A000204 -id:A000204 - cp's OEIS frontend

A003263 Number of representations of n as a sum of distinct Lucas numbers 1, 3, 4, 7, 11, ... (A000204).

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

Offset: 1

N. J. A. Sloane

A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 58.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..9349
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 58.
Casey Mongoven, Sonification of multiple Fibonacci-related sequences, Annales Mathematicae et Informaticae, 41 (2013) pp. 175-192.

Cf. A054770, A000204.

n1 = 10; n2 = LucasL[n1]; Product[1 + x^LucasL[n], {n, 1, n1}] + O[x]^n2 // CoefficientList[#, x]& // Rest (* Jean-François Alcover, Feb 17 2017, after Joerg Arndt *)

L(n)=fibonacci(n+1) + fibonacci(n-1);
N = 66;  x = 'x + O('x^N);
gf = prod(n=1, 11, 1 + x^L(n) );
Vec(gf) \\ Joerg Arndt, Jul 14 2013

G.f.: Product_{n>=1} (1 + x^L(n)) where L(n) = A000204(n). - Joerg Arndt, Jul 14 2013

More terms from James Sellers, May 29 2000

A054770 Numbers that are not the sum of distinct Lucas numbers 1,3,4,7,11, ... (A000204).

Original entry on oeis.org

2, 6, 9, 13, 17, 20, 24, 27, 31, 35, 38, 42, 46, 49, 53, 56, 60, 64, 67, 71, 74, 78, 82, 85, 89, 93, 96, 100, 103, 107, 111, 114, 118, 122, 125, 129, 132, 136, 140, 143, 147, 150, 154, 158, 161, 165, 169, 172, 176, 179, 183, 187, 190, 194, 197, 201, 205, 208, 212

Offset: 1

Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), May 28 2000

Alternatively, Lucas representation of n includes L_0 = 2. - Fred Lunnon, Aug 25 2001
Conjecture: this is the sequence of numbers for which the base phi representation includes phi itself, where phi = (1 + sqrt(5))/2 = the golden ratio. Example: let r = phi; then 6 = r^3 + r + r^(-4). - Clark Kimberling, Oct 17 2012
This conjecture is proved in my paper 'Base phi representations and golden mean beta-expansions', using the formula by Wilson/Agol/Carlitz et al. - Michel Dekking, Jun 25 2019
Numbers whose minimal Lucas representation (A130310) ends with 1. - Amiram Eldar, Jan 21 2023

G. C. Greubel, Table of n, a(n) for n = 1..5000
L. Carlitz, R. Scoville, and V. E. Hoggatt, Jr., Lucas representations, Fibonacci Quart. 10 (1972), 29-42, 70, 112.
Weiru Chen and Jared Krandel, Interpolating Classical Partitions of the Set of Positive Integers, arXiv:1810.11938 [math.NT], 2018. See sequence D2 p. 4.
Michel Dekking, Base phi representations and golden mean beta-expansions, arXiv:1906.08437 [math.NT], 2019.
Jared Krandel and Weiru Chen, Interpolating classical partitions of the set of positive integers, The Ramanujan Journal (2020).

Complement of A063732.

Cf. A003263, A003622, A022342, A130310.

[Floor(n*(Sqrt(5)+5)/2)-1: n in [1..60]]; // Vincenzo Librandi, Oct 30 2018

A054770 := n -> floor(n*(sqrt(5)+5)/2)-1;

Complement[Range[220],Total/@Subsets[LucasL[Range[25]],5]] (* Harvey P. Dale, Feb 27 2012 *)
Table[Floor[n (Sqrt[5] + 5) / 2] - 1, {n, 60}] (* Vincenzo Librandi, Oct 30 2018 *)

