A001444 Bending a piece of wire of length n+1 (configurations that can only be brought into coincidence by turning the figure over are counted as different).
1, 2, 6, 15, 45, 126, 378, 1107, 3321, 9882, 29646, 88695, 266085, 797526, 2392578, 7175547, 21526641, 64573362, 193720086, 581140575, 1743421725, 5230206126, 15690618378, 47071677987, 141215033961, 423644570442, 1270933711326, 3812799539655, 11438398618965
Offset: 0
Examples
There are 2 ways to bend a piece of wire of length 2 (bend it or not). G.f. = 1 + 2*x + 6*x^2 + 15*x^3 + 45*x^4 + 126*x^5 + 378*x^6 + ...
References
- Todd Andrew Simpson, "Combinatorial Proofs and Generalizations of Weyl's Denominator Formula", Ph. D. Dissertation, Penn State University, 1994.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Index entries for sequences obtained by enumerating foldings
- Index entries for linear recurrences with constant coefficients, signature (3,3,-9).
Programs
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Haskell
a001444 n = div (3 ^ n + 3 ^ (div n 2)) 2 -- Reinhard Zumkeller, Jun 30 2013
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Maple
f := n->(3^floor(n/2)+3^n)/2;
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Mathematica
CoefficientList[Series[(1-x-3*x^2)/((1-3*x)*(1-3*x^2)),{x,0,30}],x] (* Vincenzo Librandi, Apr 15 2012 *) LinearRecurrence[{3,3,-9},{1,2,6},40] (* Harvey P. Dale, Dec 30 2012 *)
Formula
a(n) = (3^n + 3^floor(n/2))/2.
G.f.: G(0) where G(k) = 1 + x*(3*3^k + 1)*(1 + 3*x*G(k+1))/(1 + 3^k). - Sergei N. Gladkovskii, Dec 13 2011 [Edited by Michael Somos, Sep 09 2013]
E.g.f. E(x) = (exp(3*x)+cosh(x*sqrt(3))+sinh(x*sqrt(3))/sqrt(3))/2 = G(0); G(k) = 1 + x*(3*3^k+1)/((2*k+1)*(1+3^k) - 3*x*(2*k+1)*(1+3^k)/(3*x + (2*k+2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 13 2011
From Colin Barker, Apr 02 2012: (Start)
a(n) = 3*a(n-1) + 3*a(n-2) - 9*a(n-3).
G.f.: x*(1-x-3*x^2)/((1-3*x)*(1-3*x^2)). (End)
Extensions
Interpretation in terms of bending wire from Colin Mallows.
Comments