A092297 Number of ways of 3-coloring an annulus consisting of n zones joined like a pearl necklace.
0, 6, 6, 18, 30, 66, 126, 258, 510, 1026, 2046, 4098, 8190, 16386, 32766, 65538, 131070, 262146, 524286, 1048578, 2097150, 4194306, 8388606, 16777218, 33554430, 67108866, 134217726, 268435458, 536870910, 1073741826, 2147483646
Offset: 1
Examples
a(2)=6 because we can color one zone in 3 colors and the other in 2, so 2*3=6 in all.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- K. Böhmová, C. Dalfó, and C. Huemer, On cyclic Kautz digraphs, Preprint 2016.
- Cristina Dalfó, From subKautz digraphs to cyclic Kautz digraphs, arXiv:1709.01882 [math.CO], 2017.
- Cristina Dalfó, The spectra of subKautz and cyclic Kautz digraphs, Linear Algebra and its Applications, 531 (2017), p. 210-219.
- Fern Gossow, Lyndon-like cyclic sieving and Gauss congruence, arXiv:2410.05678 [math.CO], 2024. See p. 25.
- Paul P. Martin and Siti Fatimah Zakaria, Zeros of the 4-state Potts model partition function for the square lattice revisited, J. Stat. Mech. 084003 (2019). eq. (7).
- Carlos I. Perez-Sanchez, The Spectral Action on quivers, arXiv:2401.03705 [math.RT], 2024.
- Index entries for linear recurrences with constant coefficients, signature (1,2).
Programs
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Magma
[2^n+2*(-1)^n : n in [1..40]]; // Vincenzo Librandi, Sep 27 2011
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Mathematica
nn=28;Drop[CoefficientList[Series[6x^2/(1+x)^2/(1-3x/(1+x)),{x,0,nn}],x],1] (* Geoffrey Critzer, Apr 05 2014 *) a[ n_] := 2 (2^(n - 1) + (-1)^n); (* Michael Somos, Oct 25 2014 *) a[ n_] := If[ n < 1, -(-2)^(n - 1) a[2 - n] , (-1)^n HypergeometricPFQ[ Table[ -2, {k, n}], Table[ 1, {k, n - 1}], 1]] (* Michael Somos, Oct 25 2014 *) LinearRecurrence[{1,2},{0,6},40] (* Harvey P. Dale, May 21 2024 *)
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PARI
{a(n) = 2 * (2^(n-1) - (-1)^n)}; /* Michael Somos, Oct 25 2014 */
Formula
a(n) = 2^n + 2*(-1)^n; recurrence a(1)=0, a(2)=6, a(n) = 2*a(n-2) + a(n-1).
O.g.f: -6*x^2/((1+x)*(2*x-1)) = -3 - 1/(2*x-1) + 2/(1+x). - R. J. Mathar, Dec 02 2007
a(n) = 6*A001045(n-1). - R. J. Mathar, Aug 30 2008
a(n) = (-1)^n * a(2-n) * 2^(n-1) for all n in Z. - Michael Somos, Oct 25 2014
Comments