A056854 -id:A056854 - cp's OEIS frontend

A153173 a(n) = L(5*n)/L(n) where L(n) = Lucas number A000204(n).

11, 41, 341, 2161, 15251, 103361, 711491, 4868641, 33391061, 228811001, 1568437211, 10749853441, 73681573691, 505018447961, 3461454668501, 23725145626561, 162614613425891, 1114577020834241, 7639424866266611

Offset: 1

Artur Jasinski, Dec 20 2008

All numbers in this sequence are congruent to 1 mod 10.

G. C. Greubel, Table of n, a(n) for n = 1..1000
L. Carlitz, Problem B-185, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 8, No. 3 (1970), p. 325; Lucas Ratio I, Solution to Problem B-185 by C. B. A. Peck, ibid., Vol. 9, No. 1 (1971), p. 109.
L. Carlitz, Problem B-186, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 8, No. 3 (1970), p. 326; Lucas Ratio II, Solution to Problem B-186 by John Wessner, ibid., Vol. 9, No. 1 (1971), pp. 109-110.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A000032, A000045, A000204, A005248, A056854, A085695, A110391.

I:=[11, 41, 341, 2161, 15251]; [n le 5 select I[n] else 5*Self(n-1)+15*Self(n-2)-15*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[5*n]/LucasL[n], {n, 1, 50}]
CoefficientList[Series[x*(11-14*x-29*x^2+6*x^3+x^4)/((1-x)*(x^2-7*x+1)*(x^2+3*x+1)), {x,0,50}], x] (* G. C. Greubel, Dec 21 2017 *)
a[ n_] := 1 + 5*Fibonacci[n]*Fibonacci[3*n]; (* Michael Somos, Apr 23 2022 *)

{L(n)=fibonacci(n-1)+fibonacci(n+1)}; a(n) = L(5*n)/L(n) \\ Charles R Greathouse IV, Jun 11 2015

my(x='x+O('x^30)); Vec(x*(11-14*x-29*x^2+6*x^3+x^4 )/((1-x)*(x^2-7*x +1)*(x^2+3*x+1))) \\ G. C. Greubel, Dec 21 2017

{a(n) = 1 + 5*fibonacci(n)*fibonacci(3*n)}; /* Michael Somos, Apr 23 2022 */

From R. J. Mathar, Oct 22 2010: (Start)
a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: -x*(11-14*x-29*x^2+6*x^3+x^4)/( (x-1)*(x^2-7*x+1)*(x^2+3*x+1) ).
a(n) = 1 + A056854(n) - (-1)^n*A005248(n). (End)
From Amiram Eldar, Feb 02 2022: (Start)
a(n) = Lucas(2*n)^2 - (-1)^n*Lucas(2*n) - 1 (Carlitz, Problem B-185).
a(n) = (Lucas(2*n) - 3*(-1)^n)^2 + (-1)^n*(5*Fibonacci(n))^2 (Carlitz, Problem B-186). (End)
a(n) = a(-n) = 1 + 10*A085695(n) = 5 + L(n-1)*L(n)^2*L(n+1) for all n in Z. - Michael Somos, Apr 23 2022

A246453 Lucas numbers (A000204) of the form n^2 + 2.

Original entry on oeis.org

3, 11, 18, 123, 843, 5778, 39603, 271443, 1860498, 12752043, 87403803, 599074578, 4106118243, 28143753123, 192900153618, 1322157322203, 9062201101803, 62113250390418, 425730551631123, 2918000611027443, 20000273725560978, 137083915467899403, 939587134549734843

Offset: 1

Michel Lagneau, Aug 26 2014

a(n) = {11} union {A000204(2+4*n)} for n=0,1,...

Intersection of A000204 and A059100. - Michel Marcus, Aug 26 2014

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000204 (Lucas), A059100 (n^2+2).

Cf. quadrisection of A000032: A056854 (first), A056914 (second), this sequence (third, without 11), A288913 (fourth).

