A061084 -id:A061084 - cp's OEIS frontend

A132148 Triangular array T(n,k) = C(n,k)*Lucas(n-k), 0 <= k <= n.

2, 1, 2, 3, 2, 2, 4, 9, 3, 2, 7, 16, 18, 4, 2, 11, 35, 40, 30, 5, 2, 18, 66, 105, 80, 45, 6, 2, 29, 126, 231, 245, 140, 63, 7, 2, 47, 232, 504, 616, 490, 224, 84, 8, 2, 76, 423, 1044, 1512, 1386, 882, 336, 108, 9, 2, 123, 760, 2115, 3480, 3780, 2772, 1470, 480, 135, 10, 2

Offset: 0

Peter Bala, Aug 17 2007

			Triangle starts
   2;
   1,   2;
   3,   2,    2;
   4,   9,    3,   2;
   7,  16,   18,   4,   2;
  11,  35,   40,  30,   5,  2;
  18,  66,  105,  80,  45,  6,  2;
  ...

Cf. A000032 (T(n,0)), A000045, A005248 (row sums), A061084 (alter. row sums), A094440.

with(combinat): lucas := n -> fibonacci(n-1) + fibonacci(n+1): T := (n,k) -> binomial(n,k)*lucas(n-k): for n from 0 to 10 do seq( T(n,k), k = 0..n) od;

Flatten[Table[Binomial[n,k]LucasL[n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Nov 06 2011 *)

G.f.: (2 - (2x + 1)*t)/(1 - (2x + 1)*t + (x^2 + x - 1)*t^2) = 2 + (1 + 2*x)*t + (3 + 2*x + 2*x^2)*t^2 + (4 + 9*x + 3*x^2 + 2*x^3)*t^3 + ... .
The row polynomials L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k satisfy L(n,x)*F(n,x) = F(2*n,x), where F(n,x) = Sum_{k = 0..n} C(n,k)*Fibonacci(n-k)*x^k.
Other identities and formulas include:
L(n+1,x)^2 - L(n,x)*L(n+2,x) = -5*(x^2 + x - 1)^n;
L(n+1,x) - (x^2 + x - 1)*L(n-1,x) = 5*F(n,x) for n >= 1;
L(2*n,x) - 2*(x^2 + x - 1)^n = 5*F(n,x)^2;
L(n,2*x) = Sum_{ k = 0..n} C(n,k)*L(n-k,x)*x^k;
L(n,3*x) = Sum_{ k = 0..n} C(n,k)*L(n-k,2*x)*x^k etc.
Sum_{k = 0..n} C(n,k)*L(k,x)*F(n-k,x) = 2^n*F(n,x).
Row sums: L(n,1) = Lucas(2*n); Alternating row sums: L(n,-1) = (-1)^n*Lucas(n); L(n,1/phi) = (-1)^n*L(n,-phi) = sqrt(5)^n for n >= 1, where phi = (1 + sqrt(5))/2.
The polynomials L(n,-x) satisfy a Riemann hypothesis: the zeros of L(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
From Peter Bala, Jun 29 2016: (Start)
L(n,x) = (x + phi)^n + (x - 1/phi)^n, where phi = (1 + sqrt(5))/2. The zeros of L(n,x) are given by -1/2 - i*sqrt(5)/2*cot( (2*k - 1)*Pi/(2*n) ) for k = 1..n.
d/dx(L(n,x)) = n*L(n-1,x).
L(-n,x) = L(n,x)/(x^2 + x - 1)^n.
L(n,x - 1) = (-1)^n*L(n,-x).
L(n,x)^2 - L(2*n,x) = 2*(x^2 + x - 1)^n.
L(n,x)^3 - L(3*n,x) = 3*(x^2 + x - 1)^n*L(n,x).
L(n,x)^4 - L(4*n,x) = 4*(x^2 + x - 1)^n*L(n,x)^2 - 2*(x^2 + x - 1)^(2*n).
If n divides m and m/n is odd then L(n,x) divides L(m,x) in the polynomial ring Z[x].
L(n,x) = F(n+1,x) - (x^2 + x - 1)*F(n-1,x) = 2*F(n+1,x) - (2*x + 1)*F(n,x), where F(n,x) = Sum_{k = 0..n} binomial(n,k)*Fibonacci(n-k)*x^k denotes the n-th row polynomial of A094440 (taken with an offset of 0).
exp( Sum_{n >= 1} L(n,x)*z^n/n ) = Sum_{n >= 0} F(n+1,x)*z^n.
exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*z^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)* F(n+3,x)*z^n. (End)
From Peter Bala, Dec 31 2023: (Start)
For n >= 1, the n-th row polynomial L(n,x) is the numerator of 1/(n-1)! * (d/dx)^(n-1) (2*x + 1)/(1 - x - x^2).
Recurrence: for n >= 1, n*L(n+1,x) = n*(2*x + 1)*L(n,x) + (1 - x - x^2)* d/dx(L(n,x)) with L(1,x) = 2*x + 1. (End)

