A014448 -id:A014448 - cp's OEIS frontend

A065705 a(n) = Lucas(10*n).

2, 123, 15127, 1860498, 228826127, 28143753123, 3461452808002, 425730551631123, 52361396397820127, 6440026026380244498, 792070839848372253127, 97418273275323406890123, 11981655542024930675232002, 1473646213395791149646646123, 181246502592140286475862241127

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 25 2003

Lim_{n->infinity} a(n+1)/a(n) = (123 + sqrt(15125))/2 = 122.9918693812...
Lim_{n->infinity} a(n)/a(n+1) = (123 - sqrt(15125))/2 = 0.00813061875578...
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^10) = 1.0081300769... = 1 + 1/(123 + 1/(15127 + 1/(1860498 + ...))).
Also F(-Phi^10) = 0.9918699143... has the continued fraction representation 1 - 1/(123 - 1/(15127 - 1/(1860498 - ...))) and the simple continued fraction expansion 1/(1 + 1/((123 - 2) + 1/(1 + 1/((15127 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + ...))))))).
F(Phi^10)*F(-Phi^10) = 0.9999338930... has the simple continued fraction expansion 1/(1 + 1/((123^2 - 4) + 1/(1 + 1/((15127^2 - 4) + 1/(1 + 1/((1860498^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^10)/F(-Phi^10) = 1.0081967213... has the simple continued fraction expansion 1 + 1/((123 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + 1/(28143753123 - 2) + 1/(1 + ...))))). (End)

			a(4) = 228826127 = 123*a(3) - a(2) = 123*1860498 - 15127=((123+sqrt(15125))/2)^4 + ( (123-sqrt(15125))/2)^4 =228826126.99999999562986 + 0.00000000437013 = 228826127.
a(4) = L(10 * 4) = L(40) = 228826127. - _Indranil Ghosh_, Feb 08 2017

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Indranil Ghosh, Table of n, a(n) for n = 0..477
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem,Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (123,-1).

Cf. A000032: a(n) = A000032(10*n).

Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(10*n): n in [0..90]]; // Vincenzo Librandi, Apr 14 2011

LucasL[10*Range[0, 20]] (* Paolo Xausa, Mar 04 2024 *)

a(n) = 123*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 123.
a(n) = ((123 + sqrt(15125))/2)^n + ((123 - sqrt(15125))/2)^n.
a(n)^2 = a(2*n) + 2.
G.f.: (2 - 123*x)/(1 - 123*x + x^2). -  Philippe Deléham, Nov 18 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(10*n+10)/F(10) - F(10*n-10)/F(10) = A049670(n+1) - A049670(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^10 = [34, 55; 55, 89].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
121*Sum_{n >= 1} 1/(a(n) - 125/a(n)) = 1: (125 = Lucas(10) + 2 and 121 = Lucas(10) - 2)
125*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 121/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 123*x^2 + 15128*x^3 + ... is the o.g.f. for A049670. (End)
E.g.f.: exp((1/2)*(123 - 55*sqrt(5))*x)*(1 + exp(55*sqrt(5)*x)). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^5 - 5*Lucas(2*n)^3 + 5*Lucas(2*n) = 2*T(5, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (123 - sqrt(15125))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(15125) - 123)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

A089775 Lucas numbers L(12n).

Original entry on oeis.org

2, 322, 103682, 33385282, 10749957122, 3461452808002, 1114577054219522, 358890350005878082, 115561578124838522882, 37210469265847998489922, 11981655542024930675232002, 3858055874062761829426214722, 1242282009792667284144565908482, 400010949097364802732720796316482

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 09 2004

a(n+1)/a(n) converges to (322 + sqrt(103680))/2 = 321.996894379... a(0)/a(1) = 2/322; a(1)/a(2) = 322/103682; a(2)/a(3) = 103682/33385282; a(3)/a(4) = 33385282/10749957122; etc. Lim_{n -> inf} a(n)/a(n+1) = 0.00310562... = 2/(322 + sqrt(103680)) = (322 - sqrt(103680))/2.

			a(4) = 10749957122 = 322*a(3) - a(2) = 322*33385282 - 103682 = ((322 + sqrt(103680))/2)^4 + ((322 - sqrt(103680))/2)^4.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.

Nathaniel Johnston, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (322, -1).

Cf. A000032, A060964.
a(n) = A000032(12n).
Row 9 * 2 of array A188644
Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11).

