A000032 -id:A000032 - cp's OEIS frontend

A130310 Minimal (or "greedy") Lucas representation of n, in which L(0) = 2 and L(2) = 3 are not allowed in the same representation (hence the correct representation of the integer 5 is 1010 rather than 101). A binary system of integers with Lucas numbers (A000032) as a base.

0, 10, 1, 100, 1000, 1010, 1001, 10000, 10010, 10001, 10100, 100000, 100010, 100001, 100100, 101000, 101010, 101001, 1000000, 1000010, 1000001, 1000100, 1001000, 1001010, 1001001, 1010000, 1010010, 1010001, 1010100, 10000000, 10000010, 10000001, 10000100, 10001000

Offset: 0

Casey Mongoven, May 21 2007

			a(9) = 10001 because 7 + 2 = 9.
a(10) is 10100 because 7 + 3 = 10.

Richard A. Dunlap, The Golden Ratio and Fibonacci Numbers, Singapore, World Scientific, 1997, pp. 73-77.
Edouard Zeckendorf, Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège, Vol. 41 (1972), pp. 179-182.

Amiram Eldar, Table of n, a(n) for n = 0..10000
Ron Knott, Using Powers of Phi to represent Integers.

Cf. A000032, A014417, A116543, A130311.

a[n_] := Module[{s = {}, m = n, k = 1}, While[m > 0, If[m == 1, k = 1; AppendTo[s, k]; m = 0, If[m == 2, k = 0; AppendTo[s, k]; m = 0, While[LucasL[k] <= m, k++]; k--; AppendTo[s, k]; m -= LucasL[k]; k = 1]]]; FromDigits @ IntegerDigits[Total[2^s], 2]]; Array[a, 30, 0] (* Amiram Eldar, Feb 17 2022 *)

Definition corrected by Casey Mongoven, May 29 2010

a(0) and more terms from Amiram Eldar, Feb 17 2022

A240926 a(n) = 2 + L(2*n) = 2 + A005248(n), n >= 0, with the Lucas numbers (A000032).

Original entry on oeis.org

4, 5, 9, 20, 49, 125, 324, 845, 2209, 5780, 15129, 39605, 103684, 271445, 710649, 1860500, 4870849, 12752045, 33385284, 87403805, 228826129, 599074580, 1568397609, 4106118245, 10749957124, 28143753125, 73681302249, 192900153620, 505019158609

Offset: 0

Kival Ngaokrajang, Aug 03 2014

This sequence also gives the curvature of touching circles inscribed in a special way in the smaller segment of a circle of radius 5/4 cut by a chord of length 2.
Consider a circle C of radius 5/4 (in some length units) with a chord of length 2. This has been chosen so that the larger sagitta also has length 2. The smaller sagitta has length 1/2. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/4, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle, C_n = 1/R_n, n >= 0, is conjectured to be a(n). See an illustration given in the link. As found by Wolfdieter Lang (see part II of the proof given by W. Lang in the link), this circle problem is related to the nonnegative solutions of the Pell equation X^2 - 5*Y^2 = 4: a(n) = 2 + X(n) = 2 + A005248(n). For the larger segment below the chord (with sagitta length 2) the sequence would be A115032, see W. Lang's proof given in part I of the link.
If the circle radius and the sagitta length were both equal to 1, the curvature sequence would be A099938.
Essentially a duplicate of A092387. - R. J. Mathar, Jul 07 2023

G. C. Greubel, Table of n, a(n) for n = 0..2375
Wolfdieter Lang, Proof of the coincidence of a(n) with the touching circle problem (part II).
Wolfdieter Lang, Figures for various touching circle problems.
Kival Ngaokrajang, Illustration of initial terms.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000032, A005248, A005592, A099938, A115032, A246638, A246639, A246640, A246641, A246642, A338303.

