A109115 -id:A109115 - cp's OEIS frontend

A108300 a(n+2) = 3*a(n+1) + a(n), with a(0) = 1, a(1) = 5.

1, 5, 16, 53, 175, 578, 1909, 6305, 20824, 68777, 227155, 750242, 2477881, 8183885, 27029536, 89272493, 294847015, 973813538, 3216287629, 10622676425, 35084316904, 115875627137, 382711198315, 1264009222082, 4174738864561, 13788225815765, 45539416311856

Offset: 0

Creighton Dement, Jul 24 2005

Binomial transform is A109114.
Invert transform is A109115.
Inverse invert transform is A016777.
Inverse binomial transform is A006130.

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Sergio Falcon, The k-Fibonacci difference sequences, Chaos, Solitons & Fractals, Volume 87, June 2016, Pages 153-157.
Tanya Khovanova, Recursive Sequences
Vincent Vatter, Growth rates of permutation classes: from countable to uncountable, arXiv:1605.04297 [math.CO], 2016. (Mentions a signed version.)
Index entries for linear recurrences with constant coefficients, signature (3,1).

Row sums and main diagonal of A143972. - Gary W. Adamson, Sep 06 2008

Cf. A109114, A109115, A016777, A006130, A000004, A052924, A228916, A374439.

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

LinearRecurrence[{3,1},{1,5},40] (* Harvey P. Dale, Jul 04 2013 *)

Vec((1 + 2*x)/(1 - 3*x - x^2) + O(x^30)) \\ Andrew Howroyd, Jun 05 2021

G.f.: (1 + 2*x)/(1 - 3*x - x^2).
a(n) = A052924(n+1) - A052924(n).
a(n)*a(n-2) = a(n-1)^2 + 9*(-1)^n. - Roger L. Bagula, May 17 2010
a(n) = 3^n*Sum_{k=0..n} A374439(n, k)*(1/3)^k. - Peter Luschny, Jul 26 2024

A154245 a(n) = ( (4 + sqrt(7))^n - (4 - sqrt(7))^n )/(2*sqrt(7)).

Original entry on oeis.org

1, 8, 55, 368, 2449, 16280, 108199, 719072, 4778785, 31758632, 211059991, 1402652240, 9321678001, 61949553848, 411701328775, 2736064645568, 18183205205569, 120841059834440, 803079631825399, 5337067516093232

Offset: 1

Al Hakanson (hawkuu(AT)gmail.com), Jan 05 2009

Second binomial transform of A109115.

Lim_{n -> infinity} a(n)/a(n-1) = 4 + sqrt(7) = 6.6457513110....

Vincenzo Librandi, Table of n, a(n) for n = 1..100
Index entries for linear recurrences with constant coefficients, signature (8,-9).

Equals (A094432 without initial term 0)/3.

Cf. A010465 (decimal expansion of square root of 7), A109115.

a:=[1,8];; for n in [3..30] do a[n]:=8*a[n-1]-9*a[n-2]; od; a; # G. C. Greubel, May 21 2019

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-7); S:=[ ((4+r)^n-(4-r)^n)/(2*r): n in [1..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jan 07 2009

I:=[1,8]; [n le 2 select I[n] else 8*Self(n-1)-9*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 08 2016

Table[Simplify[((4+Sqrt[7])^n -(4-Sqrt[7])^n)/(2*Sqrt[7])], {n, 30}] (* or *) LinearRecurrence[{8, -9},{1, 8}, 30] (* G. C. Greubel, Sep 07 2016 *)
Rest@ CoefficientList[Series[x/(1 -8x +9x^2), {x, 0, 30}], x] (* Michael De Vlieger, Sep 08 2016 *)

my(x='x+O('x^30)); Vec( x/(1-8*x+9*x^2) ) \\ G. C. Greubel, May 21 2019

[lucas_number1(n,8,9) for n in range(1, 21)] # Zerinvary Lajos, Apr 23 2009

From Philippe Deléham, Jan 06 2009: (Start)
a(n) = 8*a(n-1) - 9*a(n-2) for n > 1, with a(0)=0, a(1)=1.
G.f.: x/(1 - 8*x + 9*x^2). (End)
a(n) = b such that (3^(n-1)/2)*Integral_{x=0..Pi/2} (sin(n*x))/(4/3-cos(x)) dx = c + b*log(2). - Francesco Daddi, Aug 02 2011
E.g.f.: (1/sqrt(7))*exp(4*x)*sinh(sqrt(7)*x). - G. C. Greubel, Sep 07 2016

Extended beyond a(7) by Klaus Brockhaus, Jan 07 2009

Edited by Klaus Brockhaus, Oct 06 2009

A208340 Triangle of coefficients of polynomials v(n,x) jointly generated with A202390; see the Formula section.

Original entry on oeis.org

1, 2, 2, 3, 6, 3, 4, 13, 14, 5, 5, 24, 41, 30, 8, 6, 40, 96, 109, 60, 13, 7, 62, 196, 308, 262, 116, 21, 8, 91, 364, 743, 868, 590, 218, 34, 9, 128, 630, 1604, 2413, 2240, 1267, 402, 55, 10, 174, 1032, 3186, 5926, 7046, 5424, 2627, 730, 89, 11, 230, 1617

Offset: 1

Clark Kimberling, Feb 27 2012

v(n,n) = F(n+1), where F=A000045, the Fibonacci numbers.
Alternating row sums of v: (1,0,0,0,0,0,0,0,...).
As triangle T(n,k) with 0 <= k <= n, it is (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

			First five rows:
  1;
  2,  2;
  3,  6,  3;
  4, 13, 14,  5;
  5, 24, 41, 30,  8;
The first five polynomials v(n,x):
  1
  2 +  2x
  3 +  6x +  3x^2
  4 + 13x + 14x^2 +  5x^3
  5 + 24x + 41x^2 + 30x^3 + 8x^4

Cf. A202390.

u[1, x_] := 1; v[1, x_] := 1; z = 13;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1)*v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A202390 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A208340 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* u row sums *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* v row sums *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* u alt. row sums *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* v alt. row sums *)

u(n,x) = u(n-1,x) + x*v(n-1,x), v(n,x) = (x+1)*u(n-1,x) + (x+1)*v(n-1,x), where u(1,x)=1, v(1,x)=1.
From Philippe Deléham, Feb 28 2012: (Start)
As triangle T(n,k) with 0 <= k <= n:
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) + T(n-2,k-2) with T(0,0) = 1, T(1,0) = T(1,1) = 2 and T(n,k) = 0 if k < 0 or if k > n.
G.f.: (1+y*x)/(1-2*x-y*x+x^2-y^2*x^2).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000027(n+1), A003946(n), A109115(n), A180031(n) for x = -1, 0, 1, 2, 3 respectively. (End)

cp's OEIS Frontend

A108300 a(n+2) = 3*a(n+1) + a(n), with a(0) = 1, a(1) = 5.

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A154245 a(n) = ( (4 + sqrt(7))^n - (4 - sqrt(7))^n )/(2*sqrt(7)).

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A208340 Triangle of coefficients of polynomials v(n,x) jointly generated with A202390; see the Formula section.

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