PARI
```
a(n)=floor(n*(sqrt(5)+5)/2)-1
```

from math import isqrt
def A054770(n): return (n+isqrt(5*n**2)>>1)+(n<<1)-1 # Chai Wah Wu, Aug 17 2022

a(n) = floor(((5+sqrt(5))/2)*n)-1 (conjectured by David W. Wilson; proved by Ian Agol (iagol(AT)math.ucdavis.edu), Jun 08 2000)
a(n) = A000201(n) + 2*n - 1. - Michel Dekking, Sep 07 2017
G.f.: x*(x+1)/(1-x)^2 + Sum_{i>=1} (floor(i*phi)*x^i), where phi = (1 + sqrt(5))/2. - Iain Fox, Dec 19 2017
Ian Agol tells me that David W. Wilson's formula is proved in the Carlitz, Scoville, Hoggatt paper 'Lucas representations'. See Equation (1.12), and use A(A(n))+n = B(n)+n-1 = A(n)+2*n-1, the well known formulas for the lower Wythoff sequence A = A000201, and the upper Wythoff sequence B = A001950. - Michel Dekking, Jan 04 2018

More terms from James Sellers, May 28 2000

A090946 Non-Lucas numbers: complement of A000204.

Original entry on oeis.org

0, 2, 5, 6, 8, 9, 10, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 77, 78, 79

Offset: 1

N. J. A. Sloane, Feb 28 2004

G. C. Greubel, Table of n, a(n) for n = 1..10000

Cf. A000032, A000204, A022801.

L[1]:=1: L[2]:=3: for n from 3 to 10 do L[n]:=L[n-1]+L[n-2] od: A:={seq(L[n],n=1..10)}: B:={seq(n,n=0..L[10])}: C:=B minus A; # Emeric Deutsch, Dec 14 2004

Complement[Range[0,100],LucasL[Range[10]]] (* Harvey P. Dale, Mar 21 2012 *)

def A090946(n):
    if n == 1: return 0
    def f(x):
        if x<=2: return n
        a, b, c = 1, 3, 0
        while b<=x:
            a, b = b, a+b
            c += 1
        return n+c
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return m # Chai Wah Wu, Sep 10 2024

a(n) = A057854(n-2), n>2. - R. J. Mathar, Jan 29 2019

More terms from Emeric Deutsch, Dec 14 2004

Offset corrected by Michel Marcus, Aug 15 2017

A067979 Triangle read by rows of incomplete convolutions of Lucas numbers L(n+1) = A000204(n+1), n>=0.

Original entry on oeis.org

1, 3, 6, 4, 13, 17, 7, 19, 31, 38, 11, 32, 48, 69, 80, 18, 51, 79, 107, 140, 158, 29, 83, 127, 176, 220, 274, 303, 47, 134, 206, 283, 360, 432, 519, 566, 76, 217, 333, 459, 580, 706, 822, 963, 1039, 123, 351, 539, 742, 940, 1138, 1341, 1529, 1757, 1880, 199, 568, 872, 1201, 1520, 1844, 2163, 2492, 2796, 3165, 3364

Offset: 0

Wolfdieter Lang, Feb 15 2002

The diagonals d>=0 (d=0: main diagonal) give convolutions of Lucas numbers L(n+1) := A000204(n+1), n>=0, with those with d-shifted index: a(d+n,d) = Sum_{k=0..n} L(k+1)*L(d+n+1-k).
The diagonals give A004799(n-1), A067980-7 for d=n-m= 0..8, respectively. Row sums give A067989.
The row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) are generated by A(x*z)*(A(z)-x*A(x*z))/(1-x), with A(x) := (1+2*x)/(1-x-x^2) (g.f. Lucas L(n+1), n>=0).

			Triangle begins:
  {1};
  {3,6};
  {4,13,17};     p(2,x) = 4+13*x+17*x^2
  {7,19,31,38};
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A067990 (triangle with rows read backwards).

Table[Sum[LucasL[k + 1] LucasL[n - k + 1], {k, 0, m}], {n, 0, 10}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 11 2016 *)

for(n=0,10, for(k=0,n, print1(sum(m=0,k,(fibonacci(m+2) + fibonacci(m))*(fibonacci(n-m+2) + fibonacci(n-m))), ", "))) \\ G. C. Greubel, Dec 17 2017

a(n, m) = Sum_{k=0..m} L(k+1)*L(n-k+1), n>=m>=0, else 0.
a(n, m) = (m+1)*L(n-m+1)*F(m) + ((m+1)*L(n-m+1) + m*L(n-m))*F(m+1), n>=m>=0, with F(n) := A000045(n) (Fibonacci) and L(n) := A000032(n) (Lucas).
G.f. for diagonals d= n-m>=0: (x^d)*(L(d+1)+L(d)*x)*(1-2*x)/(1-x-x^2)^2.
a(n, m) = -(-1)^m*F(n-2*m-1) + m*L(n+2)+F(n+3), with F(-n) = (-1)^(n+1) * F(n), hence a(n, m) = -5*A067330(n, m)+2*(m+1)*L(n+2), n>=m>=0. - Ehren Metcalfe, Apr 11 2016

