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A110210 a(n+3) = 6*a(n) - 5*a(n+2), a(0) = -1, a(1) = 1, a(2) = -5. 3

%I

%S -1,1,-5,19,-89,415,-1961,9271,-43865,207559,-982169,4647655,

%T -21992921,104071591,-492472025,2330402599,-11027583449,52183085095,

%U -246933009881,1168499548711,-5529399232985,26165398105639,-123815993235929,585903570781735,-2772525465274841

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

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

%F Superseeker finds: a(n+1) - a(n) = ((-1)^n)*A094433(n+2) (left term in M^n * [1 0 0], M = the 3 X 3 matrix [1 -1 0 / -1 3 -2 / 0 -2 2], offset at 1); a(n+2) - a(n) = ((-1)^(n+1))*A086405(n+1) (Row T(n, 3) of number array A086404.

%F g.f.: (4*x+1)/(6*x^3-5*x-1). - _Harvey P. Dale_, Nov 09 2014

%p seriestolist(series((1+4*x)/((x-1)*(6*x^2+6*x+1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: 2baseisumseq[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},{-1,1,-5},30] (* or *) CoefficientList[ Series[ (1+4*x)/(-1-5*x+6*x^3),{x,0,30}],x] (* _Harvey P. Dale_, Nov 09 2014 *)

%Y Cf. A094433, A110211, A110212, A110213.

%K easy,sign

%O 0,3

%A _Creighton Dement_, Jul 16 2005

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Last modified April 18 20:41 EDT 2021. Contains 343089 sequences. (Running on oeis4.)