[ Lucas(12*n) : n in [0..70]]; // Vincenzo Librandi, Apr 15 2011

Table[LucasL[12n], {n, 0, 13}] (* Indranil Ghosh, Mar 15 2017 *)

Vec((2 - 322*x)/(1 - 322*x + x^2) + O(x^14)) \\ Indranil Ghosh, Mar 15 2017

a(n) = 322*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 322
a(n) = ((322 + sqrt(103680))/2)^n + ((322 - sqrt(103680))/2)^n.
(a(n))^2 = a(2n) + 2.
G.f.: (2-322*x)/(1-322*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^6 - 6*Lucas(2*n)^4 + 9*Lucas(2*n)^2 - 2 = 2*T(6, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
320*Sum_{n >= 1} 1/(a(n) - 324/a(n)) = 1: (324 = Lucas(12) + 2 and 320 = Lucas(12) - 2)
324*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 320/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (322 - sqrt(103680))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(103680) - 322)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. (End)

a(11) - a(13) from Vincenzo Librandi, Apr 15 2011

A309220 Square array A read by antidiagonals: the columns are given by A(n,1)=1, A(n,2)=n+1, A(n,3) = n^2+2n+3, A(n,4) = n^3+3n^2+6n+4, A(n,5) = n^4+4n^3+10n^2+12*n+7, ..., whose coefficients are given by A104509 (see also A118981).

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 11, 14, 1, 5, 18, 36, 34, 1, 6, 27, 76, 119, 82, 1, 7, 38, 140, 322, 393, 198, 1, 8, 51, 234, 727, 1364, 1298, 478, 1, 9, 66, 364, 1442, 3775, 5778, 4287, 1154, 1, 10, 83, 536, 2599, 8886, 19602, 24476, 14159, 2786, 1, 11, 102, 756, 4354, 18557

Offset: 1

N. J. A. Sloane, Aug 12 2019, based on R. J. Mathar's 2011 analysis of A118980

As pointed out by Peter Munn, A117938 gives the same triangle, except that it has an additional diagonal at the right. - N. J. A. Sloane, Aug 13 2019

			The first few antidiagonals are:
1,
1,2,
1,3,6,
1,4,11,14,
1,5,18,36,34,
1,6,27,76,119,82,
1,7,38,140,322,393,198,
...
The first nine rows of A are
1, 2, 6, 14, 34, 82, 198, 478, 1154, 2786, 6726, 16238, ...
1, 3, 11, 36, 119, 393, 1298, 4287, 14159, 46764, 154451, 510117, ...
1, 4, 18, 76, 322, 1364, 5778, 24476, 103682, 439204, 1860498, 7881196, ...
1, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, ...
1, 6, 38, 234, 1442, 8886, 54758, 337434, 2079362, 12813606, 78960998, 486579594, ...
1, 7, 51, 364, 2599, 18557, 132498, 946043, 6754799, 48229636, 344362251, 2458765393, ...
1, 8, 66, 536, 4354, 35368, 287298, 2333752, 18957314, 153992264, 1250895426, 10161155672, ...
1, 9, 83, 756, 6887, 62739, 571538, 5206581, 47430767, 432083484, 3936182123, 35857722591, ...
1, 10, 102, 1030, 10402, 105050, 1060902, 10714070, 108201602, 1092730090, 11035502502, 111447755110, ...

Cf. A104509, A117938, A118980, A118981, A099425 (top row), A006497 (essentially the 2nd row), A014448 (essentially the 3rd row), A087130 (essentially the 4th row).

M := 12;
A:=Array(1..2*M,1..2*M,0):
for i from 1 to M do A[i,1]:=1; od:
S := series((1 + x^2)/(1-x-x^2 + x*y), x, 120): # this is g.f. for A104509
for n from 1 to M do
R2 := expand(coeff(S, x, n));
R3 := [seq(abs(coeff(R2,y,n-i)),i=0..n)];
f := k-> add( R3[i]*k^(n-i+1), i=1..nops(R3) ): # this is the formula for the (n+1)-st column
s1 := [seq(f(i),i=1..M)];
for i from 1 to M do A[i,n+1]:=s1[i]; od:
od:
for i from 1 to M do lprint([seq(A[i,j],j=1..M)]); od:
# alternative by R. J. Mathar, Aug 13 2019 :
A104509 := proc(n,k)
    (1+x^2)/(1-x-x^2+x*y) ;
    coeftayl(%,x=0,n) ;
    coeftayl(%,y=0,k) ;
end proc:
A309220 := proc(n::integer,k::integer)
    local x;
    add( abs(A104509(k-1,i))*x^i,i=0..k-1) ;
    subs(x=n,%) ;
end proc:
seq( seq(A309220(d-k,k),k=1..d-1),d=2..13) ;

A100233 a(n) = Lucas(3*n) - 1.