A203803 G.f.: exp( Sum_{n>=1} A000204(n)^3 * x^n/n ) where A000204 is the Lucas numbers.

Original entry on oeis.org

1, 1, 14, 35, 205, 744, 3414, 13926, 60060, 252330, 1072902, 4537272, 19234463, 81452015, 345084970, 1461714517, 6192083147, 26229794928, 111111714300, 470675847900, 1993816532280, 8445939457380, 35777578796220, 151556246864400, 642002579853325, 2719566542567917

Offset: 0

Paul D. Hanna, Jan 06 2012

nonn

More generally, exp(Sum_{k>=1} A000204(k)^(2*n+1) * x^k/k) = Product_{k=0..n} 1/(1 - (-1)^(n-k)*A000204(2*k+1)*x - x^2)^binomial(2*n+1,n-k).

			G.f.: A(x) = 1 + x + 14*x^2 + 35*x^3 + 205*x^4 + 744*x^5 + 3414*x^6 +...
where
log(A(x)) = x + 3^3*x^2/2 + 4^3*x^3/3 + 7^3*x^4/4 + 11^3*x^5/5 + 18^3*x^6/6 + 29^3*x^7/7 + 47^3*x^8/8 +...+ Lucas(n)^3*x^n/n +...

G. C. Greubel, Table of n, a(n) for n = 0..1500
Index entries for linear recurrences with constant coefficients, signature (1,13,8,-20,-8,13,-1,-1).

Cf. A002571, A203804, A203805, A203806, A203807, A203808, A203809.

Cf. A203853, A203800.

CoefficientList[Series[1/((1 + x - x^2)^3*(1 - 4*x - x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 24 2017 *)

/* Subroutine used in PARI programs below: */
{Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}

{a(n)=polcoeff(exp(sum(k=1, n, Lucas(k)^3*x^k/k)+x*O(x^n)), n)}

{a(n,m=1)=polcoeff(prod(k=0,m, 1/(1 - (-1)^(m-k)*Lucas(2*k+1)*x - x^2+x*O(x^n))^binomial(2*m+1,m-k)),n)}

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

G.f.: 1/Product_{n>=1} (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^A203853(n) where A203853(n) = (1/n)*Sum_{d|n} moebius(n/d)*Lucas(d)^2.

A203804 G.f.: exp( Sum_{n>=1} A000204(n)^4 * x^n/n ) where A000204 is the Lucas numbers.

Original entry on oeis.org

1, 1, 41, 126, 1526, 7854, 63629, 400789, 2870629, 19254504, 133376760, 909578760, 6249172910, 42785312510, 293403088510, 2010553849020, 13781960765020, 94458627485820, 647442212896270, 4437595353800270, 30415849505902910, 208472981440853160, 1428896115173689560

Offset: 0

Paul D. Hanna, Jan 06 2012

nonn

More generally, exp(Sum_{k>=1} A000204(k)^(2*n) * x^k/k) = 1/(1 - (-1)^n*x)^binomial(2*n,n) * Product_{k=1..n} 1/(1 - (-1)^(n-k)*A000204(2*k)*x + x^2)^binomial(2*n,n-k).

			G.f.: A(x) = 1 + x + 41*x^2 + 126*x^3 + 1526*x^4 + 7854*x^5 + 63629*x^6 +...
where
log(A(x)) = x + 3^4*x^2/2 + 4^4*x^3/3 + 7^4*x^4/4 + 11^4*x^5/5 + 18^4*x^6/6 + 29^4*x^7/7 + 47^4*x^8/8 +...+ Lucas(n)^4*x^n/n +...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,40,45,-285,-272,1022,370,-1840,370,1022,-272,-285,45,40,1,-1).

