A006497 -id:A006497 - cp's OEIS frontend

A014176 Decimal expansion of the silver mean, 1+sqrt(2).

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

Offset: 1

N. J. A. Sloane

From Hieronymus Fischer, Jan 02 2009: (Start)
Set c:=1+sqrt(2). Then the fractional part of c^n equals 1/c^n, if n odd. For even n, the fractional part of c^n is equal to 1-(1/c^n).
c:=1+sqrt(2) satisfies c-c^(-1)=floor(c)=2, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.
1/c = sqrt(2)-1.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
Other examples of constants x satisfying the relation x-x^(-1)=floor(x) include A001622 (the golden ratio: where floor(x)=1) and A098316 (the "bronze" ratio: where floor(x)=3). (End)
In terms of continued fractions the constant c can be described by c=[2;2,2,2,...]. - Hieronymus Fischer, Oct 20 2010
Side length of smallest square containing five circles of diameter 1. - Charles R Greathouse IV, Apr 05 2011
Largest radius of four circles tangent to a circle of radius 1. - Charles R Greathouse IV, Jan 14 2013
An analog of Fermat theorem: for prime p, round(c^p) == 2 (mod p). - Vladimir Shevelev, Mar 02 2013
n*(1+sqrt(2)) is the perimeter of a 45-45-90 triangle with hypotenuse n. - Wesley Ivan Hurt, Apr 09 2016
This algebraic integer of degree 2, with minimal polynomial x^2 - 2*x - 1, is also the length ratio diagonal/side of the second largest diagonal in the regular octagon (not counting the side). The other two diagonal/side ratios are A179260 and A121601. - Wolfdieter Lang, Oct 28 2020
c^n = A001333(n) + A000129(n) * sqrt(2). - Gary W. Adamson, Apr 26 2023
c^n = c * A000129(n) + A000129(n-1), where c = 1 + sqrt(2). - Gary W. Adamson, Aug 30 2023

			2.414213562373095...

B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.

G. C. Greubel, Table of n, a(n) for n = 1..10000
Nicholas R. Beaton, Mireille Bousquet-Mélou, Jan de Gier, Hugo Duminil-Copin, and Anthony J. Guttmann, The critical fugacity for surface adsorption of self-avoiding walks on the honeycomb lattice is 1+sqrt(2), arXiv:1109.0358 [math-ph], 2011-2013.
Eric Weisstein's World of Mathematics, Silver Ratio
Wikipedia, Exact trigonometric constants
Wikipedia, Metallic mean
Wikipedia, Silver ratio
Index entries for algebraic numbers, degree 2

Apart from initial digit the same as A002193.
Cf. A000032, A006497, A080039, A179260, A121601.
See A098316 for [3;3,3,...]; A098317 for [4;4,4,...]; A098318 for [5;5,5,...]. - Hieronymus Fischer, Oct 20 2010
Cf. A000129, A049310.

Digits:=100: evalf(1+sqrt(2)); # Wesley Ivan Hurt, Apr 09 2016

RealDigits[1 + Sqrt@ 2, 10, 111] (* Or *)
RealDigits[Exp@ ArcSinh@ 1, 10, 111][[1]] (* Robert G. Wilson v, Aug 17 2011 *)
Circs[n_] := With[{r = Sin[Pi/n]/(1 - Sin[Pi/n])}, Graphics[Append[
  Table[Circle[(r + 1) {Sin[2 Pi k/n], Cos[2 Pi k/n]}, r], {k, n}],   {Blue, Circle[{0, 0}, 1]}]]] Circs[4] (* Charles R Greathouse IV, Jan 14 2013 *)

1+sqrt(2) \\ Charles R Greathouse IV, Jan 14 2013

Conjecture: 1+sqrt(2) = lim_{n->oo} A179807(n+1)/A179807(n).
Equals cot(Pi/8) = tan(Pi*3/8). - Bruno Berselli, Dec 13 2012, and M. F. Hasler, Jul 08 2016
Silver mean = 2 + Sum_{n>=0} (-1)^n/(P(n-1)*P(n)), where P(n) is the n-th Pell number (A000129). - Vladimir Shevelev, Feb 22 2013
Equals exp(arcsinh(1)) which is exp(A091648). - Stanislav Sykora, Nov 01 2013
Limit_{n->oo} exp(asinh(cos(Pi/n))) = sqrt(2) + 1. - Geoffrey Caveney, Apr 23 2014
exp(asinh(cos(Pi/2 - log(sqrt(2)+1)*i))) = exp(asinh(sin(log(sqrt(2)+1)*i))) = i. - Geoffrey Caveney, Apr 23 2014
Equals Product_{k>=1} A047621(k) / A047522(k) = (3/1) * (5/7) * (11/9) * (13/15) * (19/17) * (21/23) * ... . - Dimitris Valianatos, Mar 27 2019
From Wolfdieter Lang, Nov 10 2023:(Start)
Equals lim_{n->oo} A000129(n+1)/A000129(n) (see A000129, Pell).
Equals lim_{n->oo} S(n+1, 2*sqrt(2))/S(n, 2*sqrt(2)), with the Chebyshev S(n,x) polynomial (see A049310). (End)
From  Peter Bala, Mar 24 2024: (Start)
An infinite family of continued fraction expansions for this constant can be obtained from Berndt, Entry 25, by setting n = 1/2 and x = 8*k + 6 for k >= 0.
For example, taking k = 0 and k = 1 yields
sqrt(2) + 1 = 15/(6 + (1*3)/(12 + (5*7)/(12 + (9*11)/(12 + (13*15)/(12 + ... + (4*n + 1)*(4*n + 3)/(12 + ... )))))) and
sqrt(2) + 1 = (715/21) * 1/(14 + (1*3)/(28 + (5*7)/(28 + (9*11)/(28 + (13*15)/(28 + ... + (4*n + 1)*(4*n + 3)/(28 + ... )))))). (End)