A061083 Fibonacci-type sequence based on division: a(0) = 1, a(1) = 2 and a(n) = a(n-2)/a(n-1) but ignore decimal point.

Original entry on oeis.org

1, 2, 5, 4, 125, 32, 390625, 8192, 476837158203125, 17179869184, 277555756156289135105907917022705078125, 618970019642690137449562112

Offset: 0

Ulrich Schimke (ulrschimke(AT)aol.com)

			a(6) = 390625, since a(4)/a(5) = 125/32 = 3.90625

Reinhard Zumkeller, Table of n, a(n) for n = 0..17

Cf. A061084 for subtraction, A000301 for multiplication and A000045 for addition - the common Fibonacci numbers

a061083 n = a061083_list !! n
a061083_list = 1 : 2 : zipWith divIgnPnt a061083_list (tail a061083_list)
   where divIgnPnt x y = ddiv (10 * m) x' where
            ddiv u w | r == 0    = 10 * w + q
                     | otherwise = ddiv (10 * r) (10 * w + q)
                     where (q,r) = divMod u y
            (x',m) = divMod x y
-- Reinhard Zumkeller, Dec 29 2011

a(n) = k^(n-th Fibonacci number) with k=2 if n is odd, k=5 if n is even

A083564 a(n) = L(n)*L(2n), where L(n) are the Lucas numbers (A000204).

Original entry on oeis.org

3, 21, 72, 329, 1353, 5796, 24447, 103729, 439128, 1860621, 7880997, 33385604, 141421803, 599075421, 2537719272, 10749959329, 45537545553, 192900159396, 817138154247, 3461452823129, 14662949371128, 62113250430021

Offset: 1

Gary W. Adamson, Jun 12 2003

a(n+1)/a(n) -> (phi)^3 = ((1 + sqrt(5))/2)^3 = 4.236067...

			a(4) = Lucas(4)*Lucas(8) = 7*47 = 329.

Vincenzo Librandi, Table of n, a(n) for n = 1..160
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7
Index entries for linear recurrences with constant coefficients, signature (3, 6, -3, -1).

Third row of array A028412.

[Lucas(n)*Lucas(2*n): n in [1..25]]; // Vincenzo Librandi, May 03 2011

Table[Fibonacci[n*4]/Fibonacci[n],{n,50}] (* Vladimir Joseph Stephan Orlovsky, May 02 2011 *)

From Benoit Cloitre, Aug 30 2003: (Start)
a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4);
a(n) = Fibonacci(4*n)/Fibonacci(n) = A000045(4*n)/A000045(n). (End)
a(n) = Lucas(3*n) + (-1)^n*Lucas(n).
From R. J. Mathar, Oct 27 2008: (Start)
G.f.: x*(3+12*x-9*x^2-4*x^3)/((1+x-x^2)*(1-4*x-x^2)).
a(n) = A061084(n+1) + 2*A001077(n). (End)
a(n) = (1+phi)^n + (-phi)^n + (2*phi+1)^n + (3-2*phi)^n, phi = (1+sqrt(5))/2. - Gary Detlefs, Dec 09 2012

A099255 Expansion of g.f. (7+6x-6x^2-3x^3)/((x^2+x-1)(x^2-x-1)).

