A106853 -id:A106853 - cp's OEIS frontend

A272931 a(n) = 2^(n+1)cos(narctan(sqrt(15))).

2, 1, -7, -11, 17, 61, -7, -251, -223, 781, 1673, -1451, -8143, -2339, 30233, 39589, -81343, -239699, 85673, 1044469, 701777, -3476099, -6283207, 7621189, 32754017, 2269261, -128746807, -137823851, 377163377, 928458781, -580194727, -4294029851, -1973250943

Offset: 0

Peter Luschny, May 11 2016

For n >= 1, |a(n)| is the unique odd positive solution y to 4^(n+1) = 15*x^2 + y^2. The value of x is |A106853(n-1)|. - Jianing Song, Jan 22 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (1,-4).

Cf. A087204, A002249, A106853.

seq(simplify(((1-I*sqrt(15))^n + (1+I*sqrt(15))^n)/2^n), n=0..32);

Mathematica
```
LinearRecurrence[{1, -4}, {2, 1}, 33]
```

Vec((2 - x) / (1 - x + 4*x^2) + O(x^40)) \\ Colin Barker, Jan 22 2019

[lucas_number2(i, 1, 4) for i in range(33)]

Let a(x) = x/2 - i*sqrt(15)*x/2 and b(x) = x/2 + i*sqrt(15)*x/2, then:
a(n) = a(1)^n + b(1)^n.
a(n) = n! [x^n] exp(a(x)) + exp(b(x)).
a(n) = [x^n] (2 - x)/(4*x^2 - x + 1).
a(n) = Sum_{k=0..floor(n/2)} (-4)^k*n*(n - k - 1)!/(k!*(n - 2*k)!) for n >= 1.
For n >= 1, 15*a(n)^2 + A106853(n-1)^2 = 4^(n+1). - Jianing Song, Jan 22 2019
a(n) = a(n-1) - 4*a(n-2) for n>1. - Colin Barker, Jan 22 2019
a(n) = 2*A106853(n) - A106853(n-1). - R. J. Mathar, Aug 19 2022

A128100 Triangle read by rows: T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 5, 1, 8, 10, 3, 13, 20, 9, 1, 21, 38, 22, 4, 34, 71, 51, 14, 1, 55, 130, 111, 40, 5, 89, 235, 233, 105, 20, 1, 144, 420, 474, 256, 65, 6, 233, 744, 942, 594, 190, 27, 1, 377, 1308, 1836, 1324, 511, 98, 7, 610, 2285, 3522, 2860, 1295, 315, 35, 1, 987, 3970

Offset: 0

Emeric Deutsch, Feb 18 2007

Row sums are the Jacobsthal numbers (A001045). Column 0 yields the Fibonacci numbers (A000045); the other columns yield convolved Fibonacci numbers (A001629, A001628, A001872, A001873, etc.). Sum_{k=0..floor(n/2)} k*T(n,k) = A073371(n-2).
Triangle T(n,k), with zeros omitted, given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 24 2012
Riordan array (1/(1-x-x^2), x^2/(1-x-x^2)), with zeros omitted. - Philippe Deléham, Feb 06 2012
Diagonal sums are A000073(n+2) (tribonacci numbers). - Philippe Deléham, Feb 16 2014
Number of induced subgraphs of the Fibonacci cube Gamma(n-1) that are isomorphic to the hypercube Q_k. Example: row n=4 is 5, 5, 1; indeed, the Fibonacci cube Gamma(3) is a square with an additional pendant edge attached to one of its vertices; it has 5 vertices (i.e., Q_0's), 5 edges (i.e., Q_1's) and 1 square (i.e., Q_2). - Emeric Deutsch, Aug 12 2014
Row n gives the coefficients of the polynomial p(n,x) defined as the numerator of the rational function given by f(n,x) = 1 + (x + 1)/f(n-1,x), where f(x,0) = 1. Conjecture: for n > 2, p(n,x) is irreducible if and only if n is a (prime - 2). - Clark Kimberling, Oct 22 2014