A086902 a(n) = 7*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 7.

Original entry on oeis.org

2, 7, 51, 364, 2599, 18557, 132498, 946043, 6754799, 48229636, 344362251, 2458765393, 17555720002, 125348805407, 894997357851, 6390330310364, 45627309530399, 325781497023157, 2326097788692498, 16608466017870643, 118585359913786999, 846705985414379636

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 18 2003

a(n+1)/a(n) converges to (7+sqrt(53))/2 = 7.14005... = A176439.
Lim a(n)/a(n+1) as n approaches infinity = 0.1400549... = 2/(7+sqrt(53)) = (sqrt(53)-7)/2 = 1/A176439 = A176439 - 7.
From Johannes W. Meijer, Jun 12 2010: (Start)
In general sequences with recurrence a(n) = k*a(n-1)+a(n-2) with a(0)=2 and a(1)=k [and a(-1)=0] have generating function (2-k*x)/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n+1) = b(5n), a(3n+2)=b(5*n+3), a(3n+3)=2*b(5n+4) where b(n) is the sequence of numerators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n)/2, for n>=1, is the sequence of numerators of continued fraction convergents to sqrt(k^2/4+1).]
For the sequence given above k=7 which implies that it is associated with A041090.
For a similar statement about sequences with recurrence a(n) = k*a(n-1)+a(n-2) but with a(0)=1 [and a(-1)=0] see A054413; a sequence that is associated with A041091.
For more information follow the Khovanova link and see A087130, A140455 and A178765.
(End)

			a(4) = 7*a(3) + a(2) = 7*364 + 51 = 2599.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,1).

Cf. A058316, A176439.

Cf. A000032 (k=1), A006497 (k=3), A087130 (k=5), A086902 (k=7), A087798 (k=9), A001946 (k=11), A088316 (k=13), A090301 (k=15), A090306 (k=17). - Johannes W. Meijer, Jun 12 2010

I:=[2,7]; [n le 2 select I[n] else 7*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 7, a[n] == 7 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
LinearRecurrence[{7,1},{2,7},30] (* Harvey P. Dale, May 25 2023 *)

a(n)=([0,1; 1,7]^n*[2;7])[1,1] \\ Charles R Greathouse IV, Apr 06 2016

a(n) = ((7+sqrt(53))/2)^n + ((7-sqrt(53))/2)^n.
E.g.f. : 2exp(7x/2)cosh(sqrt(53)x/2); a(n)=2^(1-n)sum{k=0..floor(n/2), C(n, 2k)53^k7^(n-2k)}. a(n)=2T(n, 7i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
G.f.: (2-7x)/(1-7x-x^2). - Philippe Deléham, Nov 16 2008
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 7*A097837(n), a(2n) = A099368(n).
a(3n+1) = A041090(5n), a(3n+2) = A041090(5*n+3), a(3n+3) = 2*A041090(5n+4).
Limit(a(n+k)/a(k), k=infinity) = (A086902(n) + A054413(n-1)*sqrt(53))/2.
Limit(A086902(n)/A054413(n-1), n=infinity) = sqrt(53). (End)

A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

Original entry on oeis.org

2, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, 384238402, 1995189565, 10360186227, 53796120700, 279340789727, 1450500069335, 7531841136402, 39109705751345, 203080369893127

Offset: 0

Paul Barry, Aug 16 2003

Sequence is related to the fifth metallic mean [5;5,5,5,5,...] (see A098318).
The solution to the general recurrence b(n) = (2*k+1)*b(n-1)+b(n-2) with b(0)=2, b(1) = 2*k+1 is b(n) = ((2*k+1)+sqrt(4*k^2+4*k+5))^n+(2*k+1)-sqrt(4*k^2+4*k+5))^n)/2; b(n) = 2^(1-n)*Sum_{j=0..n} C(n, 2*j)*(4*k^2+4*k+5)^j*(2*k+1)^(n-2*j); b(n) = 2*T(n, (2*k+1)*x/2)(-1)^i with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
Primes in this sequence include a(0) = 2; a(1) = 5; a(4) = 727; a(8) = 528527 (3) semiprimes in this sequence include a(7) = 101785; a(13) = 1995189565; a(16) = 279340789727; a(19) = 39109705751345; a(20) = 203080369893127 - Jonathan Vos Post, Feb 09 2005
a(n)^2 - 29*A052918(n-1)^2 = 4*(-1)^n, with n>0 - Gary W. Adamson, Oct 07 2008
For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A072263. - Johannes W. Meijer, Aug 01 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence L_{5,n}.
Tanya Khovanova, Recursive Sequences
Wikipedia, Metallic mean
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A006497, A014448, A085447.