I:=[3,11,18,123]; [n le 4 select I[n] else 7*Self(n-1)-Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 21 2017

with(combinat,fibonacci):lst:={}:lst1:={}:nn:=5000:
  for n from 1 to nn do:
    lst:=lst union {2*fibonacci(n-1)+fibonacci(n)}:
  od:
   for m from 1 to nn do:
    if {m^2+2} intersect lst = {m^2+2}
    then
    lst1:=lst1 union {m^2+2}:
    else
    fi:
   od:
   print(lst1):

CoefficientList[Series[x*(3-10*x-56*x^2+8*x^3)/(1-7*x+x^2), {x,0,50}], x] (* or *) LinearRecurrence[{7,-1}, {3, 11, 18, 123}, 30] (* G. C. Greubel, Dec 21 2017 *)
Select[LucasL[Range[100]],IntegerQ[Sqrt[#-2]]&] (* Harvey P. Dale, Dec 31 2018 *)

lista(nn) = for (n=0, nn, luc = fibonacci(n+1) + fibonacci(n-1); if (issquare(luc-2), print1(luc, ", "))); \\ Michel Marcus, Mar 29 2016

Vec(x*(3 - 10*x - 56*x^2 + 8*x^3) / (1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017

From Colin Barker, Jun 20 2017: (Start)
G.f.: x*(3 - 10*x - 56*x^2 + 8*x^3) / (1 - 7*x + x^2).
a(n) = (2^(-n)*((7+3*sqrt(5))^n*(-20+9*sqrt(5)) + (7-3*sqrt(5))^n*(20+9*sqrt(5)))) / sqrt(5) for n>2.
a(n) = 7*a(n-1) - a(n-2) for n>4. (End)
E.g.f.: 2*exp(7*x/2)*(9*cosh(3*sqrt(5)*x/2) - 4*sqrt(5)*sinh(3*sqrt(5)*x/2)) + 4*x^2 - 18. - Stefano Spezia, Apr 14 2025

Corrected by Michel Marcus, Mar 29 2016

A153175 a(n) = L(7*n)/L(n) where L(n) = Lucas number A000204(n).

Original entry on oeis.org

29, 281, 6119, 101521, 1875749, 33281921, 599786069, 10745088481, 192933544679, 3461223997001, 62114818827629, 1114566304366081, 20000347407134669, 358889844987430121, 6440029487834912999, 115561554399692896321

Offset: 1

Artur Jasinski, Dec 20 2008

nonn

All numbers in this sequence are:
congruent to 9 mod 10 (iff n is odd),
congruent to 1 mod 10 (iff n is even).

G. C. Greubel, Table of n, a(n) for n = 1..795
Index entries for linear recurrences with constant coefficients, signature (13,104,-260,-260,104,13,-1).

Cf. A000032, A000204, A110391, A153173.

Cf. A153177, A153179, A153180. [From R. J. Mathar, Oct 22 2010]

[Lucas(7*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[7*n]/LucasL[n], {n, 1, 50}]

{lucas(n) = fibonacci(n+1) + fibonacci(n-1)};
for(n=0,30, print1( lucas(7*n)/lucas(n), ", ")) \\ G. C. Greubel, Dec 21 2017

From R. J. Mathar, Oct 22 2010: (Start)
a(n) = +13*a(n-1) +104*a(n-2) -260*a(n-3) -260*a(n-4) +104*a(n-5) +13*a(n-6) -a(n-7).
G.f.: -x*(-29+96*x+550*x^2-290*x^3-200*x^4+16*x^5+x^6) / ( (1+x)*(x^2-3*x+1)*(x^2-18*x+1)*(x^2+7*x+1) ).
a(n) = A005248(n) +A087215(n) -(-1)^n*A056854(n) - (-1)^n. (End)

A153177 a(n) = L(9*n)/L(n) where L(n) = Lucas number A000204(n).

Original entry on oeis.org

76, 1926, 109801, 4769326, 230701876, 10716675201, 505618944676, 23714405408926, 1114769987764201, 52357935173823126, 2459933168462154076, 115560463558534156801, 5428954301161174383676, 255043991670277234750326

Offset: 1

Artur Jasinski, Dec 20 2008

nonn

All numbers in this sequence are:
congruent to 1 mod 100 (iff n is congruent to 0 mod 3),
congruent to 26 mod 100 (iff n is congruent to 2 or 4 mod 6),
congruent to 76 mod 100 (iff n is congruent to 1 or 5 mod 6).

G. C. Greubel, Table of n, a(n) for n = 1..595
Index entries for linear recurrences with constant coefficients, signature (34,714,-4641,-12376,12376,4641,-714,-34,1).

Cf. A000032, A000204, A110391, A153173, A153175.

