^2 + 1. (See related sequences with formula below). - Richard R. Forberg, Sep 05 2013

Colin Barker, Table of n, a(n) for n = 0..750
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000032, A000045, A049660, A134492, A134494, A134495, A103134, A134497.

[Fibonacci(6*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Mathematica
```
Table[Fibonacci[6n+1], {n, 0, 30}]
```

a(n)=fibonacci(6*n+1) \\ Charles R Greathouse IV, Jul 15 2011

Vec((1-5*x)/(1-18*x+x^2) + O(x^100)) \\ Altug Alkan, Jan 24 2016

From R. J. Mathar, Apr 17 2011: (Start)
G.f.: ( 1-5*x ) / ( 1-18*x+x^2 ).
a(n) = A049660(n+1) - 5*A049660(n). (End)
a(n) = Fibonacci(3*n+1)^2 + Fibonacci(3*n)^2. - Gary Detlefs, Oct 12 2011
a(n) = 18*a(n-1) - a(n-2). - Richard R. Forberg, Sep 05 2013
a(n) = A060645(n) + A023039(n), as derives from comment above. - Richard R. Forberg, Sep 05 2013
a(n) = ((5-sqrt(5)+(5+sqrt(5))*(9+4*sqrt(5))^(2*n)))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
2*a(n) = Fibonacci(6*n) + Lucas(6*n). - Bruno Berselli, Oct 13 2017
a(n) = S(n, 18) - 5*S(n-1, 18), n >= 0, with the Chebyshev S-polynomials S(n-1, 18) = A049660(n). (See the g.f.) - Wolfdieter Lang, Jul 10 2018

Offset changed to 0 by Vincenzo Librandi, Apr 16 2011

A167808 Numerator of x(n), where x(n) = x(n-1) + x(n-2) with x(0)=0, x(1)=1/2.

Original entry on oeis.org

0, 1, 1, 1, 3, 5, 4, 13, 21, 17, 55, 89, 72, 233, 377, 305, 987, 1597, 1292, 4181, 6765, 5473, 17711, 28657, 23184, 75025, 121393, 98209, 317811, 514229, 416020, 1346269, 2178309, 1762289, 5702887, 9227465, 7465176, 24157817, 39088169, 31622993

Offset: 0

Reinhard Zumkeller, Nov 12 2009

Define a sequence c(n) by c(0)=0, c(1)=1; thereafter c(n) = (c(n-2)*c(n-1)-1)/(c(n-2)+c(n-1)+2). Then it appears that (apart from signs), a(n) is the denominator of c(n). - Jonas Holmvall, Jun 21 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Fibonacci number
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,1).

Cf. A000045, A130196 (denominator).

The a(2*n) appear in A179135. - Johannes W. Meijer, Jul 01 2010

a:=[0,1,1,1,3,5];; for n in [7..40] do a[n]:=4*a[n-3]+a[n-6]; od; a; # Muniru A Asiru, Oct 16 2018

nmax:=39; x(0):=0: x(1):=1/2:for n from 2 to nmax do x(n):=x(n-1)+x(n-2) od: for n from 0 to nmax do a(n):= numer(x(n)) od: seq(a(n),n=0..nmax); # Johannes W. Meijer, Jul 01 2010
with(combinat):f:=n->fibonacci(n):L:=n->f(n)+2*f(n-1):seq(numer(f(n)/L(n)), n=0..39); # Gary Detlefs, Dec 11 2010

f[n_]:=Numerator[Fibonacci[n]/Fibonacci[n+3]];Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011*)
Numerator[LinearRecurrence[{1,1},{0,1/2},40]] (* Harvey P. Dale, Aug 08 2014 *)
CoefficientList[Series[-x (1 + x + x^2 - x^3 + x^4)/((x^2 + x - 1) (x^4 - x^3 + 2 x^2 + x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 08 2014 *)
LinearRecurrence[{0, 0, 4, 0, 0, 1},{0, 1, 1, 1, 3, 5},40] (* Ray Chandler, Aug 03 2015 *)
a[n_]:=If[Mod[n,3]==0, Fibonacci[n]/2, Fibonacci[n]]; Array[a, 40, 0] (* Stefano Spezia, Oct 16 2018 *)