Cf. A086902, A000032, A052918.

I:=[2,5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 5, a[n] == 5 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)

{a(n) = if( n<0, (-1)^n * a(-n), polsym(x^2 - 5*x -1, n) [n + 1])} /* Michael Somos, Nov 04 2008 */

[lucas_number2(n,5,-1) for n in range(0, 21)] # Zerinvary Lajos, May 14 2009

a(n) = ((5+sqrt(29))/2)^n+((5-sqrt(29))/2)^n.
a(n) = A100236(n) + 1.
E.g.f. : 2*exp(5*x/2)*cosh(sqrt(29)*x/2); a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2k)*29^k*5^(n-2*k). a(n) = 2T(n, 5i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
O.g.f.: (-2+5*x)/(-1+5*x+x^2). - R. J. Mathar, Dec 02 2007
a(-n) = (-1)^n * a(n). - Michael Somos, Nov 01 2008
A090248(n) = a(2*n). 5 * A097834(n) = a(2*n + 1). - Michael Somos, Nov 01 2008
Limit(a(n+k)/a(k), k=infinity) = (A087130(n) + A052918(n-1)*sqrt(29))/2. Limit(A087130(n)/A052918(n-1), n= infinity) = sqrt(29). - Johannes W. Meijer, Jun 12 2010
a(3n+1) = A041046(5n), a(3n+2) = A041046(5n+3) and a(3n+3) = 2*A041046 (5n+4). - Johannes W. Meijer, Jun 12 2010
a(n) = 2*A052918(n) - 5*A052918(n-1). - R. J. Mathar, Oct 02 2020
From Peter Bala, Jul 09 2025 : (Start)
The following series telescope (Cf. A000032):
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A098316 Decimal expansion of [3, 3, ...] = (3 + sqrt(13))/2.

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein, Sep 02 2004

For reasons following from the formula section, this constant could be called "the bronze ratio". For this, compare with A001622 and A014176.
If c is this constant and n > 0, then for n even, c^n = [A100230(n), 1, A100230(n)-1, 1, A100230(n)-1, 1, A100230(n)-1, 1, ...], for n odd, c^n = [A100230(n)+1, A100230(n)+1, A100230(n)+1, ...]. - Gerald McGarvey, Dec 15 2007
This is the shape of a 3-extension rectangle; see A188640 for definitions. - Clark Kimberling, Apr 10 2011
From Vladimir Shevelev, Mar 02 2013: (Start)
An analog of Fermat theorem: for prime p, round(c^p) == 3 (mod p).
A generalization for "metallic" constants c_N = (N+sqrt(N^2+4))/2, N>=1: for prime p, round((c_N)^p) == N (mod p). (End)
This is the positive real algebraic number c of degree 2 with minimal polynomial x^3 - x - 1. The other negative root is 3 - c. - Wolfdieter Lang, Aug 29 2022
c^n = c*A006190(n) + A006190(n-1). - Gary W. Adamson, Apr 02 2024

			3.30277563...

G. C. Greubel, Table of n, a(n) for n = 1..10000
Wikipedia, Metallic mean
Index entries for algebraic numbers, degree 2

Cf. A001622, A014176, A098317, A098318, A000032, A006497, A080039, A049310.

RealDigits[(3 + Sqrt[13])/2, 10, 100][[1]] (* G. C. Greubel, Apr 16 2017 *)

(3 + sqrt(13))/2 \\ Charles R Greathouse IV, Jul 24 2013

3 plus the constant in A085550. - R. J. Mathar, Sep 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
Set c:=(3+sqrt(13))/2. Then the fractional part of c^n equals 1/c^n, if n odd. For even n, the fractional part of c^n is equal to 1-(1/c^n).
c:=(3+sqrt(13))/2 satisfies c-c^(-1)=floor(c)=3, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.
1/c=(sqrt(13)-3)/2.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
Other examples of constants x satisfying the relation x-x^(-1)=floor(x) include A001622 (the golden ratio: where floor(x)=1) and A014176 (the silver ratio: where floor(x)=2). (End)
c=3+sum{k>=1}(-1)^(k-1)/(A006190(k)*A006190(k+1)). - Vladimir Shevelev, Feb 23 2013
A generalization for "metallic" constants c_N = (N+sqrt(N^2+4))/2, N>=1. Let {A_N(n), n>=0} be the sequence 0, 1, N, N^2+1, N^3+2*N, N^4+3*N^2+1,..., a(N) = N*a(N-1) + a(N-2). Then c_N = N + sum_{n>=1} (-1)^(n-1)/(A_N(n)*A_N(n+1)) (cf. A001622, A014176, A098316, A098317, A098318). - Vladimir Shevelev, Feb 23 2013
Equals lim_{n->oo} S(n, sqrt(13))/S(n-1, sqrt(13)), with the S-Chebyshev polynomial (see A049310). - Wolfdieter Lang, Nov 15 2023

A114525 Triangle of coefficients of the Lucas (w-)polynomials.