[2+Lucas(2*n): n in [0..40]]; // Vincenzo Librandi, Oct 08 2015

Table[2 + LucasL[2 n], {n, 0, 50}] (* Vincenzo Librandi, Oct 08 2015 *)

vector(100, n, n--; 2 + fibonacci(2*n-1) + fibonacci(2*n+1)) \\ Altug Alkan, Oct 08 2015

Conjectures (proved in the next entry) from Colin Barker, Aug 25 2014 (and Aug 27 2014): (Start)
a(n) = (2 + ((1/2)*(3-sqrt(5)))^n + ((1/2)*(3+sqrt(5)))^n).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
G.f.: -(5*x^2-11*x+4) / ((x-1)*(x^2-3*x+1)). (End)
From Wolfdieter Lang, Aug 26 2014: (Start)
a(n) = 2 + S(n, 3) - S(n-2, 3) = 2 + 2*S(n, 3)  - 3*S(n-1, 3).
a(n) = 3*a(n-1) - a(n-2) - 2, n >= 1, with a(-1)= 5 and a(0) = 4 (from the S(n, 3) recurrence or from A005248).
The first of the Colin Barker conjectures above is true because of the Binet-de Moivre formula for L(2*n) (see the Jul 24 2003 Dennis P. Walsh comment on A005248). With phi = (1+sqrt(5))/2, use 1/phi = phi-1, phi^2 = phi+1, (phi-1)^2 = 2 - phi.
His third conjecture (the g.f.) follows from the g.f. of A005248 by adding 2/(1-x).
His second conjecture (recurrence) with input a(-3) = 20, a(-2) = 9 and a(-1) = 5 (from the above given recurrence) leads to his g.f. with the expanded denominator. Thus all three conjectures are true. (End)
a(n) = A005592(n) + 3, with n > 0. - Zino Magri, Feb 16 2015
a(n) = (phi^n + phi^(-n))^2, where phi = A001622 = (1 + sqrt(5))/2. - Diego Rattaggi, Jun 10 2020
Sum_{k>=0} 1/a(k) = A338303. - Amiram Eldar, Oct 22 2020

Edited: name changed (after proof has been given in part II of the W. Lang link), comments rewritten, cross refs. and link to Chebyshev index added. - Wolfdieter Lang, Aug 26 2014

A130245 Number of Lucas numbers (A000032) <= n.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10

Offset: 0

Hieronymus Fischer, May 19 2007, Jul 02 2007

nonn

Partial sums of the Lucas indicator sequence A102460.

For n>=2, we have a(A000032(n)) = n + 1.

			a(9)=5 because there are 5 Lucas numbers <=9 (2,1,3,4 and 7).

Antti Karttunen, Table of n, a(n) for n = 0..64079
Dorin Andrica, Ovidiu Bagdasar, and George Cătălin Tųrcąs, On some new results for the generalised Lucas sequences, An. Şt. Univ. Ovidius Constanţa (Romania, 2021) Vol. 29, No. 1, 17-36.

Partial sums of A102460.
For partial sums of this sequence, see A130246. Other related sequences: A000032, A130241, A130242, A130247, A130249, A130253, A130255, A130259.
For Fibonacci inverse, see A130233 - A130240, A104162, A108852.

[0] cat [1+Floor(Log((2*n+1)/2)/Log((1+Sqrt(5))/2)): n in [1..100]]; // G. C. Greubel, Sep 09 2018

Join[{0}, Table[1+Floor[Log[GoldenRatio, (2*n+1)/2]], {n,1,100}]] (* G. C. Greubel, Sep 09 2018 *)

A102460(n) = { my(u1=1,u2=3,old_u1); if(n<=2,sign(n),while(n>u2,old_u1=u1;u1=u2;u2=old_u1+u2);(u2==n)); };
A130245(n) = if(!n,n,A102460(n)+A130245(n-1));
\\ Or just as:
c=0; for(n=0,123,c += A102460(n); print1(c,", ")); \\ Antti Karttunen, May 13 2018

from itertools import count, islice
def A130245_gen(): # generator of terms
    yield from (0, 1, 2)
    a, b = 3,4
    for i in count(3):
        yield from (i,)*(b-a)
        a, b = b, a+b
A130245_list = list(islice(A130245_gen(),40)) # Chai Wah Wu, Jun 08 2022

a(n) = 1 +floor(log_phi((n+sqrt(n^2+4))/2)) = 1 +floor(arcsinh(n/2)/log(phi)) for n>=2, where phi = (1+sqrt(5))/2.
a(n) = A130241(n)+1 = A130242(n+1) for n>=2.
G.f.: g(x) = 1/(1-x)*sum{k>=0, x^Lucas(k)}.
a(n) = 1 +floor(log_phi(n+1/2)) for n>=1, where phi is the golden ratio.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3/2 - Pi/(6*sqrt(3)) - log(3)/2. - Amiram Eldar, Jul 25 2025

A070825 One half of product of first n+1 Lucas numbers A000032.