Original entry on oeis.org

1, 3, 17, 75, 321, 1363, 5777, 24475, 103681, 439203, 1860497, 7881195, 33385281, 141422323, 599074577, 2537720635, 10749957121, 45537549123, 192900153617, 817138163595, 3461452808001, 14662949395603, 62113250390417, 263115950957275, 1114577054219521

Offset: 0

Paul D. Hanna, Nov 29 2004

Main diagonal of triangle A100232.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A000032, A100230, A100232, A016064.

I:=[1, 3, 17]; [n le 3 select I[n] else 5*Self(n-1) -3*Self(n-2) -Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[3*n] - 1, {n,0,50}] (* or *) LinearRecurrence[{5,-3,-1}, {1,3,17}, 30] (* G. C. Greubel, Dec 21 2017 *)

a(n)=if(n==0,1,n*polcoeff(log((1-x)/(1-4*x-x^2)+x*O(x^n)),n))

Vec((1-2*x+5*x^2)/((1-x)*(1-4*x-x^2)) + O(x^40)) \\ Colin Barker, Jun 02 2016

a(n) = A014448(n) - 1.
a(n) = 4*a(n-1) + a(n-2) + 4 for n>1, with a(0)=1, a(1)=3.
G.f.: Sum_{n>=1} a(n)*x^n/n = log((1-x)/(1-4*x-x^2)).
a(n) = [x^n] ( 1 + 2*x + sqrt(1 + 2*x + 5*x^2) )^n. Cf. A016064. - Peter Bala, Jun 23 2015
From Colin Barker, Jun 02 2016: (Start)
a(n) = -1+(2-sqrt(5))^n+(2+sqrt(5))^n.
a(n) = 5*a(n-1)-3*a(n-2)-a(n-3) for n>2.
G.f.: (1-2*x+5*x^2) / ((1-x)*(1-4*x-x^2)).
(End)

New definition from Ralf Stephan, Dec 01 2004

A163070 a(n) = ((4+sqrt(5))(2+sqrt(5))^n + (4-sqrt(5))(2-sqrt(5))^n)/2.

Original entry on oeis.org

4, 13, 56, 237, 1004, 4253, 18016, 76317, 323284, 1369453, 5801096, 24573837, 104096444, 440959613, 1867934896, 7912699197, 33518731684, 141987625933, 601469235416, 2547864567597, 10792927505804, 45719574590813

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jul 20 2009

nonn

Binomial transform of A163069. Second binomial transform of A163141. Inverse binomial transform of A163071.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A000045, A001076, A014448, A163069, A163071, A163141.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-5); S:=[ ((4+r)*(2+r)^n+(4-r)*(2-r)^n)/2: n in [0..21] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jul 21 2009

LinearRecurrence[{4,1},{4,13},30] (* Harvey P. Dale, Sep 19 2011 *)

x='x+O('x^30); Vec((4-3*x)/(1-4*x-x^2)) \\ G. C. Greubel, Jan 08 2018

a(n) = 4*a(n-1) + a(n-2) for n > 1; a(0) = 4, a(1) = 13.
G.f.: (4-3*x)/(1-4*x-x^2).
a(n) = 2*A000032(3*n) + 5*A000045(3*n)/2 = 2*A014448(n) + 5*A001076(n). - Diego Rattaggi, Aug 09 2020

Edited and extended beyond a(5) by Klaus Brockhaus, Jul 21 2009

A352362 Array read by ascending antidiagonals. T(n, k) = L(k, n) where L are the Lucas polynomials.