Cf. A002571, A203803, A203805, A203806, A203807, A203808, A203809.

Cf. A203854, A203800.

CoefficientList[Series[1/((1 - x)^6*(1 + 3*x + x^2)^4*(1 - 7*x + x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 24 2017 *)

/* Subroutine used in PARI programs below: */
{Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}

{a(n)=polcoeff(exp(sum(k=1, n, Lucas(k)^4*x^k/k)+x*O(x^n)), n)}

{a(n,m=2)=polcoeff(1/(1 - (-1)^m*x+x*O(x^n))^binomial(2*m,m) * prod(k=1,m,1/(1 - (-1)^(m-k)*Lucas(2*k)*x + x^2+x*O(x^n))^binomial(2*m,m-k)),n)}

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

G.f.: 1/Product_{n>=1} (1 - Lucas(n)*x^n + (-1)^n*x^(2*n))^A203854(n) where A203854(n) = (1/n)*Sum_{d|n} moebius(n/d)*Lucas(d)^3.

A101032 Table (read by rows) giving the coefficients of sum formulas of n-th Lucas numbers (A000204). The k-th row (k>=1) contains T(i,k) for i=1 to k, where k=[2n+1+(-1)^(n-1)]/4 and T(i,k) satisfies L(n) = Sum_{i=1..k} T(i,k) n^(k-i) / (k-1)!.

Original entry on oeis.org

1, 1, 1, 1, -1, 2, 1, -6, 17, 6, 1, -14, 83, -142, 24, 1, -25, 265, -1235, 2314, 120, 1, -39, 655, -5565, 24184, -41556, 720, 1, -56, 1372, -18200, 137599, -556304, 944628, 5040, 1, -76, 2562, -48664, 560049, -3884524, 15021068, -24875376, 40320, 1, -99, 4398, -113022, 1829793, -19043451

Offset: 1

André F. Labossière, Nov 30 2004

			L(13)=521; substituting n=13 in the formula of the k-th row we obtain k=7 and the coefficients
T(i,7) will be the following: 1,-39,655,-5565,24184,-41556,720,
=> L(13) = [13^6-39*13^5+655*13^4-5565*13^3+24184*13^2-41556*13+720]/6! = 521.

Ron Knott, The Lucas Numbers in Pascal's Triangle.

Cf. A101033, A000204, A100492, A099731, A000045, A094216, A094638, A000108.

A127210 a(n) = 3^n*Lucas(n), where Lucas = A000204.

Original entry on oeis.org

3, 27, 108, 567, 2673, 13122, 63423, 308367, 1495908, 7263027, 35252253, 171124002, 830642283, 4032042867, 19571909148, 95004113247, 461159522073, 2238515585442, 10865982454983, 52744587633927, 256027604996628, 1242784103695227, 6032600756055333, 29282859201423042

Offset: 1

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 1..1445
Ivica Martinjak, Two Extensions of the Sury's Identity, arXiv:1508.01444 [math.CO], 2015.
Index entries for linear recurrences with constant coefficients, signature (3, 9).

Cf. A000204, A087131, A127211, A127212, A127213, A127214, A127215, A127216.

[3^n*Lucas(n): n in [1..30]]; // Vincenzo Librandi, Aug 07 2015

Table[3^n Tr[MatrixPower[{{1, 1}, {1, 0}}, x]], {x, 1, 20}]
Table[3^n LucasL[n], {n, 25}] (* Vincenzo Librandi, Aug 07 2015 *)

lucas(n) = fibonacci(n-1) + fibonacci(n+1);
vector(30, n, 3^n*lucas(n)) \\ Michel Marcus, Aug 07 2015

a(n) = Trace of matrix [({3,3},{3,0})^n] = 3^n * Trace of matrix [({1,1},{1,0})^n].
From R. J. Mathar, Oct 27 2008: (Start)
a(n) = 3*a(n-1) + 9*a(n-2).
G.f.: 3*x*(1 + 6*x)/(1 - 3*x - 9*x^2).
a(n) = 3*A099012(n) +18*A099012(n-1). (End)

More terms from Michel Marcus, Aug 07 2015

A127212 a(n) = 5^n*Lucas(n), where Lucas = A000204.

