A106853 Expansion of 1/(1 - x + 4*x^2).
1, 1, -3, -7, 5, 33, 13, -119, -171, 305, 989, -231, -4187, -3263, 13485, 26537, -27403, -133551, -23939, 510265, 606021, -1435039, -3859123, 1881033, 17317525, 9793393, -59476707, -98650279, 139256549, 533857665, -23168531, -2158599191, -2065925067
Offset: 0
Examples
G.f. = 1 + x - 3*x^2 - 7*x^3 + 5*x^4 + 33*x^5 + 13*x^6 - 119*x^7 - 171*x^8 + ... - _Michael Somos_, Oct 24 2023
Links
- T. D. Noe, Table of n, a(n) for n = 0..1000
- Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
- Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 15.
- Index entries for linear recurrences with constant coefficients, signature (1,-4).
Programs
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Magma
I:=[1,1]; [n le 2 select I[n] else Self(n-1) - 4*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 14 2018
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Maple
f:= gfun:-rectoproc({a(n)=a(n-1)-4*a(n-2), a(0)=1,a(1)=1},a(n),remember): map(f, [$0..100]); # Robert Israel, Jan 15 2018 -
Mathematica
Join[{a=1,b=1},Table[c=b-4*a;a=b;b=c,{n,80}]] (* Vladimir Joseph Stephan Orlovsky, Jan 22 2011 *) CoefficientList[Series[1/(1-x*(1-4x)),{x,0,40}],x] (* or *) LinearRecurrence[ {1,-4},{1,1},40] (* Harvey P. Dale, May 26 2013 *) a[ n_] := 2^n * ChebyshevU[n, 1/4]; (* Michael Somos, Oct 24 2023 *) -
PARI
x='x+O('x^30); Vec(1/(1-x+4*x^2)) \\ G. C. Greubel, Jan 14 2018 -
PARI
{a(n) = 2^n*polchebyshev(n, 2, 1/4)}; /* Michael Somos, Oct 24 2023 */ -
Sage
[lucas_number1(n,1,4) for n in range(1, 36)] # Zerinvary Lajos, Apr 22 2009
Formula
G.f.: 1/(1 - x + 4*x^2).
a(n) = 2^n*(cos(2*n*arctan(sqrt(15)/5))+sqrt(15)*sin(2*n*arctan(sqrt(15)/5))/15).
a(n) = ((1 + sqrt(-15))^(n+1) - (1 - sqrt(-15))^(n+1))/(2^(n+1)*sqrt(-15)).
a(n) = Sum_{k=0..n} ((-1)^(n-k)*binomial(k, n-k)*4^(n-k)).
a(n) = a(n-1) - 4*a(n-2), a(0) = 1, a(1) = 1. - Philippe Deléham, Oct 21 2008
a(n) = Sum_{k=0..n} A109466(n,k)*4^(n-k). - Philippe Deléham, Oct 25 2008
G.f.: 1/(1 - 2*x)^2/(1 + 3*x*G(0)/2), where G(k) = 1 + 1/(1 - x/(x + (k + 1)/(2*k + 4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
For n >= 1, 15*A272931(n)^2 + a(n-1)^2 = 4^(n+1). - Jianing Song, Jan 22 2019
a(n) = Product_{k=1..n} (1 + 4*cos(k*Pi/(n+1))). - Peter Luschny, Nov 28 2019
a(n) = 2^n * U(n, 1/4), where U(n, x) is the Chebyshev polynomial of the second kind. - Federico Provvedi, Mar 28 2022
Comments