Original entry on oeis.org

2, 0, 1, 2, 0, 1, 0, 3, 0, 1, 2, 0, 4, 0, 1, 0, 5, 0, 5, 0, 1, 2, 0, 9, 0, 6, 0, 1, 0, 7, 0, 14, 0, 7, 0, 1, 2, 0, 16, 0, 20, 0, 8, 0, 1, 0, 9, 0, 30, 0, 27, 0, 9, 0, 1, 2, 0, 25, 0, 50, 0, 35, 0, 10, 0, 1, 0, 11, 0, 55, 0, 77, 0, 44, 0, 11, 0, 1, 2, 0, 36, 0, 105, 0, 112, 0, 54, 0, 12, 0, 1

Offset: 0

Eric W. Weisstein, Dec 06 2005

Unsigned version of A108045.

The row reversed triangle is A162514. - Paolo Bonzini, Jun 23 2016

			2, x, 2 + x^2, 3*x + x^3, 2 + 4*x^2 + x^4, 5*x + 5*x^3 + x^5, ... give triangle
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  0:    2
  1:    0  1
  2:    2  0  1
  3:    0  3  0  1
  4:    2  0  4  0  1
  5:    0  5  0  5  0  1
  6:    2  0  9  0  6  0  1
  7:    0  7  0 14  0  7  0  1
  8:    2  0 16  0 20  0  8  0  1
  9:    0  9  0 30  0 27  0  9  0  1
  10:   2  0 25  0 50  0 35  0 10  0  1
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  .... reformatted by _Wolfdieter Lang_, Feb 10 2023

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
B. Johnson, Fibonacci numbers and matrices, 2009.
Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4.
M. Pétréolle, Characterization of Cyclically Fully commutative elements in finite and affine Coxeter Groups, arXiv preprint arXiv:1403.1130 [math.GR], 2014.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Eric Weisstein's World of Mathematics, Lucas Polynomial

Cf. A108045 (signed version).
Cf. A034807, A162514.
Cf. Sequences L(n,x): A000032(x = 1), A002203 (x = 2), A006497 (x = 3), A014448 (x = 4), A087130 (x = 5), A085447 (x = 6), A086902 (x = 7), A086594 (x = 8), A087798 (x = 9), A086927 (x = 10), A001946 (x = 11), A086928 (x = 12), A088316 (x = 13), A090300 (x = 14), A090301 (x = 15), A090305 (x = 16), A090306 (x = 17), A090307 (x = 18), A090308 (x = 19), A090309 (x = 20), A090310 (x = 21), A090313 (x = 22), A090314 (x = 23), A090316 (x = 24), A087281 (x = 29), A087287 (x = 76), A089772 (x = 199).

Lucas := proc(n,x)
    option remember;
    if  n=0 then
        2;
    elif n =1 then
        x ;
    else
        x*procname(n-1,x)+procname(n-2,x) ;
    end if;
end proc:
A114525 := proc(n,k)
    coeftayl(Lucas(n,x),x=0,k) ;
end proc:
seq(seq(A114525(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Aug 16 2019

row[n_] := CoefficientList[LucasL[n, x], x];
Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Aug 11 2018 *)

