A077854 Expansion of 1/((1-x)*(1-2*x)*(1+x^2)).
1, 3, 6, 12, 25, 51, 102, 204, 409, 819, 1638, 3276, 6553, 13107, 26214, 52428, 104857, 209715, 419430, 838860, 1677721, 3355443, 6710886, 13421772, 26843545, 53687091, 107374182, 214748364, 429496729, 858993459, 1717986918, 3435973836, 6871947673
Offset: 0
Examples
The sequence in hexadecimal shows the pattern 1, 3, 6, c, 19, 33, 66, cc, 199, 333, 666, ccc, 1999, 3333, 6666, cccc, 19999, 33333, 66666, ccccc, 199999, 333333, 666666, cccccc, 1999999, 3333333, 6666666, ccccccc, 19999999, 33333333, 66666666, cccccccc, ... - _Armands Strazds_, Oct 09 2014
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,-3,3,-2).
Crossrefs
Programs
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Haskell
import Data.Bits (xor) a077854 n = a077854_list !! n a077854_list = scanl1 xor $ tail a000975_list :: [Integer] -- Reinhard Zumkeller, Jan 04 2013
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Magma
[Round((2^(n+4)-5)/10): n in [0..40]]; // Vincenzo Librandi, Jun 25 2011
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Maple
a := proc(n) option remember; if n=0 then RETURN(1); fi; if n=1 then RETURN(3); fi; if n=2 then RETURN(6); fi; if n=3 then RETURN(12); fi; 3*a(n-1)-3*a(n-2)+3*a(n-3)-2*a(n-4); end; seq(iquo(2^n,5),n=3..35); # Zerinvary Lajos, Apr 20 2008
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Mathematica
CoefficientList[Series[1/((1 - x) (1 - 2 x) (1 + x^2)), {x, 0, 32}], x] (* Michael De Vlieger, Mar 29 2016 *) LinearRecurrence[{3,-3,3,-2},{1,3,6,12},40] (* Harvey P. Dale, Feb 06 2019 *) -
PARI
a(n)=(16<
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PARI
Vec(1/(1-3*x+3*x^2-3*x^3+2*x^4)+O(x^99)) \\ Derek Orr, Oct 26 2014
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Python
print([2**(n+3)//5 for n in range(50)]) # Karl V. Keller, Jr., Sep 26 2021
Formula
a(n) = 3*a(n-1) - 3*a(n-2) + 3*a(n-3) - 2*a(n-4), with initial values a(0) = 1, a(1) = 3, a(2) = 6, a(3) = 12.
a(n) = (1/10)*(2^(n+4) + (-1)^floor(n/2) - 2*(-1)^floor((n+1)/2) - 5).
Row sums of A130306. - Gary W. Adamson, May 20 2007
a(n) = floor(2^(n+3)/5). - Gary Detlefs, Sep 06 2010
a(n) = round((2^(n+4)-5)/10) = floor((2^(n+3)-1)/5) = ceiling((2^(n+3)-4)/5) = round((2^(n+3)-2)/5); a(n) = a(n-4) + 3*2^(n-1), n > 3. - Mircea Merca, Dec 27 2010
a(n) = 2^(n+1) - 1 - a(n-2); a(n) = a(n-1)/2 for n == 2, 3 (mod 4); a(n) = (a(n-1)-1)/2 for n == 0, 1 (mod 4). - Arie Bos, Apr 06 2013
a(n) = floor(A000975(n+2)*3/5). - Armands Strazds, Oct 18 2014
a(n) = Sum_{k=1..n+3} floor(1 + sin(k*Pi/2 + 3*Pi/4))*2^(n-k+3). - Andres Cicuttin, Mar 28 2016
a(n) = (-15 + 3*2^(3+n) + 2^(1 + n - 4*floor((1+n)/4)) + 2^(2 + n - 4*floor((2+n)/4)))/15. - Andres Cicuttin, Mar 28 2016
a(n) = (16*2^n+(-1)^((2*n-1+(-1)^n)/4)-2*(-1)^((2*n+1-(-1)^n)/4)-5)/10. - Wesley Ivan Hurt, Apr 01 2016
Comments