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Expansion of 1/((1-2*x)*(1+x^2)).
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%I #64 Oct 15 2022 14:08:56

%S 1,2,3,6,13,26,51,102,205,410,819,1638,3277,6554,13107,26214,52429,

%T 104858,209715,419430,838861,1677722,3355443,6710886,13421773,

%U 26843546,53687091,107374182,214748365,429496730,858993459,1717986918,3435973837,6871947674

%N Expansion of 1/((1-2*x)*(1+x^2)).

%C Also describes the location a(n) of the minimal scaling factor when rescaling an FFT of order 2^{n+2} in order to (currently) minimize the arithmetic operation count (Johnson & Frigo, 2007). - Steven G. Johnson (stevenj(AT)math.mit.edu), Dec 27 2006

%D M. E. Larsen, Summa Summarum, A. K. Peters, Wellesley, MA, 2007; see p. 38.

%H M. F. Hasler, <a href="/A007910/b007910.txt">Table of n, a(n) for n = 0..1000</a> (in replacement of a(0..999) indexed 1..1000 by Vincenzo Librandi)

%H M. H. Cilasun, <a href="http://arxiv.org/abs/1412.3265">An Analytical Approach to Exponent-Restricted Multiple Counting Sequences</a>, arXiv preprint arXiv:1412.3265 [math.NT], 2014.

%H M. H. Cilasun, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Cilasun/cila5.html">Generalized Multiple Counting Jacobsthal Sequences of Fermat Pseudoprimes</a>, Journal of Integer Sequences, Vol. 19, 2016, #16.2.3.

%H I. Gessel, <a href="http://www.jstor.org/stable/2974863">Problem 10424</a>, Amer. Math. Monthly, 102 (1995), 70.

%H S. G. Johnson and M. Frigo, <a href="http://www.fftw.org/newsplit.pdf">A modified split-radix FFT with fewer arithmetic operations</a>, IEEE Trans. Signal Processing 55 (2007), 111-119.

%H Kyu-Hwan Lee, Se-jin Oh, <a href="http://arxiv.org/abs/1601.06685">Catalan triangle numbers and binomial coefficients</a>, arXiv:1601.06685 [math.CO], 2016.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,2).

%F a(0) = 1, a(2n+1) = 2*a(2n) and a(2n) = 2*a(2n-1) + (-1)^n. [Corrected by _M. F. Hasler_, Feb 22 2018]

%F a(n) = (4*2^n+cos(Pi*n/2)+2*sin(Pi*n/2))/5. - _Paul Barry_, Dec 17 2003

%F a(n) = 2a(n-1)-a(n-2)+2a(n-3). Sequence equals half its second differences with first term dropped. a(n) + a(n+2) = 2^(n+2). - _Paul Curtz_, Dec 17 2007

%F a(n) = round(2^(n+2)/5). - _Mircea Merca_, Dec 27 2010

%F a(n) = Sum_{k=0..floor(n/2)} (-1)^k*2^(n-2*k). - _Gerry Martens_, Oct 15 2022

%p A007910:=n->(1/5)*(2^(n-1)+2*cos(n*Pi/2)-sin(n*Pi/2)); [seq(V(n),n=0..12)];

%p seq(round(2^(n+2)/5),n=1..25) # _Mircea Merca_, Dec 27 2010

%t CoefficientList[Series[1/((1 - 2 x) (1 + x^2)), {x, 0, 50}], x] (* _Vladimir Joseph Stephan Orlovsky_, Jun 20 2011 *)

%t LinearRecurrence[{2,-1,2},{1,2,3},40] (* _Harvey P. Dale_, Feb 22 2016 *)

%o (Magma) [Round(2^(n+2)/5): n in [0..40]]; // _Vincenzo Librandi_, Jun 21 2011

%o (PARI) a(n)=2^(n+2)\/5 \\ _Charles R Greathouse IV_, Jun 21 2011

%Y Cf. A007909, A007679.

%K nonn,easy

%O 0,2

%A Mogens Esrom Larsen (mel(AT)math.ku.dk)

%E Entry revised by _N. J. A. Sloane_, Feb 24 2004

%E Offset corrected and minor edits by _M. F. Hasler_, Feb 22 2018