			Triangle starts:
   1;
   1;
   2,  1;
   3,  2;
   5,  5,  1;
   8, 10,  3;
  13, 20,  9,  1;
  21, 38, 22,  4;
From _Philippe Deléham_, Jan 24 2012: (Start)
Triangle (1, 1, -1, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, ...) begins:
   1;
   1,  0;
   2,  1,  0;
   3,  2,  0,  0;
   5,  5,  1,  0,  0;
   8, 10,  3,  0,  0,  0;
  13, 20,  9,  1,  0,  0,  0;
  21, 38, 22,  4,  0,  0,  0,  0; (End)
From _Clark Kimberling_, Oct 22 2014: (Start)
Here are the first 4 polynomials p(n,x) as in Comment and generated by Mathematica program:
  1
  2 +  x
  3 + 2x
  5 + 5x + x^2. (End)

C.-P. Chou and H. A. Witek, An Algorithm and FORTRAN Program for Automatic Computation of the Zhang-Zhang Polynomial of Benzenoids, MATCH: Commun. Math. Comput. Chem, 68 (2012) 3-30. See Eq. (9). - From _N. J. A. Sloane_, Dec 23 2012
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, preprint.
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, Acta Appl. Math. 117, 2012, 93-105. - _Emeric Deutsch_, Aug 12 2014

Cf. A001045, A000045, A001629, A001628, A001872, A001873, A073371.

Cf. A109466, A119473.

G:=1/(1-z-(1+t)*z^2): Gser:=simplify(series(G,z=0,19)): for n from 0 to 16 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 16 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form

p[x_, n_] := 1 + (x + 1)/p[x, n - 1]; p[x_, 1] = 1;
Numerator[Table[Factor[p[x, n]], {n, 1, 20}]]  (* Clark Kimberling, Oct 22 2014 *)

G.f.: 1/(1-z-(1+t)z^2).
Sum_{k=0..n} T(n,k)*x^k = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, and -13, respectively. - Philippe Deléham, Jan 24 2012
T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - Philippe Deléham, Jan 24 2012
G.f.: T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x/((2*k+2+ x*(1+y))*x + 1/T(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 06 2013
T(n,k) = Sum_{i=k..floor(n/2)} binomial(n-i,i)*binomial(i,k). See Corollary 3.3 in the Klavzar et al. link. - Emeric Deutsch, Aug 12 2014

A133631 a(n) = a(n-1) - 4*a(n-2), a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, -2, -10, -2, 38, 46, -106, -290, 134, 1294, 758, -4418, -7450, 10222, 40022, -866, -160954, -157490, 486326, 1116286, -829018, -5294162, -1978090, 19198558, 27110918, -49683314, -158126986, 40606270, 673114214, 510689134, -2181767722, -4224524258

Offset: 0

Philippe Deléham, Dec 28 2007

			G.f. = 1 + 2*x - 2*x^2 - 10*x^3 - 2*x^4 + 38*x^5 + 46*x^6 - 106*x^7 + ... - _Michael Somos_, Oct 24 2023

Index entries for linear recurrences with constant coefficients, signature (1,-4).

Cf. A106853, A133607.

a[ n_] := 2^n * ChebyshevU[n, 1/4] + 2^(n-1) * ChebyshevU[n-1, 1/4]; (* Michael Somos, Oct 24 2023 *)
LinearRecurrence[{1,-4},{1,2},50] (* Harvey P. Dale, Feb 01 2025 *)

{a(n) = 2^n*polchebyshev(n, 2, 1/4) + 2^(n-1)*polchebyshev(n-1, 2, 1/4)}; /* Michael Somos, Oct 24 2023 */

G.f.: (1+x)/(1-x+4*x^2).
a(n) = Sum_{k=0..n} A133607(n,k)*2^k. - Philippe Deléham, Dec 29 2007
a(n) = 2^n*U(n, 1/4) + 2^(n-1)*U(n-1, 1/4) = A106853(n) + A106853(n-1) where U is the Chebyshev polynomial of the 2nd kind. - Michael Somos, Oct 24 2023

A156857 Expansion of (1+2x)/(1+x+4x^2)^2.