From Peter Bala, Mar 18 2015: (Start)
The Lucas polynomials L(n,x) satisfy the recurrence L(n+1,x) = x*L(n,x) + L(n-1,x) with L(0,x) = 2 and L(1,x) = x.
O.g.f.: Sum_{n >= 0} L(n,x)*t^n = (2 - x*t)/(1 - t^2 - x*t) = 2 + x*t + (x^2 + 2)*t^2 + (3*x + x^3)*t^3 + ....
L(n,x) = trace( [ x, 1; 1, 0 ]^n ).
exp( Sum_{n >= 1} L(n,x)*t^n/n ) = Sum_{n >= 0} F(n+1,x)*t^n, where F(n,x) denotes the n-th Fibonacci polynomial. (see Appendix A3 in Johnson).
exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*t^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)*t^n.
exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n.
L(n,1) = Lucas(n) = A000032(n); L(n,4) = Lucas(3*n) = A014448(n); L(n,11) = Lucas(5*n) = A001946(n); L(n,29) = Lucas(7*n) = A087281(n); L(n,76) = Lucas(9*n) = A087287(n); L(n,199) = Lucas(11*n) = A089772(n). The general result is L(n,Lucas(2*k + 1)) = Lucas((2*k + 1)*n). (End)
From Jeremy Dover, Jun 10 2016: (Start)
Read as a triangle T(n,k), n >= 0, n >= k >= 0, T(n,k) = (Binomial((n+k)/2,k) + Binomial((n+k-2)/2,k))*(1+(-1)^(n-k))/2.
T(n,k) = A046854(n-1,k-1) + A046854(n-1,k) + A046854(n-2,k) for even n+k with n+k > 0, assuming A046854(n,k) = 0 for n < 0, k < 0, k > n.
T(n,k) is the number of binary strings of length n with exactly k pairs of consecutive 0's and no pair of consecutive 1's, where the first and last bits are considered consecutive. (End)
From Peter Bala, Sep 03 2022: (Start)
L(n,x) = 2*(i)^n*T(n,-i*x/2), where i = sqrt(-1) and T(n,x) is the n-th Chebyshev polynomial of the first kind.
d/dx(L(n,x)) = n*F(n,x), where F(n,x) denotes the n-th Fibonacci polynomial.
Let P_n(x,y) = (L(n,x) - L(n,y))/(x - y). Then {P_n(x,y): n >= 1} is a fourth-order linear divisibility sequence of polynomials in the ring Z[x,y]: if m divides n in Z then P_m(x,y) divides P_n(x,y) in Z[x,y].
P_n(1,1) = A045925(n); P_n(1,4) = A273622; P_n(2,2) = A093967(n).
L(2*n,x)^2 - L(2*n-1,x)*L(2*n+1,x) = x^2 + 4 for n >= 1.
Sum_{n >= 1} L(2*n,x)/( L(2*n-1,x) * L(2*n+1,x) ) = 1/x^2 and
Sum_{n >= 1} (-1)^(n+1)/( L(2*n,x) + x^2/L(2*n,x) ) = 1/(x^2 + 4), both valid for all nonzero real x. (End)
From Peter Bala, Nov 18 2022: (Start)
L(n,x) = Sum_{k = 0..floor(n/2)} (n/(n-k))*binomial(n-k,k)*x^(n-2*k) for n >= 1.
For odd m, L(n, L(m,x)) = L(n*m, x).
For integral x, the sequence {u(n)} := {L(n,x)} satisfies the Gauss congruences: u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all positive integers m and r and all primes p.
Let p be an odd prime and let 0 <= k <= p - 1. Let alpha_k = the p-adic limit_{n -> oo} L(p^n,k). Then alpha_k is a well-defined p-adic-integer and the polynomial L(p,x) - x of degree p factorizes as L(p,x) - x = Product_{k = 0..p-1} (x - alpha_k). For example, L(5,x) - x = x^5 + 5*x^3 + 4*x = x*(x - A269591)*(x - A210850)*(x - A210851)*(x - A269592) in the ring of 5-adic integers. (End)
The formula for L(n,x) given in the first line of the preceding section, with L(0, x) = 2, is rewritten L(n, x) = Sum_{k = 0..floor(n/2)} A034807(n, k)*x^(n - 2*k). See the formula by Alexander Elkins in A034807. - Wolfdieter Lang, Feb 10 2023

A052924 Expansion of g.f.: (1-x)/(1 - 3*x - x^2).

Original entry on oeis.org

1, 2, 7, 23, 76, 251, 829, 2738, 9043, 29867, 98644, 325799, 1076041, 3553922, 11737807, 38767343, 128039836, 422886851, 1396700389, 4612988018, 15235664443, 50319981347, 166195608484, 548906806799, 1812916028881

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Euler encountered this sequence when finding the largest root of z^2 - 3z - 1 = 0. - V. Frederick Rickey (fred-rickey(AT)usma.edu), Aug 20 2003
Let M = a triangle with the Pell series A000129 (1, 2, 5, 12, ...) in each column, with the leftmost column shifted upwards one row. A052924 starting (1, 2, 7, 23, ...) = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 31 2010
a(n) is the number of compositions of n when there are 2 types of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
Equals partial sums of A108300 prefaced with a 1: (1, 1, 5, 16, 53, 175, 578, ...). - Gary W. Adamson, Feb 15 2012

L. Euler, Introductio in analysin infinitorum, 1748, section 338. English translation by John D. Blanton, Introduction to Analysis of the Infinite, 1988, Springer, p. 286.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Sergio Falcón, The k-Fibonacci difference sequences, Chaos, Solitons & Fractals, Volume 87, June 2016, Pages 153-157.
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 909
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,1).

A108300 (first differences), A006190 (partial sums), A355981 (primes).

