A105963 Expansion of (1+4*x)/(1-x-3*x^2).

1, 5, 8, 23, 47, 116, 257, 605, 1376, 3191, 7319, 16892, 38849, 89525, 206072, 474647, 1092863, 2516804, 5795393, 13345805, 30731984, 70769399, 162965351, 375273548, 864169601, 1989990245, 4582499048, 10552469783, 24299966927

Offset: 0

Author

Creighton Dement, Apr 28 2005

Keywords

• nonn
• easy

Comments

Inversion of the periodic sequence with initial period (1,4,-1,-4). Sequence appears to have the property: for m > n, if s divides both a(n) and a(m) then s also divides a(2*m-n). E.g., 23 divides both a(3) = 23 and a(25) = 1989990245; 23 also divides a(2*25-3) = a(47) = 185518234185384428 = (2)^2*(23)*(131)*(15393149202239).

Floretion Algebra Multiplication Program, FAMP Code: 1jesforseq[.5'k + .5k' + 2'kk' + 2e]

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,3).

Crossrefs

Cf. A006130, A105476, A274977.

Programs

GAP

a:=[1,5];; for n in [3..40] do a[n]:=a[n-1]+3*a[n-2]; od; a; # G. C. Greubel, Jan 15 2020

Magma

I:=[ 1,5]; [n le 2 select I[n] else Self(n-1)+3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jul 20 2013

Maple

seq(coeff(series((1+4*x)/(1-x-3*x^2), x, n+1), x, n), n = 0 .. 40); # G. C. Greubel, Jan 15 2020

Mathematica

CoefficientList[Series[(1+4x)/(1-x-3x^2), {x,0,40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
Table[Round[3^((n-1)/2)*(Sqrt[3]*Fibonacci[n+1, 1/Sqrt[3]] + 4*Fibonacci[n, 1/Sqrt[3]] )], {n,0,40}] (* G. C. Greubel, Jan 15 2020 *)

PARI

Vec((1+4*x)/(1-x-3*x^2)+O(x^40)) \\ Charles R Greathouse IV, Sep 27 2012

SageMath

def A077952_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1+4*x)/(1-x-3*x^2) ).list()
A077952_list(30) # G. C. Greubel, Jan 15 2020

Formula

a(n) = A006130(n) + 4*A006130(n-1) = A006130(n+1) + A006130(n-1). - R. J. Mathar, Dec 12 2009

From Colin Barker, May 01 2019: (Start)

a(n) = (2^(-1-n)*((1-sqrt(13))^n*(-9+sqrt(13)) + (1+sqrt(13))^n*(9+sqrt(13)))) / sqrt(13).

a(n) = a(n-1) + 3*a(n-2) for n > 1. (End)

a(n) = 3^((n-1)/2)*( sqrt(3)*Fibonacci(n+1, 1/sqrt(3)) + 4*Fibonacci(n, 1/sqrt(3)) ). - G. C. Greubel, Jan 15 2020