Original entry on oeis.org

1, 0, -9, 10, 45, -108, -125, 702, -135, -3320, 4239, 11250, -31931, -18180, 165915, -92762, -651375, 1101168, 1747495, -6710310, -694179, 30182500, -28394829, -101934450, 229069225, 203510232, -1198850625, 364506562, 4767453045

Offset: 0

Paul Barry, Feb 17 2009

Hankel transform of A091526.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-2,-9,-8,-16).

Cf. A091526, A106853.

I:=[1,0,-9,10]; [n le 4 select I[n] else (-1)*(2*Self(n-1) +9*Self(n-2) +8*Self(n-3) +16*Self(n-4)): n in [1..41]]; // G. C. Greubel, Jan 28 2022

LinearRecurrence[{-2,-9,-8,-16}, {1,0,-9,10}, 41] (* or *)
A156857[n_]:= (-2)^n*Sum[ChebyshevU[n-j, 1/4]*(ChebyshevU[j, 1/4] - ChebyshevU[j-1, 1/4]), {j,0,n}];
Table[A156857[n], {n, 0, 40}] (* G. C. Greubel, Jan 28 2022 *)

def A156857(n): return (-2)^n*sum( chebyshev_U(n-j, 1/4)*(chebyshev_U(j, 1/4) - chebyshev_U(j-1, 1/4)) for j in (0..n))
[A156857(n) for n in (0..40)] # G. C. Greubel, Jan 28 2022

G.f.: (1 +2*x)/(1 +2*x +9*x^2 +8*x^3 +16*x^4).

a(n) = (-2)*Sum_{j=0..n} ChebyshevU(n-j, 1/4)*(ChebyshevU(j, 1/4) - ChebyshevU(j-1, 1/4)). - G. C. Greubel, Jan 28 2022

A231114 Numbers k dividing u(k), where the Lucas sequence is defined u(i) = u(i-1) - 4*u(i-2) with initial conditions u(0)=0, u(1)=1.

Original entry on oeis.org

1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 125, 135, 171, 225, 243, 375, 405, 435, 465, 513, 625, 675, 729, 855, 1125, 1215, 1305, 1395, 1539, 1875, 2025, 2175, 2187, 2325, 2565, 3125, 3249, 3375, 3645, 3725, 3915, 4005, 4185, 4275, 4617, 5625, 6075, 6327, 6525, 6561

Offset: 1

Thomas M. Bridge, Nov 06 2013

nonn

Every term (except leading term) is divisible by at least one of 3 or 5.

Furthermore, this sequence contains 3^i*5^j for all i, j >= 0, that is, A003593 is a subsequence.

			The sequence u(i) begins 0, 1, 1, -3, -7, 5, 33. Only for k = 1, 3, 5 does k divides u(k).

Amiram Eldar, Table of n, a(n) for n = 1..1000
C. Smyth, The terms in Lucas sequences divisible by their indices, Journal of Integer Sequences, Vol.13 (2010), Article 10.2.4.
Wikipedia, Lucas sequence

Cf. A003593 (subsequence), A106853 (Lucas sequence).

nn = 10000; s = LinearRecurrence[{1, -4}, {1, 1}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* T. D. Noe, Nov 06 2013 *)

cp's OEIS Frontend

A272931 a(n) = 2^(n+1)*cos(n*arctan(sqrt(15))).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Sage

Formula

A128100 Triangle read by rows: T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A133631 a(n) = a(n-1) - 4*a(n-2), a(0)=1, a(1)=2.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A156857 Expansion of (1+2*x)/(1+x+4*x^2)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Sage

Formula

A231114 Numbers k dividing u(k), where the Lucas sequence is defined u(i) = u(i-1) - 4*u(i-2) with initial conditions u(0)=0, u(1)=1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A272931 a(n) = 2^(n+1)cos(narctan(sqrt(15))).

A156857 Expansion of (1+2x)/(1+x+4x^2)^2.