Cf. A000032, A001077, A078343, A164581, A179237, A180148, A329723.

a:=[1,2];; for n in [3..30] do a[n]:=3*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Jun 09 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-x)/(1-3*x-x^2) )); // G. C. Greubel, Jun 09 2019

spec:= [S,{S=Sequence(Prod(Sequence(Z),Union(Z,Z,Prod(Z,Z))))}, unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
seq(coeff(series((1-x)/(1-3*x-x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 16 2019

CoefficientList[Series[(1-x)/(1-3*x-x^2), {x,0,30}], x] (* G. C. Greubel, Jun 09 2019 *)

Vec((1-x)/(1-3*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011

((1-x)/(1-3*x-x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jun 09 2019

a(n) = 3*a(n-1) + a(n-2).
a(n) = Sum_{alpha=RootOf(-1+3*x+x^2)} (1/13)*(1+5*alpha)*alpha^(-1-n).
With offset 1: a(1)=1; for n > 1, a(n) = Sum_{i=1..3*n-4} a(ceiling(i/3)). - Benoit Cloitre, Jan 04 2004
Binomial transform of A006130. a(n) = (1/2 - sqrt(13)/26)*(3/2 - sqrt(13)/2)^n + (1/2 + sqrt(13)/26)*(3/2 + sqrt(13)/2)^n. - Paul Barry, Jul 20 2004
From Creighton Dement, Nov 04 2004: (Start)
a(n) = A006190(n+1) - A006190(n);
4*a(n) = 9*A006190(n+1) - A006497(n+1) - 2*A003688(n+1). (End)
Numerators in continued fraction [1, 2, 3, 3, 3, ...], where the latter = 0.69722436226...; the length of an inradius of a right triangle with legs 2 and 3. - Gary W. Adamson, Dec 19 2007
If p[1]=2, p[i]=3, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i<=j), A[i,j] = -1, (i=j+1), and A[i,j]=0 otherwise. Then, for n >= 1, a(n-1) = det A. - Milan Janjic, Apr 29 2010
a(n) = A006190(n) + A003688(n). - R. J. Mathar, Jul 06 2012
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*3^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*3^k, n>0. - R. J. Mathar, Feb 14 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) + 3*(-1)^(n+1)/a(n)) = 1/2, since 1/(a(n) + 3*(-1)^(n+1)/a(n)) = b(n) - b(n+1), where b(n) = (1/3) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 3*(-1)^(n+1)/a(n)) = 1/6, since 1/(a(n) + 3*(-1)^(n+1)/a(n)) = c(n) + c(n+1), where c(n) = (1/3) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

More terms from James Sellers, Jun 06 2000

A090301 a(n) = 15*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 15.

Original entry on oeis.org

2, 15, 227, 3420, 51527, 776325, 11696402, 176222355, 2655031727, 40001698260, 602680505627, 9080209282665, 136805819745602, 2061167505466695, 31054318401746027, 467875943531657100, 7049193471376602527

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 25 2004

Lim_{n-> infinity} a(n)/a(n+1) = 0.066372... = 2/(15+sqrt(229)) = (sqrt(229)-15)/2.
Lim_{n-> infinity} a(n+1)/a(n) = 15.066372... = (15+sqrt(229))/2 = 2/(sqrt(229)-15).
For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010

			a(4) = 15*a(3) + a(2) = 15*3420 + 227 = ((15+sqrt(229))/2)^4 + ((15-sqrt(229))/2)^4 = 51526.9999805 + 0.0000194 = 51527.

G. C. Greubel, Table of n, a(n) for n = 0..500
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 (15,1).

Cf. A058087, A071416.
Lucas polynomials: A114525.
Lucas polynomials Lucas(n,m): A000032 (m=1), A002203 (m=2), A006497 (m=3), A014448 (m=4), A087130 (m=5), A085447 (m=6), A086902 (m=7), A086594 (m=8), A087798 (m=9), A086927 (m=10), A001946 (m=11), A086928 (m=12), A088316 (m=13), A090300 (m=14), this sequence (m=15), A090305 (m=16), A090306 (m=17), A090307 (m=18), A090308 (m=19), A090309 (m=20), A090310 (m=21), A090313 (m=22), A090314 (m=23), A090316 (m=24), A330767 (m=25), A087281 (m=29), A087287 (m=76), A089772 (m=199).

m:=15;; a:=[2,m];; for n in [3..20] do a[n]:=m*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Dec 31 2019

m:=15; I:=[2,m]; [n le 2 select I[n] else m*Self(n-1) +Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 31 2019

seq(simplify(2*(-I)^n*ChebyshevT(n, 15*I/2)), n = 0..20); # G. C. Greubel, Dec 31 2019

LucasL[Range[20]-1, 15] (* G. C. Greubel, Dec 31 2019 *)

vector(21, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 15*I/2) ) \\ G. C. Greubel, Dec 31 2019