a(n) = (a(n-1)*A131534(n) + a(n-2)*A131534(n+2))/A131534(n+1) for n > 1.
a(3*n) = A001076(n) = (a(3*n-1) + a(3*n-2))/2;
a(3*n+1) = A033887(n) = 2*a(3*n-1) + a(3*n-2);
a(3*n+2) = A015448(n+1) = a(3*n-1) + 2*a(3*n-2).
From Johannes W. Meijer, Jul 01 2010: (Start)
a(2*n) = A001906(n)/A131534(n+1) for n >= 0 and a(2*n+1) = A179131(n)/5 for n >= 1.
a(6*n+2) - 2*a(6*n) = A134493(n);
2*a(6*n+1) - a(6*n+3) = A023039(n);
2*a(6*n+4) - a(6*n+2) = A134497(n);
a(6*n+5) - 2*a(6*n+3) = A103134(n);
2*a(6*n+4) - a(6*n+6) = A075796(n).
(End)
From Gary Detlefs, Dec 11 2010: (Start)
a(n) = numerator(A000045(n)/A000032(n)).
If n mod 3 = 0 then a(n) = Fibonacci(n)/2, else a(n)= Fibonacci(n). (End)
G.f.: -x*(1 + x + x^2 - x^3 + x^4) / ( (x^2 + x - 1)*(x^4 - x^3 + 2*x^2 + x + 1) ). - R. J. Mathar, Mar 08 2011
a(n) = 4*a(n-3) + a(n-6). - Muniru A Asiru, Oct 16 2018

Typo in title corrected by Johannes W. Meijer, Jun 26 2010

A322836 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{n}(x), evaluated at x=k.

Original entry on oeis.org

1, 1, 0, 1, 1, -1, 1, 2, 1, 0, 1, 3, 7, 1, 1, 1, 4, 17, 26, 1, 0, 1, 5, 31, 99, 97, 1, -1, 1, 6, 49, 244, 577, 362, 1, 0, 1, 7, 71, 485, 1921, 3363, 1351, 1, 1, 1, 8, 97, 846, 4801, 15124, 19601, 5042, 1, 0, 1, 9, 127, 1351, 10081, 47525, 119071, 114243, 18817, 1, -1

Offset: 0

Seiichi Manyama, Dec 28 2018

			Square array begins:
   1, 1,    1,     1,      1,      1,       1, ...
   0, 1,    2,     3,      4,      5,       6, ...
  -1, 1,    7,    17,     31,     49,      71, ...
   0, 1,   26,    99,    244,    485,     846, ...
   1, 1,   97,   577,   1921,   4801,   10081, ...
   0, 1,  362,  3363,  15124,  47525,  120126, ...
  -1, 1, 1351, 19601, 119071, 470449, 1431431, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Mirror of A101124.
Columns 0-20 give A056594, A000012, A001075, A001541, A001091, A001079, A023038, A011943(n+1), A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203, A322888, A056771, A322889, A078986, A322890.
Rows 0-10 give A000012, A001477, A056220, A144129, A144130, A243131, A243132, A243133, A243134, A243135, A243136.
Main diagonal gives A115066.
Cf. A323182 (Chebyshev polynomial of the second kind).

Table[ChebyshevT[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 28 2018 *)

T(n,k) = polchebyshev(n,1,k);
matrix(7, 7, n, k, T(n-1,k-1)) \\ Michel Marcus, Dec 28 2018

T(n, k) = round(cos(n*acos(k)));\\ Seiichi Manyama, Mar 05 2021

T(n, k) = if(n==0, 1, n*sum(j=0, n, (2*k-2)^j*binomial(n+j, 2*j)/(n+j))); \\ Seiichi Manyama, Mar 05 2021

A(0,k) = 1, A(1,k) = k and A(n,k) = 2 * k * A(n-1,k) - A(n-2,k) for n > 1.
A(n,k) = cos(n*arccos(k)). - Seiichi Manyama, Mar 05 2021
A(n,k) = n * Sum_{j=0..n} (2*k-2)^j * binomial(n+j,2*j)/(n+j) for n > 0. - Seiichi Manyama, Mar 05 2021

A128052 a(n) = (F(2n-1) + F(2n+1))(5/6 - cos(2Pi*n/3)/3), where F(n) = Fibonacci(n).

