A055271 -id:A055271 - cp's OEIS frontend

A108306 Expansion of (3x+1)/(1-3x-3*x^2).

1, 6, 21, 81, 306, 1161, 4401, 16686, 63261, 239841, 909306, 3447441, 13070241, 49553046, 187869861, 712268721, 2700415746, 10238053401, 38815407441, 147160382526, 557927369901, 2115263257281, 8019571881546, 30404505416481, 115272231894081

Offset: 0

Creighton Dement, Jul 24 2005

Binomial transform is A055271. May be seen as a ibasefor-transform of the zero-sequence A000004 with respect to the floretion given in the program code.
The sequence is the INVERT transform of (1, 5, 10, 20, 40, 80, 160, ...) and can be obtained by extracting the upper left terms of matrix powers of [(1,5); (1,2)]. These results are a case (a=5, b=2) of the conjecture:  The INVERT transform of a sequence starting (1, a, a*b, a*b^2, a*b^3, ...) is equivalent to extracting the upper left terms of powers of the  2x2 matrix [(1,a); (1,b)]. - Gary W. Adamson, Jul 31 2016
From Klaus Purath, Mar 09 2023: (Start)
For any terms (a(n), a(n+1)) = (x, y), -3*x^2 - 3*x*y + y^2 = 15*(-3)^n = A082111(2)*(-3)^n. This is valid in general for all recursive sequences (t) with constant coefficients (3,3) and t(0) = 1: -3*x^2 - 3*x*y + y^2 = A082111(t(1)-4)*(-3)^n.
By analogy to this, for three consecutive terms (x, y, z) of any sequence (t) of the form (3,3) with t(0) = 1: y^2 - x*z = A082111(t(1)-4)*(-3)^n. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,3).

Cf. A055271.

Cf. A084057.

I:=[1,6]; [n le 2 select I[n] else 3*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 01 2016

seriestolist(series((3*x+1)/(1-3*x-3*x^2), x=0,25));

CoefficientList[Series[(3 x + 1) / (1 - 3 x - 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 01 2016 *)

Recurrence: a(0)=1; a(1)=6; a(n) = 3a(n-1) + 3a(n-2) - N-E. Fahssi, Apr 20 2008

A207607 Triangle of coefficients of polynomials v(n,x) jointly generated with A207606; see Formula section.

Original entry on oeis.org

1, 1, 2, 1, 5, 2, 1, 9, 9, 2, 1, 14, 25, 13, 2, 1, 20, 55, 49, 17, 2, 1, 27, 105, 140, 81, 21, 2, 1, 35, 182, 336, 285, 121, 25, 2, 1, 44, 294, 714, 825, 506, 169, 29, 2, 1, 54, 450, 1386, 2079, 1716, 819, 225, 33, 2, 1, 65, 660, 2508, 4719, 5005, 3185, 1240, 289, 37, 2

Offset: 1

Clark Kimberling, Feb 19 2012

Subtriangle of the triangle T(n,k) given by (1, 0, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 03 2012

			First five rows:
  1;
  1,  2;
  1,  5,  2;
  1,  9,  9,  2;
  1, 14, 25, 13,  2;
Triangle (1, 0, 1/2, 1/2, 0, 0, 0, ...) DELTA (0, 2, -1, 0, 0, 0, 0, ...) begins:
  1;
  1,  0;
  1,  2,  0;
  1,  5,  2,  0;
  1,  9,  9,  2,  0;
  1, 14, 25, 13,  2,  0;
  1, 20, 55, 49, 17,  2,  0;
  ...
1 = 2*1 - 1, 20 = 2*14 + 1 - 9, 55 = 2*25 + 14 - 9, 49 = 2*13 + 25 - 2, 17 = 2*2 + 1 - 0, 2 = 2*0 + 2 - 0. - _Philippe Deléham_, Mar 03 2012

G. C. Greubel, Rows n = 1..100 of the triangle, flattened

Cf. A207606.