Original entry on oeis.org

5, 75, 500, 4375, 34375, 281250, 2265625, 18359375, 148437500, 1201171875, 9716796875, 78613281250, 635986328125, 5145263671875, 41625976562500, 336761474609375, 2724456787109375, 22041320800781250, 178318023681640625

Offset: 1

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,25).

Cf. A000204, A087131, A127210, A127211, A127213, A127214, A127215, A127216.

[5^n*Lucas(n): n in [1..30]]; // G. C. Greubel, Dec 18 2017

Table[5^n Tr[MatrixPower[{{1, 1}, {1, 0}}, x]], {x, 1, 20}]
Table[5^n*LucasL[n], {n,1,50}] (* G. C. Greubel, Dec 18 2017 *)
LinearRecurrence[{5,25},{5,75},20] (* Harvey P. Dale, Jan 11 2024 *)

x='x+O('x^30); Vec(-5*x*(10*x+1)/(25*x^2+5*x-1)) \\ G. C. Greubel, Dec 18 2017

a(n) = Trace of matrix [({5,5},{5,0})^n].
a(n) = 5^n * Trace of matrix [({1,1},{1,0})^n].
From Colin Barker, Sep 02 2013: (Start)
a(n) = 5*a(n-1) + 25*a(n-2).
G.f.: -5*x*(10*x+1)/(25*x^2+5*x-1). (End)

A127213 a(n) = 6^n*Lucas(n), where Lucas = A000204.

Original entry on oeis.org

6, 108, 864, 9072, 85536, 839808, 8118144, 78941952, 765904896, 7437339648, 72196614144, 700923912192, 6804621582336, 66060990332928, 641332318961664, 6226189565755392, 60445100877152256, 586813429630107648

Offset: 1

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,36).

Cf. A000204, A087131, A127210, A127211, A127212, A127214, A127215, A127216.

[6^n*Lucas(n): n in [1..30]]; // G. C. Greubel, Dec 18 2017

Table[6^n Tr[MatrixPower[{{1, 1}, {1, 0}}, x]], {x, 1, 20}]
Table[6^n*LucasL[n], {n,1,50}] (* G. C. Greubel, Dec 18 2017 *)
LinearRecurrence[{6,36},{6,108},20] (* Harvey P. Dale, Jan 20 2024 *)

x='x+O('x^30); Vec(-6*x*(12*x+1)/(36*x^2+6*x-1)) \\ G. C. Greubel, Dec 18 2017

a(n) = Trace of matrix [({6,6},{6,0})^n].
a(n) = 6^n * Trace of matrix [({1,1},{1,0})^n].
From Colin Barker, Sep 02 2013: (Start)
a(n) = 6*a(n-1) + 36*a(n-2).
G.f.: -6*x*(12*x+1)/(36*x^2+6*x-1). (End)

cp's OEIS Frontend

A003263 Number of representations of n as a sum of distinct Lucas numbers 1, 3, 4, 7, 11, ... (A000204).

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A054770 Numbers that are not the sum of distinct Lucas numbers 1,3,4,7,11, ... (A000204).

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A090946 Non-Lucas numbers: complement of A000204.

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A067979 Triangle read by rows of incomplete convolutions of Lucas numbers L(n+1) = A000204(n+1), n>=0.

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A203803 G.f.: exp( Sum_{n>=1} A000204(n)^3 * x^n/n ) where A000204 is the Lucas numbers.

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A203804 G.f.: exp( Sum_{n>=1} A000204(n)^4 * x^n/n ) where A000204 is the Lucas numbers.

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A101032 Table (read by rows) giving the coefficients of sum formulas of n-th Lucas numbers (A000204). The k-th row (k>=1) contains T(i,k) for i=1 to k, where k=[2n+1+(-1)^(n-1)]/4 and T(i,k) satisfies L(n) = Sum_{i=1..k} T(i,k) n^(k-i) / (k-1)!.