A176738 Expansion of 1 / ((1+x)*(1-x-4*x^2)). (5,4)-Padovan sequence.
1, 0, 5, 4, 25, 40, 141, 300, 865, 2064, 5525, 13780, 35881, 91000, 234525, 598524, 1536625, 3930720, 10077221, 25800100, 66108985, 169309384, 433745325, 1110982860, 2845964161, 7289895600, 18673752245, 47833334644, 122528343625, 313861682200, 803975056701
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (0,5,4).
Crossrefs
Cf. A176737 ((4,3)-Padovan).
Programs
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Mathematica
LinearRecurrence[{0,5,4},{1,0,5},40] (* Harvey P. Dale, May 27 2016 *) f[n_] := Simplify[((-1)^(1 +n) + (2^(-1 -n)*((1 + Sqrt[17])^n*(-5 +3Sqrt[17]) + (1 -Sqrt[17])^n*(5 + 3Sqrt[17])))/Sqrt[17])/2]; Array[f, 31, 0] (* or *) CoefficientList[Series[1/(1 -5x^2 -4x^3), {x, 0, 30}], x] (* or *) RecurrenceTable[{a[n] == 5 a[n - 2] + 4 a[n - 3], a[0] == 1, a[1] == 0, a[2] == 5}, a, {n, 30}] (* Robert G. Wilson v, Dec 25 2017 *) -
PARI
Vec(1 / ((1 + x)*(1 - x - 4*x^2)) + O(x^40)) \\ Colin Barker, Dec 25 2017
Formula
O.g.f.: 1/((1-x-4*x^2)*(1+x)) = ((3-4*x)/(1-x-4*x^2) -1/(1+x))/2.
a(n) = (3*b(n) - 4*b(n-1) - (-1)^n)/2, n>=0, with b(n):=A006131(n) ((1,4)-Fibonacci), b(-1):=0.
From Colin Barker, Dec 25 2017: (Start)
a(n) = ((-1)^(1+n) + (2^(-1-n)*((1+sqrt(17))^n*(-5+3*sqrt(17)) + (1-sqrt(17))^n*(5+3*sqrt(17)))) / sqrt(17)) / 2.
a(n) = 5*a(n-2) + 4*a(n-3) for n>2.
(End)
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