A207607:= (n,k) -> `if`(k=1, 1, binomial(n+k-3, 2*k-2) + 2*binomial(n+k-3, 2*k-3) ); seq(seq(A207607(n, k), k = 1..n), n = 1..10); # G. C. Greubel, Mar 15 2020

(* First program *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A207606 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A207607 *)
(* Second program *)
Table[If[k==1, 1, Binomial[n+k-3, 2*k-2] + 2*Binomial[n+k-3, 2*k-3]], {n, 10}, {k, n}]//Flatten (* G. C. Greubel, Mar 15 2020 *)

from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else x*u(n - 1, x) + (x + 1)*v(n - 1, x)
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

def T(n, k):
    if k == 1: return 1
    else: return binomial(n+k-3, 2*k-2) + 2*binomial(n+k-3, 2*k-3)
[[T(n, k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Mar 15 2020

u(n,x) = u(n-1,x) + v(n-1,x), v(n,x) = x*u(n-1,x) + (x+1)v(n-1,x), where u(1,x)=1, v(1,x)=1.
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Mar 03 2012
G.f.: (1-x+y*x)/(1-(y+2)*x+x^2). - Philippe Deléham, Mar 03 2012
For n >= 1, Sum{k=0..n} T(n,k)*x^k = A000012(n), A001906(n), A001834(n-1), A055271(n-1), A038761(n-1), A056914(n-1) for x = 0, 1, 2, 3, 4, 5 respectively. - Philippe Deléham, Mar 03 2012
T(n,k) = C(n+k-1,2*k) + 2*C(n+k-1,2*k-1). where C is binomial. - Yuchun Ji, May 23 2019
T(n,k) = T(n-1,k) + A207606(n,k-1). - Yuchun Ji, May 28 2019
Sum_{k=1..n} T(n, k)*x^k = { 4*(-1)^(n-1)*A016921(n-1) (x=-4), 3*(-1)^(n-1) * A130815(n-1) (x=-3), 2*(-1)^(n-1)*A010684(n-1) (x=-2), A057079(n+1) (x=-1), 0 (x=0), A001906(n) = Fibonacci(2*n) (x=1), 2*A001834(n-1) (x=2), 3*A055271(n-1) (x=3), 4*A038761(n-1) (x=4) }. - G. C. Greubel, Mar 15 2020

A107402 a(n)= -a(n-1) +5a(n-2) +5a(n-3) -a(n-4) -a(n-5).

Original entry on oeis.org

0, 1, 1, 2, 3, 11, 12, 55, 55, 266, 261, 1277, 1248, 6121, 5977, 29330, 28635, 140531, 137196, 673327, 657343, 3226106, 3149517, 15457205, 15090240, 74059921, 72301681, 354842402, 346418163, 1700152091, 1659789132, 8145918055

Offset: 0

Roger L. Bagula, May 25 2005

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

m = 5 M = {{0, 1, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}, {0, 0, 0, 0, 1}, {-1, -1, m, m, -1}} Expand[Det[M - x*IdentityMatrix[5]]] NSolve[Det[M - x*IdentityMatrix[5]] == 0, x] v[1] = {0, 1, 1, 2, 3} digits = 50 v[n_] := v[n] = M.v[n - 1] a = Table[v[n][[1]], {n, 1, digits}]
LinearRecurrence[{-1,5,5,-1,-1},{0,1,1,2,3},40] (* Harvey P. Dale, Sep 23 2012 *)

G.f.: -x*(5*x^3+2*x^2-2*x-1)/((x+1)*(x^4-5*x^2+1)). [From Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009]

a(0)=0, a(2n) = (1/3)*(A055271(n)+2).

Definition replaced by recurrence by the Associate Editors of the OEIS, Sep 28 2009

cp's OEIS Frontend

A108306 Expansion of (3x+1)/(1-3x-3*x^2).

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A207607 Triangle of coefficients of polynomials v(n,x) jointly generated with A207606; see Formula section.

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A107402 a(n)= -a(n-1) +5a(n-2) +5a(n-3) -a(n-4) -a(n-5).

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cp's OEIS Frontend

A108306 Expansion of (3*x+1)/(1-3*x-3*x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A207607 Triangle of coefficients of polynomials v(n,x) jointly generated with A207606; see Formula section.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Sage

Formula

A107402 a(n)= -a(n-1) +5*a(n-2) +5*a(n-3) -a(n-4) -a(n-5).

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Extensions

A108306 Expansion of (3x+1)/(1-3x-3*x^2).

A107402 a(n)= -a(n-1) +5a(n-2) +5a(n-3) -a(n-4) -a(n-5).