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%I #22 Oct 03 2013 21:37:04
%S 1,1,1,5,4,3,25,23,22,149,130,110,785,693,623,4389,3880,3397,23977,
%T 21115,18684,131893,116502,102680,724705,638985,563949,3980357,
%U 3512812,3098935,21873593,19295871,17024690
%N Expand 1/(8 - 8 x + 3 x^3 - 2 x^4) in powers of x, then multiply coefficient of x^n by 8^(1 + floor(n/3)) to get integers.
%C Bob Hanlon (hanlonr(AT)cox.net) helped convert the expansion to a recursion.
%H Alois P. Heinz, <a href="/A206568/b206568.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: (-4*x^8-6*x^7-9*x^6-4*x^5-5*x^4-6*x^3-x^2-x-1) / (64*x^12 +69*x^9 +21*x^6 -x^3-1).
%p a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <64|69|21|-1>>^ iquo(n, 3, 'r'). `if`(r=0, <<1, 5, 25, 149>>, `if`(r=1, <<1, 4, 23, 130>>, <<1, 3, 22, 110>>)))[1, 1]: seq (a(n), n=0..40); # _Alois P. Heinz_, Feb 11 2012
%t (* expansion*)
%t Table[8^(1 + Floor[n/3])*SeriesCoefficient[Series[1/(8 - 8 x + 3 x^3 - 2 x^4), {x, 0, 50}], n], {n, 0,50}]
%t (*recursion*)
%t a[1] = 1; a[2] = 1; a[3] = 1; a[4] = 5; a[5] = 4; a[6] = 3;
%t a[7] = 25; a[8] = 23; a[9] = 22; a[10] = 149; a[11] = 130;
%t a[12] = 110;
%t a[n_Integer?Positive] := a[n] = 64*a[-12 + n] + 69*a[-9 + n] + 21*a[-6 +n] - a[-3 + n]
%t Table[a[n], {n, 1, 50}]
%Y Cf. A202907, A167602, A167602, A117791, A107293, A204631, A185357, A205961.
%K nonn,easy
%O 0,4
%A _Roger L. Bagula_, Feb 09 2012
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