Original entry on oeis.org

1, 1, 3, 12, 84, 924, 16632, 482328, 22669416, 1722875616, 211913700768, 42170826452832, 13579006117811904, 7074662187380001984, 5963940223961341672512, 8134814465483270041306368, 17953535525321576981163154176, 64112075360923351399733623562496, 370439571435415124387660876944101888

Offset: 0

Wolfdieter Lang, May 10 2002

Vincenzo Librandi, Table of n, a(n) for n = 0..90
R. Grünwald, E. Heidel, A. Strätz, M. Sünkel and R. Terbach, Induction on Number Series, Fakultät fur Wirtschaftsinformatik und Angewandte Informatik, Otto-Friedrich-Universität Bamberg, 2012. - _N. J. A. Sloane_, Feb 07 2013

Cf. A000032, A003266 (for Fibonacci), A003046 (for Catalan), A101690, A135407, A218490.

[1] cat [&*[Lucas(i+1): i in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Sep 15 2016

c := arccsch(2) - I*Pi/2:
A070825 := n -> local j; 2^n*mul(I^j*cosh(c*j), j = 1..n):
seq(simplify(A070825(n)), n = 0..18);  # Peter Luschny, Jul 07 2025

FoldList[Times, LucasL[Range[0, 20]]]/2 (* or *)
Table[Round[GoldenRatio^(n(n+1)/2) QPochhammer[-1, GoldenRatio-2, n+1]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

a(n) = prod(k=0, n, fibonacci(k+1)+fibonacci(k-1))/2; \\ Michel Marcus, Mar 18 2016

a(n) = (Product_{k=0..n} L(k))/2 with L = A000032.
Sum_{n>=0} 1/a(n) = 1 + A101690. - Amiram Eldar, Nov 09 2020
a(n) = 2^n*Product_{j=1..n} i^j*cosh(c*j), where c = arccsch(2) - i*Pi/2. - Peter Luschny, Jul 07 2025

A203800 a(n) = (1/n) * Sum_{d|n} moebius(n/d) * Lucas(d)^(d-1), where Lucas(n) = A000032(n).

Original entry on oeis.org

1, 1, 5, 85, 2928, 314925, 84974760, 63327890015, 123670531939440, 644385861467631972, 8853970669063185618000, 321538767413685546538468385, 30768712746239178236068160093280, 7755868453482819803691622493685140880, 5144106193113274410507722020733942141881664

Offset: 1

Paul D. Hanna, Jan 06 2012

nonn

			G.f.: F(x) = 1/((1-x-x^2) * (1-3*x^2+x^4) * (1-4*x^3-x^6)^5 * (1-7*x^4+x^8)^85 * (1-11*x^5-x^10)^2928 * (1-18*x^6+x^12)^314925 * (1-29*x^7-x^14)^84974760 * (1-47*x^8+x^16)^63327890015 * (1-76*x^9-x^18)^123670531939440 *...).
where F(x) = exp( Sum_{n>=1} Lucas(n)^n * x^n/n ) = g.f. of A156216:
F(x) = 1 + x + 5*x^2 + 26*x^3 + 634*x^4 + 32928*x^5 + 5704263*x^6 +...
so that the logarithm of F(x) begins:
log(F(x)) = x + 3^2*x^2/2 + 4^3*x^3/3 + 7^4*x^4/4 + 11^5*x^5/5 + 18^6*x^6/6 + 29^7*x^7/7 + 47^8*x^8/8 + 76^9*x^9/9 + 123^10*x^10/10 +...+ Lucas(n)^n*x^n +...

G. C. Greubel, Table of n, a(n) for n = 1..69

Cf. A156216, A000032 (Lucas), A203318, A203803, A203804, A203805, A203806, A203807, A203808.

a[n_] := 1/n DivisorSum[n, MoebiusMu[n/#] LucasL[#]^(#-1)&]; Array[a, 15] (* Jean-François Alcover, Dec 23 2015 *)

{a(n)=if(n<1, 0, sumdiv(n, d, moebius(n/d)*(fibonacci(d-1)+fibonacci(d+1))^(d-1))/n)}

{Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=local(F=exp(sum(m=1, n, Lucas(m)^m*x^m/m)+x*O(x^n)));if(n==1,1,polcoeff(F*prod(k=1,n-1,(1 - Lucas(k)*x^k + (-1)^k*x^(2*k) +x*O(x^n))^a(k)),n)/Lucas(n))}

G.f.: 1/Product_{n>=1} (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^a(n) = exp(Sum_{n>=1} Lucas(n)^n * x^n/n), which is the g.f. of A156216.