[2*(-I)^n*chebyshev_T(n, 15*I/2) for n in (0..20)] # G. C. Greubel, Dec 31 2019

a(n) = 15*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 15.
a(n) = ((15+sqrt(229))/2)^n + ((15-sqrt(229))/2)^n.
(a(n))^2 = a(2n) - 2 if n=1, 3, 5...
(a(n))^2 = a(2n) + 2 if n=2, 4, 6...
G.f.: (2-15*x)/(1-15*x-x^2). - Philippe Deléham, Nov 02 2008
Contribution from Johannes W. Meijer, Jun 12 2010: (Start)
Lim_{k-> infinity} a(n+k)/a(k) = (A090301(n) + A154597(n)*sqrt(229))/2.
Lim_{n-> infinity} A090301(n)/ A154597(n) = sqrt(229).
a(2n+1) = 15*A098246(n).
a(3n+1) = A041426(5n), a(3n+2) = A041426(5n+3), a(3n+3) = 2*A041426(5n+4).
(End)
a(n) = Lucas(n, 15) = 2*(-i)^n * ChebyshevT(n, 15*i/2). - G. C. Greubel, Dec 31 2019
E.g.f.: 2*exp(15*x/2)*cosh(sqrt(229)*x/2). - Stefano Spezia, Jan 01 2020

More terms from Ray Chandler, Feb 14 2004

A090306 a(n) = 17*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 17.

Original entry on oeis.org

2, 17, 291, 4964, 84679, 1444507, 24641298, 420346573, 7170533039, 122319408236, 2086600473051, 35594527450103, 607193567124802, 10357885168571737, 176691241432844331, 3014108989526925364, 51416544063390575519

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 25 2004

Lim_{n-> infinity} a(n)/a(n+1) = 0.058621... = 2/(17+sqrt(293)) = (sqrt(293)-17)/2.
Lim_{n-> infinity} a(n+1)/a(n) = 17.058621... = (17+sqrt(293))/2 = 2/(sqrt(293)-17).
For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010

			a(4) = 17*a(3) + a(2) = 17*4964 + 291=((17+sqrt(293))/2)^4 + ((17-sqrt(293))/2)^4 = 84678.999988190 + 0.000011809 = 84679.

G. C. Greubel, Table of n, a(n) for n = 0..500
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 (17,1).

Cf. A005074.

Lucas polynomials Lucas(n,m): A000032 (m=1), A002203 (m=2), A006497 (m=3), A014448 (m=4), A087130 (m=5), A085447 (m=6), A086902 (m=7), A086594 (m=8), A087798 (m=9), A086927 (m=10), A001946 (m=11), A086928 (m=12), A088316 (m=13), A090300 (m=14), A090301 (m=15), A090305 (m=16), this sequence (m=17), A090307 (m=18), A090308 (m=19), A090309 (m=20), A090310 (m=21), A090313 (m=22), A090314 (m=23), A090316 (m=24), A330767 (m=25).

m:=17;; a:=[2,m];; for n in [3..20] do a[n]:=m*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Dec 30 2019

m:=17; I:=[2,m]; [n le 2 select I[n] else m*Self(n-1) +Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 30 2019

seq(simplify(2*(-I)^n*ChebyshevT(n, 17*I/2)), n = 0..20); # G. C. Greubel, Dec 30 2019

LinearRecurrence[{17,1},{2,17},30] (* Harvey P. Dale, Jan 24 2018 *)
LucasL[Range[20]-1, 17] (* G. C. Greubel, Dec 30 2019 *)

vector(21, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 17*I/2) ) \\ G. C. Greubel, Dec 30 2019

[2*(-I)^n*chebyshev_T(n, 17*I/2) for n in (0..20)] # G. C. Greubel, Dec 30 2019

a(n) = 17*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 17.
a(n) = ((17+sqrt(293))/2)^n + ((17-sqrt(293))/2)^n.
(a(n))^2 = a(2n) - 2 if n=1, 3, 5, ...
(a(n))^2 = a(2n) + 2 if n=2, 4, 6, ...
G.f.: (2-17*x)/(1-17*x-x^2). - Philippe Deléham, Nov 02 2008
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 17*A098249(n).
a(3n+1) = A041550(5n), a(3n+2) = A041550(5n+3), a(3n+3) = 2*A041550(5n+4).
Lim_{k-> infinity} a(n+k)/a(k) = (A090306(n) + A178765(n)*sqrt(293))/2.
Lim_{n-> infinity} A090306(n)/A178765(n) = sqrt(293). (End)
a(n) = Lucas(n, 17) = 2*(-i)^n * ChebyshevT(n, 17*i/2). - G. C. Greubel, Dec 30 2019
E.g.f.: 2*exp(17*x/2)*cosh(sqrt(293)*x/2). - Stefano Spezia, Dec 31 2019

More terms from Ray Chandler, Feb 14 2004

A041018 Numerators of continued fraction convergents to sqrt(13).

Original entry on oeis.org

3, 4, 7, 11, 18, 119, 137, 256, 393, 649, 4287, 4936, 9223, 14159, 23382, 154451, 177833, 332284, 510117, 842401, 5564523, 6406924, 11971447, 18378371, 30349818, 200477279, 230827097, 431304376, 662131473

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,36,0,0,0,0,1).

Cf. A010122 (continued fraction for sqrt(13)).