Original entry on oeis.org

2, 2, 0, 2, 1, 2, 2, 2, 3, 0, 2, 3, 6, 4, 2, 2, 4, 11, 14, 7, 0, 2, 5, 18, 36, 34, 11, 2, 2, 6, 27, 76, 119, 82, 18, 0, 2, 7, 38, 140, 322, 393, 198, 29, 2, 2, 8, 51, 234, 727, 1364, 1298, 478, 47, 0, 2, 9, 66, 364, 1442, 3775, 5778, 4287, 1154, 76, 2

Offset: 0

Peter Luschny, Mar 18 2022

			Array starts:
n\k 0, 1,  2,   3,    4,     5,      6,       7,        8, ...
--------------------------------------------------------------
[0] 2, 0,  2,   0,    2,     0,      2,       0,        2, ... A010673
[1] 2, 1,  3,   4,    7,    11,     18,      29,       47, ... A000032
[2] 2, 2,  6,  14,   34,    82,    198,     478,     1154, ... A002203
[3] 2, 3, 11,  36,  119,   393,   1298,    4287,    14159, ... A006497
[4] 2, 4, 18,  76,  322,  1364,   5778,   24476,   103682, ... A014448
[5] 2, 5, 27, 140,  727,  3775,  19602,  101785,   528527, ... A087130
[6] 2, 6, 38, 234, 1442,  8886,  54758,  337434,  2079362, ... A085447
[7] 2, 7, 51, 364, 2599, 18557, 132498,  946043,  6754799, ... A086902
[8] 2, 8, 66, 536, 4354, 35368, 287298, 2333752, 18957314, ... A086594
[9] 2, 9, 83, 756, 6887, 62739, 571538, 5206581, 47430767, ... A087798
A007395|A059100|
    A001477 A079908

Eric Weisstein's World of Mathematics, Lucas Polynomial

Rows: A010673, A000032, A002203, A006497, A014448, A087130, A085447, A086902, A086594, A087798.
Columns: A007395, A001477, A059100, A079908.
Cf. A320570 (main diagonal), A114525, A309220 (variant), A117938 (variant), A352361 (Fibonacci polynomials), A350470 (Jacobsthal polynomials).

T := (n, k) -> (n/2 + sqrt((n/2)^2 + 1))^k + (n/2 - sqrt((n/2)^2 + 1))^k:
seq(seq(simplify(T(n - k, k)), k = 0..n), n = 0..10);

Table[LucasL[k, n], {n, 0, 9}, {k, 0, 9}] // TableForm
(* or *)
T[ 0, k_] := 2 Mod[k+1, 2]; T[n_, 0] := 2;
T[n_, k_] := n^k Hypergeometric2F1[1/2 - k/2, -k/2, 1 - k, -4/n^2];
Table[T[n, k], {n, 0, 9}, {k, 0, 8}] // TableForm

T(n, k) = ([0, 1; 1, k]^n*[2; k])[1, 1] ;
export(T)
for(k = 0, 9, print(parvector(10, n, T(n - 1, k))))

T(n, k) = Sum_{j=0..floor(k/2)} binomial(k-j, j)*(k/(k-j))*n^(k-2*j) for k >= 1.
T(n, k) = (n/2 + sqrt((n/2)^2 + 1))^k + (n/2 - sqrt((n/2)^2 + 1))^k.
T(n, k) = [x^k] ((2 - n*x)/(1 - n*x - x^2)).
T(n, k) = n^k*hypergeom([1/2 - k/2, -k/2], [1 - k], -4/n^2) for n,k >= 1.

A191347 Array read by antidiagonals: ((floor(sqrt(n)) + sqrt(n))^k + (floor(sqrt(n)) - sqrt(n))^k)/2 for columns k >= 0 and rows n >= 0.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 1, 1, 0, 4, 3, 1, 1, 0, 8, 7, 4, 2, 1, 0, 16, 17, 10, 8, 2, 1, 0, 32, 41, 28, 32, 9, 2, 1, 0, 64, 99, 76, 128, 38, 10, 2, 1, 0, 128, 239, 208, 512, 161, 44, 11, 2, 1, 0, 256, 577, 568, 2048, 682, 196, 50, 12, 3, 1