Cf. A153179, A153180. - R. J. Mathar, Oct 22 2010

[Lucas(9*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[9*n]/LucasL[n], {n, 1, 50}]
LinearRecurrence[{34,714,-4641,-12376,12376,4641,-714,-34,1},{76,1926,109801,4769326,230701876,10716675201,505618944676,23714405408926,1114769987764201},20] (* Harvey P. Dale, Aug 12 2012 *)

{lucas(n) = fibonacci(n+1) + fibonacci(n-1)};
for(n=0,30, print1( lucas(9*n)/lucas(n), ", ")) \\ G. C. Greubel, Dec 21 2017

From R. J. Mathar, Oct 22 2010: (Start)
a(n) = 34*a(n-1) +714*a(n-2) -4641*a(n-3) -12376*a(n-4) +12376*a(n-5) +4641*a(n-6) -714*a(n-7) -34*a(n-8) +a(n-9).
G.f.: -x*(76-658*x-9947*x^2+13644*x^3+26020*x^4-5306*x^5-1372*x^6+42*x^7 +x^8) / ((x-1)*(x^2+18*x+1)*(x^2-47*x+1)*(x^2+3*x+1)*(x^2-7*x+1)).
a(n) = 1-(-1)^n*A087215(n) -(-1)^n*A005248(n) +A056854(n) +A087265(n). (End)

A153179 a(n) = L(11*n)/L(n) where L(n) = A000204(n).

Original entry on oeis.org

199, 13201, 1970299, 224056801, 28374454999, 3450736132801, 426236170575799, 52337681992411201, 6441140796368008699, 792018481913198430001, 97420733208491869044199, 11981539981561372141075201

Offset: 1

Artur Jasinski, Dec 20 2008

nonn

All numbers in this sequence are:
congruent to 99 mod 100 (iff n is odd),
congruent to 1 mod 100 (iff n is even).

G. C. Greubel, Table of n, a(n) for n = 1..475
Index entries for linear recurrences with constant coefficients, signature (89,4895,-83215,-582505,1514513,1514513,-582505,-83215,4895,89,-1).

Cf. A000032, A000204, A110391, A153173, A153175, A153177.

[Lucas(11*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[11*n]/LucasL[n], {n, 1, 50}]

{lucas(n) = fibonacci(n+1) + fibonacci(n-1)};
for(n=0,30, print1( lucas(11*n)/lucas(n), ", ")) \\ G. C. Greubel, Dec 21 2017

From R. J. Mathar, Oct 22 2010: (Start)
a(n) = +89*a(n-1) +4895*a(n-2) -83215*a(n-3) -582505*a(n-4) +1514513*a(n-5) +1514513*a(n-6) -582505*a(n-7) -83215*a(n-8) +4895*a(n-9) +89*a(n-10) -a(n-11).
G.f.: -1 -1/(1+x) +(-2-47*x)/(x^2+47*x+1) +(2-3*x)/(x^2-3*x+1) +(-2-7*x)/(x^2+7*x+1) +(2-123*x)/(x^2-123*x+1) +(2-18*x)/(x^2-18*x+1).
a(n) = -(-1)^n -(-1)^n*A087265(n) +A005248(n) -(-1)^n*A056854(n) +A065705(n) +A087215(n). (End)

A156094 5 F(2n) (F(2n) - 1) + 1 where F(n) denotes the n-th Fibonacci number.

Original entry on oeis.org

1, 1, 31, 281, 2101, 14851, 102961, 708761, 4865911, 33372361, 228792301, 1568309051, 10749725281, 73680695281, 505017569551, 3461448647801, 23725139605861, 162614572159411, 1114576979567761, 7639424583421961, 52361395886149351

Offset: 0

Stuart Clary, Feb 04 2009

Natural bilateral extension (brackets mark index 0): ..., 15401, 2311, 361, 61, 11, [1], 1, 31, 281, 2101, 14851, ... This is A156095-reversed followed by A156094, without repeating the central 1. That is, A156094(-n) = A156095(n).

