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a(n) = -(A(n) - A(n-1)) where A(n) = A057597(n+1), for n >= 0.
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%I #18 Oct 25 2018 21:36:23

%S 0,-1,2,-1,-2,5,-4,-3,12,-13,-2,27,-38,9,56,-103,56,103,-262,215,150,

%T -627,692,85,-1404,2011,-522,-2893,5426,-3055,-5264,13745,-11536,

%U -7473,32754,-36817,-3410,72981,-106388,29997,149372,-285757,166382,268747,-720886,618521,371112,-1710519,1957928,123703,-3792150

%N a(n) = -(A(n) - A(n-1)) where A(n) = A057597(n+1), for n >= 0.

%C This sequence appears in the reduction formula for negative powers of the tribonacci constant t = A058265: t^(-n) = A(n)*t^2 + a(n)*t + A(n+1)*1, with A(n) = A057597(n+1), for n >= 0. This follows from t^3 = t^2 + t + 1, or 1/t = t^2 - t - 1 = A192918, leading to the recurrence: A(n) = -A(n) - A(n-1) + A(n-2), with inputs A(-3) = 1, A(-2) = 1 and A(-1) = 0 and a(n) = -(A(n) - A(n-1)). See the example below.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-1,-1,1).

%F a(n) = -(A057597(n+1) - A057597(n)), for n >= 0.

%F Recurrence a(n) = -a(n-1) - a(n-2) + a(n-3), for n >=0, with a(-3) = 1, a(-2) = 0 and a(-1) = 1.

%F G.f.: (1 + 1/x)/(1 + x + x^2 - x^3).

%e The coefficients of t^2, t, 1 for t^(-n) begin, for n >= -3:

%e n t^2 t 1

%e -----------------

%e -3 1 1 1

%e -2 1 0 0

%e -1 0 1 0

%e ----------------

%e +0 0 0 1

%e +1 1 -1 -1

%e +2 -1 2 0

%e +3 0 -1 2

%e +4 2 -2 -3

%e +5 -3 5 1

%e +6 1 -4 4

%e +7 4 -3 -8

%e +8 -8 12 5

%e +9 5 -13 7

%e 10 7 -2 -20

%e ...

%o (PARI) a057597(n) = polcoeff( if( n<0, x / ( 1 - x - x^2 - x^3), x^2 / ( 1 + x + x^2 - x^3) ) + x*O(x^abs(n)), abs(n)) \\ after _Michael Somos_ in A057597

%o a(n) = -(a057597(n+1)-a057597(n)) \\ _Felix Fröhlich_, Oct 23 2018

%Y Cf. A057597, A058265, A078016(n+1) (different signs), A192918.

%K sign,easy

%O 0,3

%A _Wolfdieter Lang_, Oct 23 2018