Offset: 0

Charles L. Hohn, May 31 2011

			1, 0,  0,   0,    0,    0,     0,      0,       0,        0,        0, ...
1, 1,  2,   4,    8,   16,    32,     64,     128,      256,      512, ...
1, 1,  3,   7,   17,   41,    99,    239,     577,     1393,     3363, ...
1, 1,  4,  10,   28,   76,   208,    568,    1552,     4240,    11584, ...
1, 2,  8,  32,  128,  512,  2048,   8192,   32768,   131072,   524288, ...
1, 2,  9,  38,  161,  682,  2889,  12238,   51841,   219602,   930249, ...
1, 2, 10,  44,  196,  872,  3880,  17264,   76816,   341792,  1520800, ...
1, 2, 11,  50,  233, 1082,  5027,  23354,  108497,   504050,  2341691, ...
1, 2, 12,  56,  272, 1312,  6336,  30592,  147712,   713216,  3443712, ...
1, 3, 18, 108,  648, 3888, 23328, 139968,  839808,  5038848, 30233088, ...
1, 3, 19, 117,  721, 4443, 27379, 168717, 1039681,  6406803, 39480499, ...
1, 3, 20, 126,  796, 5028, 31760, 200616, 1267216,  8004528, 50561600, ...
1, 3, 21, 135,  873, 5643, 36477, 235791, 1524177,  9852435, 63687141, ...
1, 3, 22, 144,  952, 6288, 41536, 274368, 1812352, 11971584, 79078912, ...
1, 3, 23, 153, 1033, 6963, 46943, 316473, 2133553, 14383683, 96969863, ...
...

Row 1 is A000007, row 2 is A011782, row 3 is A001333, row 4 is A026150, row 5 is A081294, row 6 is A001077, row 7 is A084059, row 8 is A108851, row 9 is A084128, row 10 is A081341, row 11 is A005667, row 13 is A141041.
Row 3*2 is A002203, row 4*2 is A080040, row 5*2 is A155543, row 6*2 is A014448, row 8*2 is A080042, row 9*2 is A170931, row 11*2 is A085447.
Cf. A191348 which uses ceiling() in place of floor().

T(n, k) = if (n==0, k==0, my(x=sqrtint(n)); sum(i=0, (k+1)\2, binomial(k, 2*i)*x^(k-2*i)*n^i));
matrix(9,9, n, k, T(n-1,k-1)) \\ Michel Marcus, Aug 22 2019

T(n, k) = if (k==0, 1, if (k==1, sqrtint(n), T(n,k-2)*(n-T(n,1)^2) + T(n,k-1)*T(n,1)*2));
matrix(9, 9, n, k, T(n-1, k-1)) \\ Charles L. Hohn, Aug 22 2019

For each row n>=0 let T(n,0)=1 and T(n,1)=floor(sqrt(n)), then for each column k>=2: T(n,k)=T(n,k-2)*(n-T(n,1)^2) + T(n,k-1)*T(n,1)*2. - Charles L. Hohn, Aug 22 2019

T(n, k) = Sum_{i=0..floor((k+1)/2)} binomial(k, 2*i)*floor(sqrt(n))^(k-2*i)*n^i for n > 0, with T(0, 0) = 1 and T(0, k) = 0 for k > 0. - Michel Marcus, Aug 23 2019

A014731 Squares of even Lucas numbers.

Original entry on oeis.org

4, 16, 324, 5776, 103684, 1860496, 33385284, 599074576, 10749957124, 192900153616, 3461452808004, 62113250390416, 1114577054219524, 20000273725560976, 358890350005878084, 6440026026380244496, 115561578124838522884, 2073668380220713167376

Offset: 0

Mohammad K. Azarian

Colin Barker, Table of n, a(n) for n = 0..700
Index entries for linear recurrences with constant coefficients, signature (17,17,-1).

Cf. A000045, A014448.

(Table[LucasL@ n, {n, 0, 52}] /. n_ /; OddQ@ n -> Nothing)^2 (* Michael De Vlieger, Mar 04 2016 *)
LinearRecurrence[{17,17,-1},{4,16,324},20] (* Harvey P. Dale, Nov 19 2024 *)

Vec(4*(1-13*x-4*x^2)/((1+x)*(1-18*x+x^2)) + O(x^20)) \\ Colin Barker, Mar 04 2016

a(n) = Fibonacci(6*n+3) - 2*Fibonacci(6*n) + 2*(-1)^n. - Ralf Stephan, May 14 2004
G.f.: 4*(-4*x^2-13*x+1)/((1+x)*(1-18*x+x^2)). - Ralf Stephan, May 14 2004
From Colin Barker, Mar 04 2016: (Start)
a(n) = 2*(-1)^n+(9+4*sqrt(5))^(-n)+(9+4*sqrt(5))^n.
a(n) = 17*a(n-1)+17*a(n-2)-a(n-3) for n>2. (End)
a(n) = A014448(n)^2. - Sean A. Irvine, Nov 18 2018
a(n) = 5*Fibonacci(3*n)^2 + 4*(-1)^n. - Amiram Eldar, Jan 11 2022

More terms from Erich Friedman.