Original entry on oeis.org

1, 3, 7, 9, 47, 123, 161, 843, 2207, 2889, 15127, 39603, 51841, 271443, 710647, 930249, 4870847, 12752043, 16692641, 87403803, 228826127, 299537289, 1568397607, 4106118243, 5374978561, 28143753123, 73681302247, 96450076809, 505019158607, 1322157322203

Offset: 0

Paul Barry, Feb 13 2007

The a(n+1) are the numerators of A178381(4*n+3)/A178381(4*n+2). For the denominators see A179133(n). - Johannes W. Meijer, Jul 01 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,18,0,0,-1).

Cf. A128053.

Cf. A179134. Trisection: A023039.

I:=[1,3,7,9,47,123]; [n le 6 select I[n] else 18*Self(n-3)-Self(n-6): n in [1..30]]; // Vincenzo Librandi, Jul 17 2019

with(combinat): nmax:=25; for n from 0 to nmax do a(n):= (fibonacci(2*n-1)+fibonacci(2*n+1))*(5/6-cos(2*Pi*n/3)/3) od: seq(a(n),n=0..nmax); # Johannes W. Meijer, Jul 01 2010

LinearRecurrence[{0, 0, 18, 0, 0, -1}, {1, 3, 7, 9, 47, 123}, 40] (* Vincenzo Librandi, Jul 17 2019 *)

Lim_{n->infinity} A128052(n+1)/A179133(n) = 1 + cos(Pi/5). - Johannes W. Meijer, Jul 01 2010
a(n) = Lucas(2*n)*(Fibonacci(n) mod 2 + 1)/2, Lucas(n)=A000032, Fibonacci(n)=A000045. - Gary Detlefs, Jan 19 2001
From Colin Barker, Jun 27 2013: (Start)
a(n) = 18*a(n-3) - a(n-6).
G.f: -(3*x^5 + 7*x^4 + 9*x^3 - 7*x^2 - 3*x - 1) / ((x^2 - 3*x + 1)*(x^4 + 3*x^3 + 8*x^2 + 3*x + 1)). (End)
With L(n) the Lucas number A000032, a(n) = L(2*n)/2 or L(2*n) according as n is, or is not, divisible by 3. - David Callan, Jul 17 2019

A033314 Least D in the Pellian x^2 - D*y^2 = 1 for which x has least solution n.

Original entry on oeis.org

3, 2, 15, 6, 35, 12, 7, 5, 11, 30, 143, 42, 195, 14, 255, 18, 323, 10, 399, 110, 483, 33, 23, 39, 27, 182, 87, 210, 899, 60, 1023, 17, 1155, 34, 1295, 38, 1443, 95, 1599, 105, 1763, 462, 215, 506, 235, 138, 47, 96, 51, 26, 2703, 78, 2915, 21, 3135, 203, 3363

Offset: 2

Eric W. Weisstein

nonn

The i-th solution pair V(i) = [x(i), y(i)] to the Pellian x^2 - D*y^2 = 1 for a given least solution x(1) = n may be generated through the recurrence V(i+2) = 2*n*V(i+1) - V(i) taking V(0) = [1, 0] and V(1) = [n, sqrt((n^2-1)/a(n))]. V(i) stands for the numerator and denominator of the 2i-th convergent of the continued fraction expansion of sqrt(D).

Thus setting n = 3, for instance, we have D = a(3) = 2 and V(1) = [3, 2] so that along with V(0) = [1, 0] recurrence V(i+2) = 6*V(i+1) - V(i) generates [A001333(2k), A000129(2k)]. Similarly, setting n = 9 generates [A023039, A060645], respectively the numerator and denominator of the 2i-th convergent of sqrt(a(9)), i.e., sqrt(5). - Lekraj Beedassy, Feb 26 2002

Ray Chandler, Table of n, a(n) for n = 2..1001
Eric Weisstein's World of Mathematics, Pell Equation.

