A202907
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.
Original entry on oeis.org
1, 3, 9, 27, 211, 633, 1899, 5697, 52297, 156891, 470673, 1412019, 12675403, 38026209, 114078627, 342235881, 3081171505, 9243514515, 27730543545, 83191630635, 748691121283, 2246073363849, 6738220091547
Offset: 0
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,211,0,0,0,7776).
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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 *)
A205961
Expansion of 1/(-32*x^5 + 8*x^3 - 4*x^2 - x + 1).
Original entry on oeis.org
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
Offset: 0
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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 *)
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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
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.
Original entry on oeis.org
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
Offset: 0
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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
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(* 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}]
Showing 1-3 of 3 results.
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