A007179 Dual pairs of integrals arising from reflection coefficients.
0, 1, 1, 4, 6, 16, 28, 64, 120, 256, 496, 1024, 2016, 4096, 8128, 16384, 32640, 65536, 130816, 262144, 523776, 1048576, 2096128, 4194304, 8386560, 16777216, 33550336, 67108864, 134209536, 268435456, 536854528, 1073741824, 2147450880, 4294967296, 8589869056
Offset: 0
Examples
From _Gus Wiseman_, Feb 26 2022: (Start) Also the number of integer compositions of n with at least one odd part. For example, the a(1) = 1 through a(5) = 16 compositions are: (1) (1,1) (3) (1,3) (5) (1,2) (3,1) (1,4) (2,1) (1,1,2) (2,3) (1,1,1) (1,2,1) (3,2) (2,1,1) (4,1) (1,1,1,1) (1,1,3) (1,2,2) (1,3,1) (2,1,2) (2,2,1) (3,1,1) (1,1,1,2) (1,1,2,1) (1,2,1,1) (2,1,1,1) (1,1,1,1,1) (End)
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
- J. Heading, Theorem relating to the development of a reflection coefficient in terms of a small parameter, J. Phys. A 14 (1981), 357-367.
- Kyu-Hwan Lee, Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
- A. Yajima, How to calculate the number of stereoisomers of inositol-homologs, Bull. Chem. Soc. Jpn. 2014, 87, 1260-1264 | doi:10.1246/bcsj.20140204. See Tables 1 and 2 (and text). - _N. J. A. Sloane_, Mar 26 2015
- Index entries for linear recurrences with constant coefficients, signature (2,2,-4).
Crossrefs
Programs
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Magma
[Floor(2^n/2-2^(n/2)*(1+(-1)^n)/4): n in [0..40]]; // Vincenzo Librandi, Aug 20 2011
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Maple
f := n-> if n mod 2 = 0 then 2^(n-1)-2^((n-2)/2) else 2^(n-1); fi;
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Mathematica
LinearRecurrence[{2,2,-4},{0,1,1},30] (* Harvey P. Dale, Nov 30 2015 *) Table[2^(n-1)-If[EvenQ[n],2^(n/2-1),0],{n,0,15}] (* Gus Wiseman, Feb 26 2022 *)
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PARI
Vec(x*(1-x)/((1-2*x)*(1-2*x^2)) + O(x^50)) \\ Michel Marcus, Jan 28 2016
Formula
From Paul Barry, Apr 28 2004: (Start)
G.f.: x*(1-x)/((1-2*x)*(1-2*x^2)).
a(n) = 2*a(n-1)+2*a(n-2)-4*a(n-3).
a(n) = 2^n/2-2^(n/2)*(1+(-1)^n)/4. (End)
G.f.: (1+x*Q(0))*x/(1-x), where Q(k)= 1 - 1/(2^k - 2*x*2^(2*k)/(2*x*2^k - 1/(1 + 1/(2*2^k - 8*x*2^(2*k)/(4*x*2^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013