Cf. A000037, A033313, A033318.

squarefreepart[n_] :=
  Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ FactorInteger[n]);
pellminx[d_] := Module[{q, p, z}, {q, p} = ContinuedFraction[Sqrt[d]];
  If[OddQ[p // Length], p = Join[p, p]];
  z = FromContinuedFraction[Join[{q}, Drop[p, -1]]]; Numerator[z]]
NMAX = 60; a = {};
For[n = 2, n <= NMAX, n++, s = squarefreepart[n^2 - 1];
sd = s Divisors[Sqrt[(n^2 - 1)/s]]^2;
t = Sort[Transpose[{sd, pellminx[#] & /@ sd}]];
AppendTo[a, Select[t, #[[2]] == n &, 1][[1, 1]]]
]; a (* Herbert Kociemba, Jun 05 2022 *)

A099279 Squares of A001076.

Original entry on oeis.org

0, 1, 16, 289, 5184, 93025, 1669264, 29953729, 537497856, 9645007681, 173072640400, 3105662519521, 55728852710976, 1000013686278049, 17944517500293904, 322001301319012225, 5778078906241926144, 103683419011035658369, 1860523463292399924496, 33385738920252162982561

Offset: 0

Wolfdieter Lang, Oct 18 2004

For the generalized Fibonacci sequences U(n-1;a) = (ap(a)^n - am(a)^n)/(ap(a) - am(a)) with ap(a) = (a + sqrt(a^2+4))/2, am(a) = (a - sqrt(a^2+4))/2, a from the integers, one has for the squared sequences U(n-1;a)^2 = (2*T(n,(a^2+2)/2) - 2*(-1)^n)/(a^2+4). Here T(n,x) are Chebyshev's polynomials of the first kind (see A053120). Therefore the o.g.f. for the squared sequence is x*(1-x)/((1+x)*(1-(a^2+2)*x+x^2)) = x*(1-x)/(1 - (a^2+1)*x - (a^2+1)*x^2 + x^3). For this example a=4.
Unsigned member r=-16 of the family of Chebyshev sequences S_r(n) defined in A092184.
(-1)^(n+1)*a(n) = S_{-16}(n), n >= 0, defined in A092184.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 4 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 4 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Mar 12 2023

G. C. Greubel, Table of n, a(n) for n = 0..750
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (17,17,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001076, A001654, A023039, A053120, A092184.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, this sequence, A099365, A099366, A099367, A099369, A099372, A099374.

[Fibonacci(3*n)^2/4: n in [0..30]]; // G. C. Greubel, Aug 18 2022

with (combinat):seq(fibonacci(n,4)^2,n=0..16); # Zerinvary Lajos, Apr 09 2008
nmax:=48: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: for n from 1 to nmax/3 do a(n):=a(n-1)+A001654(3*n-2) od: seq(a(n),n=0..nmax/3); # Johannes W. Meijer, Sep 22 2010

LinearRecurrence[{17,17,-1},{0,1,16},30] (* Harvey P. Dale, Mar 26 2012 *)
Fibonacci[3*Range[0, 30]]^2/4 (* G. C. Greubel, Aug 18 2022 *)

numlib::fibonacci(3*n)^2/4 $ n = 0..35; // Zerinvary Lajos, May 13 2008

my(x='x+O('x^99)); concat([0], Vec(x*(1-x)/((1-18*x+x^2)*(1+x)))) \\ Altug Alkan, Dec 17 2017

[(fibonacci(3*n))^2/4 for n in range(0, 17)] # Zerinvary Lajos, May 15 2009

a(n) = A001076(n)^2.
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3), n >= 3, a(0)=0, a(1)=1, a(2)=16.
a(n) = 18*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2, a(0)=0, a(1)=1.
a(n) = (T(n, 9) - (-1)^n)/10 with Chebyshev's T(n, x) polynomials of the first kind. T(n, 9) = A023039(n).
G.f.: x*(1-x)/((1+x)*(1-18*x+x^2)) = x*(1-x)/(1-17*x-17*x^2+x^3).
a(n) = a(n-1) + A001654(3*n-2) with a(0)=0, where A001654 are the golden rectangle numbers. - Johannes W. Meijer, Sep 22 2010
a(n+1) = (1 + (-1)^n)/2 + 16*Sum_{r=1..n} ( r*a(n+1-r) ). - Michael A. Allen, Mar 12 2023
E.g.f.: exp(-x)*(exp(10*x)*cosh(4*sqrt(5)*x) - 1)/10. - Stefano Spezia, Apr 06 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (2 + sqrt(5))/4 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A115032 Expansion of (5-14x+x^2)/((1-x)(x^2-18*x+1)).