Original entry on oeis.org

7, 6, 15, 15, 38, 39, 99, 102, 259, 267, 678, 699, 1775, 1830, 4647, 4791, 12166, 12543, 31851, 32838, 83387, 85971, 218310, 225075, 571543, 589254, 1496319, 1542687, 3917414, 4038807, 10255923, 10573734, 26850355, 27682395, 70295142, 72473451

Offset: 0

Creighton Dement, Oct 09 2004

One of two sequences involving the Lucas/Fibonacci numbers.

This sequence consists of pairs of numbers more or less close to each other with "jumps" in between pairs. "pos((Ex)^n)" sums up over all floretion basis vectors with positive coefficients for each n. The following relations appear to hold: a(2n) - (a(2n-1) + a(2n-2)) = 2*Luc(2n) a(2n+1) - a(2n) = Fib(2n), apart from initial term a(2n+1)/a(2n-1) -> 2 + golden ratio phi a(2n)/a(2n-2) -> 2 + golden ratio phi An identity: (1/2)a(n) - (1/2)A099256(n) = ((-1)^n)A000032(n)

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).

Cf. A099256, A000032.

LinearRecurrence[{0,3,0,-1},{7,6,15,15},40] (* Harvey P. Dale, Dec 29 2012 *)

a(n) = 2*pos((Ex)^n)
a(0) = 7, a(1) = 6, a(2) = a(3) = 15, a(n+4) = 3a(n+2) - a(n).
a(2n) = A022097(2n+1), a(2n+1) = A022086(2n+3).
a(n) = A061084(n+1)+A013655(n+2). [R. J. Mathar, Nov 30 2008]

More terms from Creighton Dement, Apr 19 2005

A099256 Expansion of g.f. (3-x)(1+3x+x^2)/((1-x-x^2)*(1+x-x^2)).

Original entry on oeis.org

3, 8, 9, 23, 24, 61, 63, 160, 165, 419, 432, 1097, 1131, 2872, 2961, 7519, 7752, 19685, 20295, 51536, 53133, 134923, 139104, 353233, 364179, 924776, 953433, 2421095, 2496120, 6338509, 6534927, 16594432, 17108661, 43444787, 44791056, 113739929, 117264507, 297775000, 307002465, 779585071

Offset: 0

Creighton Dement, Oct 18 2004

One of two sequences involving the Lucas/Fibonacci numbers. This sequence consists of pairs of numbers more or less close to each other with "jumps" in between pairs.
a(n+3) + a(n) - a(n+2) appears to be mysteriously connected with a(n+1).
Both this sequence and A099255 were created using "Floretion dynamical symmetries" (see link for further details).

Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).

Cf. A000045, A099255, A000032, A055273 (bisection), A097134 (bisection).

LinearRecurrence[{0,3,0,-1},{3,8,9,23},40] (* Harvey P. Dale, Apr 22 2012 *)

a(2n+2) - a(2n+1) = Fibonacci(2n-1).
A099255(n)/2 - a(n)/2 = (-1)^n*A000032(n)
a(0) = 3, a(1) = 8, a(2) = 9, a(3) = 23, a(n+4) = 3a(n+2) - a(n).
a(2n) = A022086(2n+2), a(2n+1) = A022097(2n+2).
a(n) = A013655(n+2)-A061084(n+1).

Definition corrected, extended. - R. J. Mathar, Nov 13 2008

A190914 Expansion of ( 5-9x^2-2x^3 ) / ( (1+x-x^2)*(1-x-x^2-x^3) ).

