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 A206568 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. 2
 1, 1, 1, 5, 4, 3, 25, 23, 22, 149, 130, 110, 785, 693, 623, 4389, 3880, 3397, 23977, 21115, 18684, 131893, 116502, 102680, 724705, 638985, 563949, 3980357, 3512812, 3098935, 21873593, 19295871, 17024690 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Bob Hanlon (hanlonr(AT)cox.net) helped convert the expansion to a recursion. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..1000 FORMULA 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). MAPLE 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 MATHEMATICA (* expansion*) 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}] (*recursion*) a[1] = 1; a[2] = 1; a[3] = 1; a[4] = 5; a[5] = 4; a[6] = 3; a[7] = 25; a[8] = 23; a[9] = 22; a[10] = 149; a[11] = 130; a[12] = 110; a[n_Integer?Positive] := a[n] = 64*a[-12 + n] + 69*a[-9 + n] + 21*a[-6 +n] - a[-3 + n] Table[a[n], {n, 1, 50}] CROSSREFS Cf. A202907, A167602, A167602, A117791, A107293, A204631, A185357, A205961. Sequence in context: A125900 A019102 A019179 * A304656 A220248 A147533 Adjacent sequences:  A206565 A206566 A206567 * A206569 A206570 A206571 KEYWORD nonn,easy AUTHOR Roger L. Bagula, Feb 09 2012 STATUS approved

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Last modified January 20 21:46 EST 2022. Contains 350472 sequences. (Running on oeis4.)