Original entry on oeis.org

5, 81, 1445, 25921, 465125, 8346321, 149768645, 2687489281, 48225038405, 865363202001, 15528312597605, 278644263554881, 5000068431390245, 89722587501469521, 1610006506595061125, 28890394531209630721, 518417095055178291845, 9302617316461999622481

Offset: 0

Creighton Dement, Feb 26 2006

Relates squares of numerators and denominators of continued fraction convergents to sqrt(5).
Sequence is generated by the floretion A*B*C with A = + 'i - 'k + i' - k' - 'jj' - 'ij' - 'ji' - 'jk' - 'kj' ; B = - 'i + 'j - i' + j' - 'kk' - 'ik' - 'jk' - 'ki' - 'kj' ; C = - 'j + 'k - j' + k' - 'ii' - 'ij' - 'ik' - 'ji' - 'ki' (apart from a factor (-1)^n)
Floretion Algebra Multiplication Program, FAMP Code: tesseq[A*B*C].
The sequence a(n-1), n >= 0, with a(-1) = 1, is also the curvature of circles inscribed in a special way in the larger segment of a circle of radius 5/4 (in some units) cut by a chord of length 2. For the smaller segment, the analogous curvature sequence is given in A240926. For more details see comments on A240926. See also the illustration in the present sequence, and the proof of the coincidence of the curvatures with a(n-1) in part I of the W. Lang link. - Kival Ngaokrajang, Aug 23 2014

			G.f. = 5 + 81*x + 1445*x^2 + 25921*x^3 + 465125*x^4 + 8346321*x^5 + ...

G. C. Greubel, Table of n, a(n) for n = 0..795
Creighton Dement, Floretions associated with A115032.
Wolfdieter Lang, A proof for the touching circle problem (part I).
Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
Kival Ngaokrajang, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001076, A001077, A007805, A023039, A097924.

Cf. also A000045, A049660, A049310, A023039. - Wolfdieter Lang, Aug 22 2014

seq((9*combinat:-fibonacci(6*(n+1)) - combinat:-fibonacci(6*n) + 8)/16, n = 0 .. 20); # Robert Israel, Aug 25 2014

LinearRecurrence[{19,-19,1},{5,81,1445},30] (* Harvey P. Dale, Nov 14 2014 *)
CoefficientList[Series[(5 - 14*x + x^2)/((1 - x)*(x^2 - 18*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

Vec((5-14*x+x^2)/((1-x)*(x^2-18*x+1)) + O(x^20)) \\ Michel Marcus, Aug 23 2014

sqrt(a(2*n)) = sqrt(5)*A007805(n) = sqrt(5)*Fibonacci(6*n+3)/2 = sqrt(5)*A001076(2*n+1); sqrt(a(2*n+1)) = A023039(2*n+1) = A001077(2*n).
From Wolfdieter Lang, Aug 22 2014: (Start)
O.g.f.: (5-14*x+x^2)/((1-x)*(x^2-18*x+1)) (see the name).
a(n) = (9*F(6*(n+1)) - F(6*n) + 8)/16, n >= 0 with F(n) = A000045(n) (Fibonacci). From the partial fraction decomposition of the o.g.f.: (1/2)*((9 - x)/(1 - 18*x + x^2) + 1/(1 - x)). For F(6*n)/8 see A049660(n). a(-1) = 1 with F(-6) = -F(6) = -8.
a(n) = (9*S(n, 18) - S(n-1, 18) + 1)/2, with the Chebyshev S-polynomials (see A049310). From A049660.
a(n) = (A023039(n+1) + 1)/2.
(End)
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3). - Colin Barker, Aug 23 2014
From Wolfdieter Lang, Aug 24 2014: (Start)
a(n) = 18*a(n-1) - a(n-2) - 8, n >= 1, a(-1) = 1, a(0) = 5. See the Chebyshev S-polynomial formula above.
The o.g.f. for the sequence a(n-1) with a(-1) = 1, n >= 0, [1, 5,  81, 1445, ..] is (1-14*x+5*x^2)/((1-x)*(1-18*x+x^2)).
(See the Colin Barker formula from Aug 04 2014 in the history of A240926.) (End)

