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Expand 1/(1 - (3/2)*x + (2/3)*x^4 - x^5) in powers of x, then multiply coefficient of x^n by 3^floor(n/4)*2^n to get integers.
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3
1, 3, 9, 27, 211, 633, 1899, 5697, 52297, 156891, 470673, 1412019, 12675403, 38026209, 114078627, 342235881, 3081171505, 9243514515, 27730543545, 83191630635, 748691121283, 2246073363849, 6738220091547
FORMULA
a(n) = 211*a(n-4) + 7776*a(n-8) for n > 7.
G.f.: -(3*x + 1)*(9*x^2 + 1)/(7776*x^8 + 211*x^4 - 1). (End)
MATHEMATICA
Table[3^Floor[k/4]*2^k*SeriesCoefficient[ Series[1/(1 - (3/2)* x + (2/3) x^4 - x^5), {x, 0, 30}], k], {k, 0, 30}] (* Bagula *)
a[ n_] := 2^n 3^Quotient[ n, 4] SeriesCoefficient[ 1 / (1 - 3/2 x + 2/3 x^4 - x^5), {x, 0, n}] (* Michael Somos, Jan 27 2012 *)
Expansion of 1/(-32*x^5 + 8*x^3 - 4*x^2 - x + 1).
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2
1, 1, 5, 1, 13, 9, 85, 177, 477, 921, 1701, 4289, 9389, 28201, 60917, 153041, 308349, 733625, 1645125, 4062177, 9670989, 22625865, 52288405, 118067953, 276204317, 639640537, 1523941861
COMMENTS
Previous name was: Expand 1/(1 - x/2 - x^2 + x^3 - x^5) in powers of x, then multiply coefficient of x^n by 2^n to get integers.
The sequence is from -1 + x^2 - x^3 - x^4/2 + x^5 with real root 1.1647612555333289.
The limiting ratio of successive terms is 2*1.1647612555333289.
Recurrence: -32 *a (n) + 8 *a (n + 2) - 4 *a (n + 4) + a (n + 5) == 0; with a (1) == 1; a (2) == 1; a (3) == 5; a (4) == 1; a (5) == 13 (from FindSequenceFunction[]).
MATHEMATICA
CoefficientList[Series[1/(1 - x/2 - x^2 + x^3 - x^5), {x, 0, 50}], x] * 2^Range[0, 50]
LinearRecurrence[{1, 4, -8, 0, 32}, {1, 1, 5, 1, 13}, 100] (* G. C. Greubel, Nov 16 2016 *)
PROG
(PARI) for(n=0, 30, print1(2^n*polcoeff(1/(1-x/2 - x^2 + x^3 - x^5) + O(x^32), n), ", ")) \\ G. C. Greubel, Nov 16 2016
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.
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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
COMMENTS
Bob Hanlon (hanlonr(AT)cox.net) helped convert the expansion to a recursion.
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}]
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