G.f.: Product_{n>=1} G_n(x^n)^a(n) = exp(Sum_{n>=1} Lucas(n)^n * x^n/n) where G_n(x^n) = Product_{k=0..n-1} G(u^k*x) where G(x) = 1/(1-x-x^2) and u is an n-th root of unity.

A130311 Maximal (or "lazy") Lucas representation of n. Binary system for representing integers using Lucas numbers (A000032) as a base.

Original entry on oeis.org

0, 10, 1, 11, 110, 101, 111, 1011, 1110, 1101, 1111, 10110, 10101, 10111, 11011, 11110, 11101, 11111, 101011, 101110, 101101, 101111, 110110, 110101, 110111, 111011, 111110, 111101, 111111, 1010110, 1010101, 1010111, 1011011, 1011110, 1011101, 1011111, 1101011, 1101110

Offset: 0

Casey Mongoven, May 21 2007; corrected Mar 23 2008

			a(7) = 1110 because 4 + 3 + 1 = 8.
a(8) = 1101 because 4 + 3 + 2 = 9.

Edouard Zeckendorf, Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège, Vol. 41 (1972), pp. 179-182.

Amiram Eldar, Table of n, a(n) for n = 0..10000
Ron Knott, Using Fibonacci Numbers to Represent Whole Numbers.

Cf. A000032, A104326, A130310, A131343.

lazy = Select[IntegerDigits[Range[10^2], 2], SequenceCount[#, {0, 0}] == 0 &]; t = Total[# * Reverse@LucasL[Range[0, Length[#] - 1]]] & /@ lazy; Join[{0}, FromDigits /@ lazy[[TakeWhile[Flatten[FirstPosition[t, #] & /@ Range[Max[t]]], NumberQ]]]] (* Amiram Eldar, Feb 17 2022 *)

a(0) and more terms from Amiram Eldar, Feb 17 2022

A153415 Decimal expansion of Sum_{n>=1} 1/A000032(2*n).

Original entry on oeis.org

5, 6, 6, 1, 7, 7, 6, 7, 5, 8, 1, 1, 3, 8, 4, 5, 5, 0, 2, 7, 5, 9, 2, 9, 3, 2, 1, 2, 1, 2, 0, 6, 2, 0, 0, 3, 7, 3, 6, 1, 4, 4, 1, 9, 7, 8, 6, 5, 9, 0, 5, 5, 7, 0, 4, 9, 2, 3, 4, 4, 4, 1, 3, 2, 5, 4, 5, 7, 5, 5, 5, 4, 5, 3, 0, 2, 0, 8, 6, 8, 5, 6, 1, 4, 8, 5, 5, 6, 7, 8, 4, 2, 1, 8, 1, 8, 3, 2, 6, 6, 4, 6, 1, 5, 3

Offset: 0

Eric W. Weisstein, Dec 25 2008

From Peter Bala, Oct 15 2019: (Start)
c = (1/4)*(theta_3( (3-sqrt(5))/2 )^2 - 1 ), where theta_3(q) = 1 + 2*Sum_{n >= 1} q^n^2. See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A056854.
Series acceleration formulas (L(n) = A000032(n)):
c = 1 - 5*Sum_{n >= 1} 1/( L(2*n)*(L(2*n)^2 - 5) ).
c = (1/6) + 15*Sum_{n >= 1} 1/( L(2*n)*(L(2*n)^2 + 5) ).
c = (11/16) - 10*Sum_{n >= 1} (L(2*n)^2 - 10)/( L(2*n)*(L(2*n)^2 - 5)*(L(2*n)^2 - 20) ). (End)
Compare with Sum_{n >= 1} 1/(L(2*n) - sqrt(5)) = phi and Sum_{n >= 1} 1/(L(2*n) + sqrt(5)) = 2 - phi, where phi = (sqrt(5) + 1)/2. - Peter Bala, Nov 23 2019
This constant is transcendental (Duverney et al., 1997). - Amiram Eldar, Oct 30 2020

			0.56617767581138455027...