More terms from Michel Marcus, Aug 23 2014
Edited (comment by Kival Ngaokrajang rewritten, Chebyshev index link added) by Wolfdieter Lang, Aug 26 2014
Partially edited by Jon E. Schoenfield and N. J. A. Sloane, Mar 15 2024

A049683 a(n) = (Lucas(6*n) - 2)/16.

Original entry on oeis.org

0, 1, 20, 361, 6480, 116281, 2086580, 37442161, 671872320, 12056259601, 216340800500, 3882078149401, 69661065888720, 1250017107847561, 22430646875367380, 402501626648765281, 7222598632802407680, 129604273763794572961, 2325654329115499905620

Offset: 0

Clark Kimberling

This is the r = 20 member of the r-family of sequences S_r(n), n >= 1, defined in A092184 where more information can be found.

Colin Barker, Table of n, a(n) for n = 0..750
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A000032, A004146, A023039, A049660, A049682, A049684, A092184.

List([0..30], n-> (Lucas(1,-1,6*n)[2] - 2)/16 ); # G. C. Greubel, Dec 14 2019

[(Lucas(6*n) -2)/16: n in [0..30]]; // G. C. Greubel, Dec 02 2017

with(combinat); seq( (5*fibonacci(3*n)^2 -2*(1-(-1)^n))/16, n=0..30); # G. C. Greubel, Dec 14 2019

LinearRecurrence[{19,-19,1}, {0,1,30}, 20] (* or *) Table[(LucasL[6*n] -2)/16, {n,0,30}] (* G. C. Greubel, Dec 02 2017 *)

concat(0, Vec(x*(1+x)/((1-x)*(1-18*x+x^2)) + O(x^30))) \\ Colin Barker, Jun 03 2016

vector(31, n, (5*fibonacci(3*n-3)^2 -2*(1+(-1)^n))/16 ) \\ G. C. Greubel, Dec 14 2019

[(lucas_number2(6*n,1,-1) -2)/16 for n in (0..30)] # G. C. Greubel, Dec 14 2019

a(n) = (-2 + (9 + 4*sqrt(5))^n + (9 - 4*sqrt(5))^n)/16. - Ralf Stephan, Apr 14 2004
a(n) = (T(n, 9) - 1)/8 with Chebyshev's polynomials of the first kind evaluated at x = 9: T(n, 9) = A023039(n). Wolfdieter Lang, Oct 18 2004
G.f.: x*(1 + x)/((1 - x)*(1 - 18*x + x^2)) = x*(1 + x)/(1 - 19*x + 19*x^2 - x^3). (from the Stephan link, see A092184).
exp( Sum_{n >= 1} 16*a(n)*x^n/n ) = 1 + 2*Sum_{n >= 1} Fibonacci(6*n)*x^n. - Peter Bala, Jun 03 2016
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3) for n>2. - Colin Barker, Jun 03 2016

cp's OEIS Frontend

A075796 Numbers k such that 5*k^2 + 5 is a square.

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A060645 a(0) = 0, a(1) = 4, then a(n) = 18*a(n-1) - a(n-2).

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A134493 a(n) = Fibonacci(6*n+1).

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A167808 Numerator of x(n), where x(n) = x(n-1) + x(n-2) with x(0)=0, x(1)=1/2.

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A322836 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{n}(x), evaluated at x=k.

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A128052 a(n) = (F(2*n-1) + F(2*n+1))*(5/6 - cos(2*Pi*n/3)/3), where F(n) = Fibonacci(n).

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A128052 a(n) = (F(2n-1) + F(2n+1))(5/6 - cos(2Pi*n/3)/3), where F(n) = Fibonacci(n).

A115032 Expansion of (5-14x+x^2)/((1-x)(x^2-18*x+1)).