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Sequence whose G.f is given by: 1/(1-z)/(1-2*z)^2/(1-z-z^2).
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%I #9 May 24 2021 07:56:09

%S 1,6,24,79,232,632,1633,4058,9788,23063,53332,121452,273089,607534,

%T 1339376,2929951,6366480,13752880,29556545,63232370,134731956,

%U 286044711,605326044,1277246724,2687879137,5642847462,11820387528,24710992303

%N Sequence whose G.f is given by: 1/(1-z)/(1-2*z)^2/(1-z-z^2).

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (6,-12,7,4,-4).

%F a(n+5) = 6*a(n+4)-12*a(n+3)+7*a(n+2)+4*a(n+1)-4*a(n).

%F a(n) = (38/5*5^(1/2)+17)*((1+sqrt(5))/2)^n+(-38/5*5^(1/2)+17)*((1-sqrt(5))/2)^n-32*2^n-1+16*2^(n-1)*n.

%F a(n) = F(n+8)+2^(n+3)*(n-4)-1, where (F(n)) is the Fibonacci sequence for which F(0)=F(1)=1, F(2)=2, ... (related to A000045).

%p c(0):=1:c(1):=6:c(2):=24:c(3):=79:c(4):=232:for n from 0 to 30 do : c(n+5):=6*c(n+4)-12*c(n+3)+7*c(n+2)+4*c(n+1)-4*c(n): od :seq(c(n),n=0..30); taylor((-1/(-1+z)/(-1+2*z)^2/(1-z-z^2)),z=0,30); for n from 0 to 30 do a(n):=simplify((38/5*5^(1/2)+17)*((1+sqrt(5))/2)^n+(-38/5*5^(1/2)+17)*((1-sqrt(5))/2)^n-32*2^n-1+16*2^(n-1)*n):od:seq(a(n),n=0..30);

%K easy,nonn

%O 0,2

%A _Richard Choulet_, Feb 07 2010