Cf. A010470, A041019 (denominators), A041046, A041090, A041150, A041226, A041318, A041426 and A041550.

a[0]:=3: a[-1]:=1: b(0):=6: b(1):=1; b(2):=1: b(3):=1: b(4):=1:
for n from 1 to 100 do  k:=n mod 5:
   a[n]:=b(k)*a[n-1]+a[n-2]:
   printf("%12d", a[n]):
end do: # Paul Weisenhorn, Aug 17 2018

Numerator[Convergents[Sqrt[13], 30]] (* Vincenzo Librandi, Oct 27 2013 *)
CoefficientList[Series[(3 + 4*x + 7*x^2 + 11*x^3 + 18*x^4 + 11*x^5 - 7*x^6 + 4*x^7 - 3*x^8 + x^9)/(1 - 36*x^5 - x^10),{x,0,50}],x] (* Stefano Spezia, Aug 31 2018 *)

From Johannes W. Meijer, Jun 12 2010: (Start)
a(5*n) = A006497(3*n+1),
a(5*n+1) = (A006497(3*n+2)-A006497(3*n+1))/2,
a(5*n+2) = (A006497(3*n+2)+A006497(3*n+1))/2,
a(5*n+3) = A006497(3*n+2),
a(5*n+4) = A006497(3*n+3)/2.
(End)
G.f.: (3 + 4*x + 7*x^2 + 11*x^3 + 18*x^4 + 11*x^5 - 7*x^6 + 4*x^7 - 3*x^8 + x^9)/(1 - 36*x^5 - x^10). - Peter J. C. Moses, Jul 29 2013
a(n) = A010122(n)*a(n-1)+a(n-2) with a(0)=3, a(-1)=1. - Paul Weisenhorn, Aug 19 2018

A090309 a(n) = 20*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 20.

Original entry on oeis.org

2, 20, 402, 8060, 161602, 3240100, 64963602, 1302512140, 26115206402, 523606640180, 10498248010002, 210488566840220, 4220269584814402, 84615880263128260, 1696537874847379602, 34015373377210720300

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 25 2004

Lim_{n-> infinity} a(n)/a(n+1) = 0.0498756... = 1/(10+sqrt(101)) = (sqrt(101)-10).

Lim_{n-> infinity} a(n+1)/a(n) = 20.0498756... = (10+sqrt(101)) = 1/(sqrt(101)-10).

			a(4) = 20*a(3) + a(2) = 20*8060 + 402 = (10+sqrt(101))^4 + (10-sqrt(101))^4 = 161601.999993811 + 0.000006188 = 161602.

G. C. Greubel, Table of n, a(n) for n = 0..500
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 (20,1).

Cf. A002116.

Lucas polynomials Lucas(n,m): A000032 (m=1), A002203 (m=2), A006497 (m=3), A014448 (m=4), A087130 (m=5), A085447 (m=6), A086902 (m=7), A086594 (m=8), A087798 (m=9), A086927 (m=10), A001946 (m=11), A086928 (m=12), A088316 (m=13), A090300 (m=14), A090301 (m=15), A090305 (m=16), A090306 (m=17), A090307 (m=18), A090308 (m=19), this sequence (m=20), A090310 (m=21), A090313 (m=22), A090314 (m=23), A090316 (m=24), A330767 (m=25).

m:=20;; a:=[2,m];; for n in [3..20] do a[n]:=m*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Dec 30 2019

m:=20; I:=[2,m]; [n le 2 select I[n] else m*Self(n-1) +Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 30 2019

seq(simplify(2*(-I)^n*ChebyshevT(n, 10*I)), n = 0..20); # G. C. Greubel, Dec 30 2019

LinearRecurrence[{20,1},{2,20},20] (* Harvey P. Dale, Nov 19 2015 *)
LucasL[Range[20]-1,20] (* G. C. Greubel, Dec 30 2019 *)

vector(21, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 10*I) ) \\ G. C. Greubel, Dec 30 2019

[2*(-I)^n*chebyshev_T(n, 10*I) for n in (0..20)] # G. C. Greubel, Dec 30 2019

a(n) = 20*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 20.
a(n) = (10 + sqrt(101))^n + (10 - sqrt(101))^n.
(a(n))^2 = a(2n) - 2 if n=1, 3, 5, ... .
(a(n))^2 = a(2n) + 2 if n=2, 4, 6, ... .
G.f.: (2-20*x)/(1-20*x-x^2). - Philippe Deléham, Nov 02 2008
a(n) = Lucas(n, 20) = 2*(-i)^n * ChebyshevT(n, 10*i). - G. C. Greubel, Dec 30 2019

More terms from Ray Chandler, Feb 14 2004

cp's OEIS Frontend

A014176 Decimal expansion of the silver mean, 1+sqrt(2).

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A086902 a(n) = 7*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 7.

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A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

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A098316 Decimal expansion of [3, 3, ...] = (3 + sqrt(13))/2.

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A114525 Triangle of coefficients of the Lucas (w-)polynomials.

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A052924 Expansion of g.f.: (1-x)/(1 - 3*x - x^2).

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