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A110213 a(n+3) = 6*a(n) - 5*a(n+2), a(0) = 1, a(1) = -7, a(2) = 35. 4

%I

%S 1,-7,35,-169,803,-3805,18011,-85237,403355,-1908709,9032123,

%T -42740485,202250171,-957058117,4528847675,-21430737349,101411338043,

%U -479883604165,2270833596731,-10745699955397,50849198151995,-240620989179589,1138630746165563,-5388058541915845

%N a(n+3) = 6*a(n) - 5*a(n+2), a(0) = 1, a(1) = -7, a(2) = 35.

%H Harvey P. Dale, <a href="/A110213/b110213.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-5,0,6).

%F G.f. (-1+2*x)/((x-1)*(6*x^2+6*x+1))

%F a(x)=(3-Sqrt[3]+(7-8*Sqrt[3])(-3+Sqrt[3])^x+(-3-Sqrt[3])^x (-10+9*Sqrt[3]))/(13*(-3+Sqrt[3])). - _Harvey P. Dale_, Mar 01 2015

%p seriestolist(series((-1+2*x)/((x-1)*(6*x^2+6*x+1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: 2baseksumseq[A*B] with A = + 'i + 'ii' + 'ij' + 'ik' and B = + .5'i - .5'j + .5'k + .5i' + .5j' - .5k' - .5'ij' - .5'ik' + .5'ji' + .5'ki' Sumtype is set to: sum[(Y[0], Y[1], Y[2]),mod(3)

%t LinearRecurrence[{-5,0,6},{1,-7,35},30] (* _Harvey P. Dale_, Mar 01 2015 *)

%Y Cf. A110210, A110211, A110212.

%K easy,sign

%O 0,2

%A _Creighton Dement_, Jul 16 2005

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Last modified February 25 09:57 EST 2021. Contains 341606 sequences. (Running on oeis4.)