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

Daniel Duverney, Keiji Nishioka, Kumiko Nishioka and Iekata Shiokawa, Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers, Proceedings of the Japan Academy, Series A, Mathematical Sciences, Vol. 73, No. 7 (1997), pp. 140-142.
Index entries for transcendental numbers

Cf. A000032, A000122, A056854, A093540, A153416.

First[ RealDigits[ N[(EllipticTheta[3, 0, GoldenRatio^(-2)]^2 - 1)/4, 120], 10, 105]](* Jean-François Alcover, Jun 07 2012, after Eric W. Weisstein *)

th3(x)=1 + 2*suminf(n=1,x^n^2)
phi=(sqrt(5)+1)/2
(th3(phi^-2)^2-1)/4 \\ Charles R Greathouse IV, Jun 06 2016

A122869 Primes p that divide Lucas((p-1)/2), where Lucas is A000032.

Original entry on oeis.org

11, 19, 31, 59, 71, 79, 131, 139, 151, 179, 191, 199, 211, 239, 251, 271, 311, 331, 359, 379, 419, 431, 439, 479, 491, 499, 571, 599, 619, 631, 659, 691, 719, 739, 751, 811, 839, 859, 911, 919, 971, 991, 1019, 1031, 1039, 1051, 1091, 1151, 1171, 1231, 1259

Offset: 1

Alexander Adamchuk, Sep 16 2006

nonn

Final digit of a(n) is 1 or 9.
A002145 is the union of this sequence and A122870, Primes p that divide Lucas((p+1)/2).
Conjecture: This sequence is just the primes congruent to 11 or 19 mod 20. - Charles R Greathouse IV, May 25 2011 [The conjecture is correct. - Jianing Song, Jun 20 2025]
Note that F(p-1) = F((p-1)/2)*Lucas((p-1)/2), where F = A000045. Since gcd(F(n),Lucas(n)) = 1 or 2 (because Lucas(n)^2 - 5*F(n)^2 = 4*(-1)^n), this sequence lists primes p such that p divides F(p-1) but does not divides F((p-1)/2). By Propositions 1.1 and 1.2 (the k = 3 case) of my link below, this is primes p == 11, 19 (mod 20). - Jianing Song, Jun 20 2025

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Vincenzo Librandi)
Jianing Song, Lucas sequences and entry point modulo p
Eric Weisstein's World of Mathematics, Gaussian Prime.
Eric Weisstein's World of Mathematics, Lucas Number.

Cf. A000032, A000045, A053032, A076518, A122870.

Subsequence of A002145, A003626, A040105, A040147 and A064739.

Select[Prime[Range[1000]],IntegerQ[(Fibonacci[(#1-1)/2-1]+Fibonacci[(#1-1)/2+1])/#1]&]

lista(kmax) = {my(lucas1 = 1, lucas2 = 3, lucas3, p); for(k = 3, kmax, lucas3 = lucas1 + lucas2; p = 2*k + 1; if(isprime(p) && !(lucas3 % p), print1(p, ", ")); lucas1 = lucas2; lucas2 = lucas3);} \\ Amiram Eldar, Jun 06 2024

A060922 Convolution triangle for Lucas numbers A000032(n+1), n >= 0.

Original entry on oeis.org

1, 3, 1, 4, 6, 1, 7, 17, 9, 1, 11, 38, 39, 12, 1, 18, 80, 120, 70, 15, 1, 29, 158, 315, 280, 110, 18, 1, 47, 303, 753, 905, 545, 159, 21, 1, 76, 566, 1687, 2568, 2120, 942, 217, 24, 1, 123, 1039, 3612, 6666, 7043, 4311

Offset: 0

Wolfdieter Lang, Apr 20 2001

In the language of Shapiro et al. (see A053121 for the reference) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. G.f. for row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) is (1+2*z)/(1-(1+x)*z-(1+2*x)*z^2).
Row sums give A060925. Column sequences (without leading zeros) are, for m=0..6: A000032(n+1)= A000204(n+1) (Lucas), A004799(n+1), A060929-33.
Bisection of this triangle gives triangles A060923 (even part) and A060924 (odd part).
For the m-th column sequence (without leading zeros) one has: a(n+m,m)= (pL1(m,n)*L(n+2)+pL2(m,n)*L(n+1))/(m!*5^m), m >= 0, with the Lucas numbers L(n)=A000032(n), n >= 0 and the row polynomials pL1(n,x) := sum(A061188(n,m)*x^n,m=0..n) and pL2(n,x) := sum(A061189(n,m)*x^m,m=0..n).
Riordan array ((1+2*x)/(1-x-x^2), x*(1+2*x)/(1-x-x^2)). - Philippe Deléham, Jan 21 2014
T is the convolution triangle of A000204 (see A357368). - Peter Luschny, Oct 19 2022