Original entry on oeis.org

5, 0, 6, 3, 18, 10, 57, 42, 178, 165, 566, 616, 1821, 2236, 5914, 7963, 19362, 27982, 63813, 97394, 211458, 336633, 703786, 1157544, 2350597, 3964960, 7872702, 13541691, 26425522, 46147178, 88853297, 156994354, 299165378, 533410837, 1008343310, 1810544592, 3401446413, 6140811708, 11481472994, 20815538227

Offset: 0

Reikku Kulon, May 23 2011

The sequence ..., 14, 29, 10, 2, 9, 2, 0, [5], 0, 6, 3, 18, 10, 57, 42, ...
(the number in square brackets at index 0) equals the trace of:
  [ 0 0 0 0-1 ]
  [ 1 0 0 0 0 ]
  [ 0 1 0 0 1 ]^(+n)
  [ 0 0 1 0 3 ]
  [ 0 0 0 1 0 ]
or
  [ 0 0 0 0-1 ]
  [ 1 0 0 0 0 ]
  [ 0 1 0 0 3 ]^(-n)
  [ 0 0 1 0 1 ]
  [ 0 0 0 1 0 ]
Its characteristic polynomial is (x^2 +/- x - 1) * (x^3 -/+ x^2 -/+ x - 1); these factors are Fibonacci and tribonacci polynomials.  The ratio of negative terms approaches the golden ratio; the ratio of positive terms approaches the tribonacci constant.
Prime numbers p divide a(+p) and a(-p), as the trace of a matrix M^p (mod p) is constant.
Nonprimes c very rarely divide a(+c) and a(-c) simultaneously.  The only known dual pseudoprime in the sequence is 1.
The distribution of residues induces gaps between pseudoprimes having roughly the size of c.  For example, after 1034881 there is a gap of more than one million terms without either variety of pseudoprime.
Pseudoprimes appear limited to squared primes and squarefree numbers with three or more prime factors.  11 and 13 are more common than other factors.
Positive pseudoprimes: c | a(+c)
----------------------------------------------
1
3481. . . . 59^2
17143 . . . 7 31 79
105589. . . 11 29 331
635335. . . 5 283 449
2992191 . . 3 29 163 211
3659569 . . 1913^2
Negative pseudoprimes: c | a(-c)
----------------------------------------------
1
9 . . . . . 3^2
806 . . . . 2 13 31
1419. . . . 3 11 43
6241. . . . 79^2
6721. . . . 11 13 47
12749 . . . 11 19 61
21106 . . . 2 61 173
38714 . . . 2 13 1489
146689. . . 383^2
649621. . . 7 17 53 103
1034881 . . 41 43 587

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,1,0,-1).

Cf. A190913 (extended to negative indices), A000045, A000073, A001608, A000040, A005117, A125666.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (5-9*x^2 -2*x^3)/((1+x-x^2)*(1-x-x^2-x^3)) )); // G. C. Greubel, Apr 23 2019

LinearRecurrence[{0, 3, 1, 0, -1}, {5, 0, 6, 3, 18}, 40] (* G. C. Greubel, Apr 23 2019 *)

my(x='x+O('x^40)); Vec((5-9*x^2-2*x^3)/((1+x-x^2)*(1-x-x^2-x^3))) \\ G. C. Greubel, Apr 23 2019

((5-9*x^2-2*x^3)/((1+x-x^2)*(1-x-x^2-x^3))).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 23 2019

a(n) = A061084(n+1) + A001644(n). - R. J. Mathar, Jun 06 2011

A365907 Smallest nonnegative integer that is not the sum of fewer than n signed Lucas numbers.

Original entry on oeis.org

0, 1, 5, 16, 63, 262, 1105, 4676, 19803, 83882, 355325, 1505176, 6376023, 27009262, 114413065, 484661516, 2053059123, 8696898002, 36840651125, 156059502496, 661078661103, 2800374146902, 11862575248705, 50250675141716, 212865275815563, 901711778403962

Offset: 0

Mike Speciner, Sep 22 2023

Signed Lucas numbers are the union of A000032 and A061084.

			a(0) = 0, the sum of 0 Lucas numbers.
a(1) = 1 = A000032(1), the sum of 1 Lucas number.
a(2) = 5 = 1+4 = A000032(1)+A000032(3), the sum of 2 Lucas numbers. (2, 3, and 4 need only one term, since they are Lucas numbers.)
a(4) = 63 = 1+4+11+47.
For comparison, 45 is the first sum requiring 4 positive Lucas numbers (45 = 1+4+11+29, see A004146), but here 45 = 47+2-4 requires only 3 signed Lucas numbers so that a(4) != 45.

Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A000032, A061084, A004146 (analogous with only positive Lucas numbers).

