A073382 Fifth convolution of A000129(n+1) (generalized (2,1)-Fibonacci, called Pell numbers), n>=0, with itself.
1, 12, 90, 532, 2709, 12432, 52808, 211248, 805374, 2951576, 10465476, 36079848, 121412942, 400054032, 1293997416, 4117416496, 12910964103, 39956039172, 122193599822, 369685154076, 1107503284923, 3288114790112
Offset: 0
Examples
x^3 + 12*x^4 + 90*x^5 + 532*x^6 + 2709*x^7 + 12432*x^8 + 52808*x^9 + 211248*x^10 + 805374*x^11 +......+ 122193599822*x^21 + 369685154076*x^22 + 1107503284923*x^23 + 3288114790112*x^24 + etc. [_Zerinvary Lajos_, Jun 01 2009]
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (12,-54,100,-15,-168,76,168,-15,-100,-54,-12,-1).
Programs
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Magma
R
:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/(1-2*x-x^2)^6 )); // G. C. Greubel, Oct 02 2022 -
Mathematica
CoefficientList[Series[1/(1-2*x-x^2)^6, {x,0,40}], x] (* G. C. Greubel, Oct 02 2022 *) -
Sage
taylor( 1/(1-2*x-x^2)^6, x, 0, 24).list() # G. C. Greubel, Oct 02 2022
Formula
a(n) = Sum_{k=0..floor(n/2)} 2^(n-2*k) * binomial(n-k+5, 5) * binomial(n-k, k).
a(n) = ((50085 +53006*n +19594*n^2 +3016*n^3 +164*n^4)*(n+1)*U(n+1) +(11355 +16336*n +7042*n^2 +1184*n^3 +68*n^4)*(n+2)*U(n))/(3*5*2^13), with U(n) = A000129(n+1), n >= 0.
G.f.: 1/(1-(2+x)*x)^6.
a(n) = F'''''(n+6, 2)/5!, that is, 1/5! times the 5th derivative of the (n+6)th Fibonacci polynomial evaluated at x=2. - T. D. Noe, Jan 19 2006