Cf. A124296, A124297, A001603, A001604, A156095

a[n_Integer] := 5 Fibonacci[2n] (Fibonacci[2n] - 1) + 1
5(#*(#-1))&/@Fibonacci[Range[0,40,2]]+1 (* Harvey P. Dale, Jan 06 2013 *)

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
Alternate formula: a(n) = L(4n) - 5 F(2n) - 1.
Recurrence: a(n) - 10 a(n-1) + 23 a(n-2) - 10 a(n-3) + a(n-4) = -5.
Recurrence: a(n) - 11 a(n-1) + 33 a(n-2) - 33 a(n-3) + 11 a(n-4) - a(n-5) = 0.
G.f.: A(x) = (1 - 10 x + 53 x^2 - 60 x^3 + 11 x^4)/(1 - 11 x + 33 x^2 - 33 x^3 + 11 x^4 - x^5) = (1 - 10 x + 53 x^2 - 60 x^3 + 11 x^4)/((1 - x) (1 - 7 x + x^2) (1 - 3 x + x^2)).
a(n)=A056854(n)-5*A001906(n)-1. - R. J. Mathar, Feb 23 2009
a(n)=((2*sqrt(5))/2)*(((3-sqrt(5))/2)^n-((3+sqrt(5))/2)^n)+((7+3*sqrt(5))/2)^n+((7-3*sqrt(5))/2)^n-1. - Tim Monahan, Aug 15 2011

A180033 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 5x - 5x^2).

Original entry on oeis.org

1, 6, 35, 205, 1200, 7025, 41125, 240750, 1409375, 8250625, 48300000, 282753125, 1655265625, 9690093750, 56726796875, 332084453125, 1944056250000, 11380703515625, 66623798828125, 390022511718750, 2283231552734375

Offset: 0

Johannes W. Meijer, Aug 09 2010

The a(n) represent the number of n-move routes of a fairy chess piece starting in the corner and side squares (m = 1, 3, 7, 9; 2, 4, 6, 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a white queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen, see A180032.
The sequence above corresponds to 56 red queen vectors, i.e., A[5] vector, with decimal values between 47 and 488. The central squares lead for these vectors to A057088.
Inverse binomial transform of A004187 (without the leading 0).
Equals the INVERT transform of A086347 and the INVERTi transform of A180167. - Gary W. Adamson, Aug 14 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jean-Paul Allouche, Jeffrey Shallit, and Manon Stipulanti, Combinatorics on words and generating Dirichlet series of automatic sequences, arXiv:2401.13524 [math.CO], 2025. See p. 14.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 7.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (5,5).

Cf. A057088, A180032.

Cf. A086347, A180167. - Gary W. Adamson, Aug 14 2010

I:=[1,6]; [n le 2 select I[n] else 5*Self(n-1)+5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

with(LinearAlgebra): nmax:=20; m:=1; A[5]:= [0,0,0,1,0,1,1,1,1]: A:=Matrix([[0,1,1,1,1,0,1,0,1], [1,0,1,1,1,1,0,1,0], [1,1,0,0,1,1,1,0,1], [1,1,0,0,1,1,1,1,0], A[5], [0,1,1,1,1,0,0,1,1], [1,0,1,1,1,0,0,1,1], [0,1,0,1,1,1,1,0,1], [1,0,1,0,1,1,1,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);

LinearRecurrence[{5,5},{1,6}, 30] (* Vincenzo Librandi, Nov 15 2011 *)

my(x='x+O('x^30)); Vec((1+x)/(1-5*x-5*x^2)) \\ G. C. Greubel, Apr 07 2019

((1+x)/(1-5*x-5*x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 07 2019

G.f.: (1+x)/(1 - 5*x - 5*x^2).
a(n) = 5*a(n-1) + 5*a(n-2) with a(0) = 1 and a(1) = 6.
a(n) = ((7+5*A)*A^(-n-1) + (7+5*B)*B^(-n-1))/45 with A = (-5+3*sqrt(5))/10 and B = (-5-3*sqrt(5))/10.
Limit_{k->oo} a(n+k)/a(k) = 2*5^(n/2)/(L(2*n) - F(2*n)*sqrt(5)) with L(n) = A000032(n) and F(n) = A000045(n).
Limit_{k->oo} a(2*n+k)/a(k) = 2*A000351(n)/(A056854(n) - 3*A004187(n)*sqrt(5)) for n >= 1.
Limit_{k->oo} a(2*n-1+k)/a(k) = 2*A000351(n)/(3*A049685(n-1)*sqrt(5) - 5*A033890(n-1)) for n >= 1.
a(n) = A057088(n+1)/5. a(2*n) = 5^n*F(4*(n+1))/3, a(2*n+1) = 5^n*L(2*(2*n+3))/3. - Ehren Metcalfe, Apr 04 2019
E.g.f.: exp(5*x/2)*(15*cosh(3*sqrt(5)*x/2) + 7*sqrt(5)*sinh(3*sqrt(5)*x/2))/15. - Stefano Spezia, Mar 17 2025

A288913 a(n) = Lucas(4*n + 3).