LinearRecurrence[{5, -3, -1}, {0, 1, 5, 16, 63}, 26] (* Amiram Eldar, Sep 26 2023 *)

from sympy import lucas
a = lambda n: n if n<2 else (lucas(3*n-2)+3)//2

a(n) = n, for n<2.
a(n) = (A000032(3*n-2)+3)/2 = 1+4+Sum_{i=2..n-1} A000032(3*n-1), for n>1.
G.f.: x*(1 - 6*x^2 - x^3)/((1 - x)*(1 - 4*x - x^2)). - Stefano Spezia, Sep 25 2023

A226377 Lucas numbers differences triangle T(n,k), k<=n, where column k+1 holds the k-th differences of A000204, read by rows.

Original entry on oeis.org

1, 3, 2, 4, 1, -1, 7, 3, 2, 3, 11, 4, 1, -1, -4, 18, 7, 3, 2, 3, 7, 29, 11, 4, 1, -1, -4, -11, 47, 18, 7, 3, 2, 3, 7, 18, 76, 29, 11, 4, 1, -1, -4, -11, -29, 123, 47, 18, 7, 3, 2, 3, 7, 18, 47, 199, 76, 29, 11, 4, 1, -1, -4, -1, -29, -76, -199

Offset: 1

Richard R. Forberg, Jul 31 2013

Consecutive columns (i.e. k =1,2,3...) shift the Lucas sequence (A000204) down by 2 indices.
Diagonal (n=k) produces A061084, and Lucas numbers at increasingly negative indices for n=k>2.
Row sums equal A203976(n) for n=>1, which equals Lucas numbers  A000204(n) if n is odd, and 5 * A000045(2*n) (Fibonacci) if n is even.
Compare A227431 which is a differences triangle for the Fibonacci sequence A000045.

			Triangle begins:
1;
3,  2;
4,  1, -1;
7,  3,  2,  3;
11,  4,  1, -1, -4;
18,  7,  3,  2,  3,  7;
29, 11,  4,  1, -1, -4, -11;
47, 18,  7,  3,  2,  3,   7,  18;
76, 29, 11,  4,  1, -1,  -4, -11, -29;
...

Cf. A000204, A203976

T(n,1) = A000204(n) for n>0, T(n,k) = T(n,k-1) - T(n-1,k-1).

A226447 Expansion of (1-x+x^3)/(1-x^2+2*x^3-x^4).

Original entry on oeis.org

1, -1, 1, -2, 4, -5, 9, -15, 23, -38, 62, -99, 161, -261, 421, -682, 1104, -1785, 2889, -4675, 7563, -12238, 19802, -32039, 51841, -83881, 135721, -219602, 355324, -574925, 930249, -1505175, 2435423, -3940598, 6376022, -10316619, 16692641, -27009261, 43701901, -70711162, 114413064, -185124225

Offset: 0

Paul Curtz, Jun 28 2013

a(n) and its differences:
.   1,   -1,    1,   -2,    4,    -5,     9,   -15,    23,    -38, ...
.  -2,    2,   -3,    6,   -9,    14,   -24,    38,   -61,    100, ...
.   4,   -5,    9,  -15,   23,   -38,    62,   -99,   161,   -261, ...
.  -9,   14,  -24,   38,  -61,   100,  -161,   260,  -422,    682, ...
.  23,  -38,   62,  -99,  161,  -261,   421,  -682,  1104,  -1785, ...
. -61,  100, -161,  260, -422,   682, -1103,  1786, -2889,   4674, ...
. 161, -261,  421, -682, 1104, -1785,  2889, -4675,  7563, -12238, ...
The third row is the first shifted .
The main diagonal is A001077(n). The fourth is -A001077(n+1). By "shifted" antidiagonals there are one 1, two 2's (-2 of the first row and 2), generally A001651(n) (-1)^n *A001077(n).
a(n+1)/a(n) tends to A001622 (the golden ratio) as n -> infinity.

Index entries for linear recurrences with constant coefficients, signature (0,1,-2,1).