			p(2,x) = 4+6*x+x^2.
Triangle begins:
1 ;
3, 1;
4, 6, 1;
7, 17, 9, 1;
11, 38, 39, 12, 1;
18, 80, 120, 70, 15, 1;
29, 158, 315, 280, 110, 18, 1;
47, 303, 753, 905, 545, 159, 21, 1;

Cf. A000032.

# Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left.
PMatrix(10, A000204); # Peter Luschny, Oct 19 2022

a(n, m)=((n-m+1)*a(n, m-1)+2*(2*n-m)*a(n-1, m-1)+4*(n-1)*a(n-2, m-1))/(5*m), n >= m >= 1, a(n, 0)= A000204(n+1)= A000032(n+1).
G.f. for m-th column: ((1+2*x)/(1-x-x^2))* ((x*(1+2*x))/(1-x-x^2))^m.
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 21 2014

Example improved by Philippe Deléham, Jan 21 2014

A127211 a(n) = 4^n*Lucas(n), where Lucas = A000032.

Original entry on oeis.org

2, 4, 48, 256, 1792, 11264, 73728, 475136, 3080192, 19922944, 128974848, 834666496, 5402263552, 34963718144, 226291089408, 1464583847936, 9478992822272, 61349312856064, 397061136580608, 2569833552019456, 16632312393367552, 107646586405781504, 696703343917006848

Offset: 0

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 0..1225
Index entries for linear recurrences with constant coefficients, signature (4,16).

Cf. A000032, A000204, A087131, A087131, A127210, A127212, A127213, A127214, A127215, A127216.

[4^n*Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 18 2017

a:= n-> 4^n*(<<1|1>, <1|0>>^n. <<2, -1>>)[1, 1]:
seq(a(n), n=0..22);  # Alois P. Heinz, Apr 15 2024

Table[4^n Tr[MatrixPower[{{1, 1}, {1, 0}}, n]], {n, 0, 20}]
Table[4^n*LucasL[n], {n, 0, 50}] (* G. C. Greubel, Dec 18 2017 *)

my(x='x + O('x^30)); Vec(-4*x*(8*x+1)/(16*x^2+4*x-1)) \\ G. C. Greubel, Dec 18 2017

a(n) = Trace of matrix [({4,4},{4,0})^n].
a(n) = 4^n * Trace of matrix [({1,1},{1,0})^n].
From Colin Barker, Sep 02 2013: (Start)
a(n) = 4*a(n-1) + 16*a(n-2).
G.f.: 2*x*(2*x-1)/(16*x^2+4*x-1). (End)
From Peter Luschny, Apr 15 2024: (Start)
a(n) = 2^n*((1 - sqrt(5))^n + (1 + sqrt(5))^n).
a(n) = 4^n*(Fibonacci(n+1) + Fibonacci(n-1)). (End)
a(n) = 2^n*A087131(n). - Michel Marcus, Apr 15 2024

a(0)=2 prepended by Alois P. Heinz, Apr 15 2024

cp's OEIS Frontend

A130310 Minimal (or "greedy") Lucas representation of n, in which L(0) = 2 and L(2) = 3 are not allowed in the same representation (hence the correct representation of the integer 5 is 1010 rather than 101). A binary system of integers with Lucas numbers (A000032) as a base.

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A240926 a(n) = 2 + L(2*n) = 2 + A005248(n), n >= 0, with the Lucas numbers (A000032).

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A130245 Number of Lucas numbers (A000032) <= n.

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A070825 One half of product of first n+1 Lucas numbers A000032.

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A203800 a(n) = (1/n) * Sum_{d|n} moebius(n/d) * Lucas(d)^(d-1), where Lucas(n) = A000032(n).

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A130311 Maximal (or "lazy") Lucas representation of n. Binary system for representing integers using Lucas numbers (A000032) as a base.

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A153415 Decimal expansion of Sum_{n>=1} 1/A000032(2*n).

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