Original entry on oeis.org

4, 29, 199, 1364, 9349, 64079, 439204, 3010349, 20633239, 141422324, 969323029, 6643838879, 45537549124, 312119004989, 2139295485799, 14662949395604, 100501350283429, 688846502588399, 4721424167835364, 32361122672259149, 221806434537978679, 1520283919093591604

Offset: 0

Bruno Berselli, Jun 19 2017

a(n) mod 4 gives A101000.

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A101000.
Cf. A033891: fourth quadrisection of A000045.
Partial sums are in A081007 (after 0).
Positive terms of A098149, and subsequence of A001350, A002878, A016897, A093960, A068397.
Quadrisection of A000032: A056854 (first), A056914 (second), A246453 (third, without 11), this sequence (fourth).

[Lucas(4*n + 3): n in [0..30]]; // G. C. Greubel, Dec 22 2017

LucasL[4 Range[0, 21] + 3]
LinearRecurrence[{7,-1}, {4,29}, 30] (* G. C. Greubel, Dec 22 2017 *)

Vec((4 + x)/(1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017

from sympy import lucas
def a(n):  return lucas(4*n + 3)
print([a(n) for n in range(22)]) # Michael S. Branicky, Apr 29 2021

def L():
    x, y = -1, 4
    while True:
        yield y
        x, y = y, 7*y - x
r = L(); [next(r) for  in (0..21)] # _Peter Luschny, Jun 20 2017

G.f.: (4 + x)/(1 - 7*x + x^2).
a(n) = 7*a(n-1) - a(n-2) for n>1, with a(0)=4, a(1)=29.
a(n) = ((sqrt(5) + 1)^(4*n + 3) - (sqrt(5) - 1)^(4*n + 3))/(8*16^n).
a(n) = Fibonacci(4*n+4) + Fibonacci(4*n+2).
a(n) = 4*A004187(n+1) + A004187(n).
a(n) = 5*A003482(n) + 4 = 5*A081016(n) - 1.
a(n) = A002878(2*n+1) = A093960(2*n+3) = A001350(4*n+3) = A068397(4*n+3).
a(n+1)*a(n+k) - a(n)*a(n+k+1) = 15*Fibonacci(4*k). Example: for k=6, a(n+1)*a(n+6) - a(n)*a(n+7) = 15*Fibonacci(24) = 695520.

A081069 a(n) = Lucas(4n)+2 = Lucas(2n)^2.

Original entry on oeis.org

4, 9, 49, 324, 2209, 15129, 103684, 710649, 4870849, 33385284, 228826129, 1568397609, 10749957124, 73681302249, 505019158609, 3461452808004, 23725150497409, 162614600673849, 1114577054219524, 7639424778862809

Offset: 0

R. K. Guy, Mar 04 2003

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Pridon Davlianidze, Problem B-1270, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; Four Telescopic Infinite Products, Solution to Problem B-1270 by Jason L. Smith, ibid., Vol. 59, No. 2 (2021), pp. 183-184.
Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, Table 2, k=2.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000032 (Lucas numbers), A001622, A005248, A056854.

[ Lucas(2*n)^2: n in [0..70] ]; // Vincenzo Librandi, Apr 16 2011

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d,`,luc(4*n)+2) od: # James Sellers, Mar 05 2003
G:=(x,n)-> cos(x)^n +cos(3*x)^n: seq(simplify(2^(4*n)*G(Pi/5,2*n)^2), n=0..19) # Gary Detlefs, Dec 05 2010
t:= n-> sum(fibonacci(4*k+2),k=0..n):seq(5*t(n)+4,n=-1..18); # Gary Detlefs, Dec 06 2010

LucasL[4*Range[0,20]]+2 (* Harvey P. Dale, Sep 09 2012 *)

a(n) = A005248(n)^2 = A056854(n)+2.
a(n) = 8a(n-1) - 8a(n-2) + a(n-3).
a(n) = 2^(4*n)*(cos(Pi/5)^(2*n)+cos(3*Pi/5)^(2*n))^2. - Gary Detlefs, Dec 05 2010
From Gary Detlefs, Dec 06 2010: (Start)
a(n) = 7*a(n-1)-a(n-2)-10, n>1.
a(n) = 5*Sum_{k=0..n}(Fibonacci(4*k+2))+4, with offset -1. (End)
G.f.: -(9*x^2-23*x+4)/((x-1)*(x^2-7*x+1)). - Colin Barker, Jun 24 2012
Product_{n>=0} (1 + 5/a(n)) = 3*phi^2/2, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 04 2024
a(n) = Sum_{k>=0} Lucas(2*n*k)/(Lucas(2*n)^k). - Diego Rattaggi, Jan 12 2025
E.g.f.: 2*(cosh(x) + exp(7*x/2)*cosh(3*sqrt(5)*x/2) + sinh(x)). - Stefano Spezia, Jan 20 2025

A215465 a(n) = (Lucas(4n) - Lucas(2n))/4.