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-x+x^3)/(1-x^2+2*x^3-x^4))); // Bruno Berselli, Jul 04 2013

a[0] = 1; a[1] = -1; a[n_] := a[n] = a[n-2] - a[n-1] - {-1, 0, 1, 1, 0, -1}[[Mod[n+1, 6] + 1]]; Table[a[n], {n, 0, 41}] (* Jean-François Alcover, Jul 04 2013 *)

a(0)=1, a(1)=-1; for n>1, a(n) = a(n-2) - a(n-1) + A010892(n+2).
a(n) = a(n-2) -2*a(n-3) +a(n-4).
a(n) = A226956(-n).
a(n+1) = A039834(n) - (-1)^n*A094686(n).
a(n+6) - a(n) = 2*(-1)^n* A000032(n+3).
a(2n+1) = -A226956(2n+1).
G.f. ( -1+x-x^3 ) / ( (x^2-x-1)*(1-x+x^2) ). - R. J. Mathar, Jun 29 2013
2*a(n) = A010892(n+2)+A061084(n+1). - R. J. Mathar, Jun 29 2013

A367816 Number of terms in a shortest sequence of Lucas numbers that sum to n, allowing Lucas numbers with negative indices.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 2, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 2, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3, 2, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 3, 2, 3, 3, 3, 3, 4, 3, 2, 3, 3, 2, 1, 2, 2, 2

Offset: 0

Mike Speciner, Dec 01 2023

			For n = 0, the empty sequence sums to 0, so a(0) = 0.
For n = 1, 2, 3, 4, 7, 11, 18, each n is a Lucas number, so a(n) = 1.
The first n needing a negative-index Lucas number is 17 = 18 + -1; a(17) = 2.

Cf. A000032 Lucas numbers; A061084 negative index Lucas numbers.
A116543 is the similar sequence where negative index Lucas numbers are not allowed.
a(A365907(n)) is the first occurrence of n.

from itertools import count
def  a(n) :
  """For integer n, the least number of Lucas terms required to sum to n."""
  f = [2,1];    # Lucas numbers, starting with Lucas(0)
  while f[-1] <= (n or 1) :
    f.append(f[-2]+f[-1]);
  a = [0 for _ in range(f[-1]+1)];
  for i in f :
    a[i] = 1;
  for c in count(2) :
    if not all(a[4:]) :
      for i in range(4,f[-1]) :
        if not a[i] :
          for j in f :
            if j >= i :
              break;
            if a[i-j] == c-1 :
              a[i] = c;
              break;
          if not a[i]:
            for j in f[1::2] :
              if i+j >= len(a) :
                break;
              if a[i+j] == c-1 :
                a[i] = c;
                break;
    else :
      break;
  return a[n];

a(0) = 0; a(A000032(n)) = 1.

For n > 0, a(n) = 1+min(a(n-Lucas(k))) where k ranges over Z.

cp's OEIS Frontend

A132148 Triangular array T(n,k) = C(n,k)*Lucas(n-k), 0 <= k <= n.

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A061083 Fibonacci-type sequence based on division: a(0) = 1, a(1) = 2 and a(n) = a(n-2)/a(n-1) but ignore decimal point.

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A083564 a(n) = L(n)*L(2n), where L(n) are the Lucas numbers (A000204).

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Magma

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A099255 Expansion of g.f. (7+6*x-6*x^2-3*x^3)/((x^2+x-1)*(x^2-x-1)).

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A099256 Expansion of g.f. (3-x)*(1+3*x+x^2)/((1-x-x^2)*(1+x-x^2)).

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A190914 Expansion of ( 5-9*x^2-2*x^3 ) / ( (1+x-x^2)*(1-x-x^2-x^3) ).

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Magma

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PARI

SageMath

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A365907 Smallest nonnegative integer that is not the sum of fewer than n signed Lucas numbers.

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A099255 Expansion of g.f. (7+6x-6x^2-3x^3)/((x^2+x-1)(x^2-x-1)).

A099256 Expansion of g.f. (3-x)(1+3x+x^2)/((1-x-x^2)*(1+x-x^2)).

A190914 Expansion of ( 5-9x^2-2x^3 ) / ( (1+x-x^2)*(1-x-x^2-x^3) ).