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A093968
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Inverse binomial transform of n*Pell(n).
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6
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0, 1, 2, 6, 8, 20, 24, 56, 64, 144, 160, 352, 384, 832, 896, 1920, 2048, 4352, 4608, 9728, 10240, 21504, 22528, 47104, 49152, 102400, 106496, 221184, 229376, 475136, 491520, 1015808, 1048576, 2162688, 2228224, 4587520, 4718592, 9699328, 9961472, 20447232, 20971520
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OFFSET
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0,3
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COMMENTS
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Binomial transform of (-1)^(n+1)(n*Pell(n-2)) (see A093969).
a(n) is also the number of projective permutations of vertices of regular n-gons. A permutation of n vertices (AFB...CD) is considered 'projective' if there exists a line so that all the vertices can be projected onto it and the resulted points can be read in the same order: A'F'B'...C'D'. - Anton Zakharov, Jul 25 2016
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LINKS
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FORMULA
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G.f.: x(1+2x+2x^2)/(1-2x^2)^2;
a(n) = 2^((n-4)/2)n((1+sqrt(2)) + (1-sqrt(2))(-1)^n).
G.f.: x*G(0)/(1-x) where G(k) = 1 + x/(k+1 - 2*x*(k+1)*(k+2)/(2*x*(k+2) + 1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 01 2013
E.g.f.: x*(sqrt(2)*sinh(sqrt(2)*x) + 2*cosh(sqrt(2)*x))/2. - Ilya Gutkovskiy, Jul 25 2016
Sum_{n>=1} 1/a(n) = log(2) + sqrt(2)*log(1+sqrt(2)). - Amiram Eldar, Feb 13 2023
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EXAMPLE
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a(3) = 6, as there are only 6 projective permutations of vertices in a triangle ABC: ABC,CBA,ACB,BCA,CAB,BAC and it is equal to the number of simple permutations of three elements.
a(4) = 8, as there are only 8 permutations of vertices in a square, satisfying the projective criterion: ADBC,DACB,DCAB,CDBA,СBDA,BCAD,BACD,ABDC. ADCB is not allowed, cause there is no way to draw a line so that the projections A'B'C'D' of the original points form a line segment B'C' lying inside A'D' on this line. - Anton Zakharov, Jul 25 2016
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MATHEMATICA
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a[n_] := n*2^Floor[(n - 1)/2]; Array[a, 40, 0] (* Amiram Eldar, Feb 13 2023 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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