A245580 Smallest Lucas number L(m) > L(n) that is divisible by the n-th Lucas number L(n) = A000204(n).

Original entry on oeis.org

3, 18, 76, 322, 1364, 5778, 24476, 103682, 439204, 1860498, 7881196, 33385282, 141422324, 599074578, 2537720636, 10749957122, 45537549124, 192900153618, 817138163596, 3461452808002, 14662949395604, 62113250390418, 263115950957276, 1114577054219522

Offset: 1

Michel Lagneau, Jul 26 2014

Property: a(1) = L(2) and a(n) = L(3*n), for n >=2, where L = A000204 are the Lucas numbers.

			a(4) = 322 is the first Lucas number that is divisible by 7, the 4th Lucas number, so a(4) = 322. With the property a(n) = L(3*n), a(4) = A000204(12).

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000204, A237268, A014448 .

Table[k=1;While[Mod[LucasL[k],LucasL[n]] !=0||LucasL[k]==LucasL[n],k++];LucasL[k],{n,0,30}]
LinearRecurrence[{4,1},{3,18,76},30] (* Harvey P. Dale, Jan 05 2022 *)

Vec(-x*(x^2+6*x+3)/(x^2+4*x-1) + O(x^100)) \\ Colin Barker, Jul 31 2014

a(n) = A014448(n), n>1.
From Colin Barker, Jul 29 2014: (Start)
a(n) = (2-sqrt(5))^n+(2+sqrt(5))^n for n>1.
a(n) = 4*a(n-1)+a(n-2) for n>3.
G.f.: -x*(x^2+6*x+3) / (x^2+4*x-1). (End)

A245799 Lucas(3*n) - Fibonacci(n).

Original entry on oeis.org

2, 3, 17, 74, 319, 1359, 5770, 24463, 103661, 439170, 1860443, 7881107, 33385138, 141422091, 599074201, 2537720026, 10749956135, 45537547527, 192900151034, 817138159415, 3461452801237, 14662949384658, 62113250372707, 263115950928619, 1114577054173154

Offset: 0

Vincenzo Librandi, Aug 02 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000032, A000045, A005248, A014448.

[Lucas(3*n) - Fibonacci(n): n in [0..30]];

Table[LucasL[3 n] - Fibonacci[n], {n, 0, 30}] (* or *) CoefficientList[Series[(2 - 7 x + 6 x^2 + 5 x^3)/((1 - x - x^2) (1 - 4 x - x^2)), {x, 0, 40}], x]

import sympy
{print(sympy.lucas(3*n)-sympy.fibonacci(n),end=', ') for n in range(50)}
# Derek Orr, Aug 02 2014

G.f.: (2 - 7*x + 6*x^2 + 5*x^3)/((1 - x - x^2)(1 - 4*x - x^2)).

cp's OEIS Frontend

A065705 a(n) = Lucas(10*n).

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A089775 Lucas numbers L(12n).

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A309220 Square array A read by antidiagonals: the columns are given by A(n,1)=1, A(n,2)=n+1, A(n,3) = n^2+2n+3, A(n,4) = n^3+3*n^2+6*n+4, A(n,5) = n^4+4*n^3+10*n^2+12*n+7, ..., whose coefficients are given by A104509 (see also A118981).

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A100233 a(n) = Lucas(3*n) - 1.

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A163070 a(n) = ((4+sqrt(5))*(2+sqrt(5))^n + (4-sqrt(5))*(2-sqrt(5))^n)/2.

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A352362 Array read by ascending antidiagonals. T(n, k) = L(k, n) where L are the Lucas polynomials.

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A191347 Array read by antidiagonals: ((floor(sqrt(n)) + sqrt(n))^k + (floor(sqrt(n)) - sqrt(n))^k)/2 for columns k >= 0 and rows n >= 0.

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A309220 Square array A read by antidiagonals: the columns are given by A(n,1)=1, A(n,2)=n+1, A(n,3) = n^2+2n+3, A(n,4) = n^3+3n^2+6n+4, A(n,5) = n^4+4n^3+10n^2+12*n+7, ..., whose coefficients are given by A104509 (see also A118981).

A163070 a(n) = ((4+sqrt(5))(2+sqrt(5))^n + (4-sqrt(5))(2-sqrt(5))^n)/2.