Original entry on oeis.org

0, 1, 10, 76, 540, 3751, 25840, 177451, 1217160, 8344876, 57202750, 392089501, 2687463360, 18420257701, 126254611990, 865362736876, 5931286406640, 40653646980451, 278644255208560, 1909856172864751, 13090349042248500

Offset: 0

R. J. Mathar, Aug 11 2012

This is a divisibility sequence, that is, if n | m then a(n) | a(m). However, it is not a strong divisibility sequence. It is the case k = 3 of a 1-parameter family of 4th-order linear divisibility sequences with o.g.f. x*(1 - x^2)/( (1 - k*x + x^2)*(1 - (k^2 - 2)*x + x^2) ). Compare with A000290 (case k = 2) and A085695 (case k = -3). - Peter Bala, Jan 17 2014

In general, for distinct integers r and s with r = s (mod 2), the sequence Lucas(r*n) - Lucas(s*n) is a fourth-order divisibility sequence. See A273622 for the case r = 3, s = 1. - Peter Bala, May 27 2016

			a(3) = 76 because the 12th (4 * 3rd) Lucas number is 22, the 6th (2 * 3rd) Lucas number is 18, and (322 - 18)/4 = 304/4 = 76.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Peter Bala, Lucas sequences and divisibility sequences
E. L. Roettger and H. C. Williams, Appearance of Primes in Fourth-Order Odd Divisibility Sequences, J. Int. Seq., Vol. 24 (2021), Article 21.7.5.
Hugh Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory vol. 7 (5) (2011) 1255-1277
Index entries for linear recurrences with constant coefficients, signature (10,-23,10,-1).

Cf. A085695, A215466, A273622.

[(Lucas(4*n) - Lucas(2*n))/4: n in [0..20]]; // Vincenzo Librandi, Dec 23 2012

A215465 := proc(n)
    (A000032(4*n)-A000032(2*n))/4 ;
end proc:

Table[(LucasL[4n] - LucasL[2n])/4, {n, 0, 19}] (* Alonso del Arte, Aug 11 2012 *)
CoefficientList[Series[-x*(x-1)*(1+x)/((x^2 - 7*x + 1)* (x^2 - 3*x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 23 2012 *)
LinearRecurrence[{10,-23,10,-1},{0,1,10,76},50] (* G. C. Greubel, Jun 02 2016 *)

{a(n) = my(w = quadgen(5)^(2*n)); (2*real(w^2-w) + imag(w^2-w))/4}; /* Michael Somos, Dec 29 2022 */

4*a(n) = A000032(4*n) - A000032(2*n).
a(n) = A056854(n)/4 - A005248(n)/4.
G.f.: -x*(x-1)*(1+x) / ( (x^2-7*x+1)*(x^2-3*x+1) ).
a(n) = 10*a(n-1) - 23*a(n-2) + 10*a(n-3) - a(n-4). - G. C. Greubel, Jun 02 2016
a(n) = 2^(-2-n)*((7-3*sqrt(5))^n-(3-sqrt(5))^n-(3+sqrt(5))^n+(7+3*sqrt(5))^n). - Colin Barker, Jun 02 2016
a(n) = a(-n) for all n in Z. - Michael Somos, Dec 29 2022

cp's OEIS Frontend

A153173 a(n) = L(5*n)/L(n) where L(n) = Lucas number A000204(n).

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A246453 Lucas numbers (A000204) of the form n^2 + 2.

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A153175 a(n) = L(7*n)/L(n) where L(n) = Lucas number A000204(n).

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A153177 a(n) = L(9*n)/L(n) where L(n) = Lucas number A000204(n).

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A153179 a(n) = L(11*n)/L(n) where L(n) = A000204(n).

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A156094 5 F(2n) (F(2n) - 1) + 1 where F(n) denotes the n-th Fibonacci number.

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A180033 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 5*x - 5*x^2).

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A180033 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 5x - 5x^2).