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A121177 Catapolyoctagons (see Cyvin et al. for precise definition). 3

%I #34 Oct 09 2023 11:23:19

%S 0,2,12,62,312,1562,7812,39062,195312,976562,4882812,24414062,

%T 122070312,610351562,3051757812,15258789062,76293945312,381469726562,

%U 1907348632812,9536743164062,47683715820312,238418579101562,1192092895507812,5960464477539062,29802322387695312

%N Catapolyoctagons (see Cyvin et al. for precise definition).

%C From _Petros Hadjicostas_, Jul 30 2019: (Start)

%C The conjecture by _Philipp Emanuel Weidmann_ (see link below) is correct. In Cyvin et al. (1997), this sequence has a double meaning. See Eqs. (6) and (7) and Table I on p. 58 in that paper. The terms of the sequence are related to the enumeration of unbranched catapolyoctagons.

%C The number of unbranched catapolyoctagons of the symmetry C_{2h} is given by c_r = (1/2) *(5^(floor(r/2)-1) - 1) + (2/5) * binomial(1, r), where r is the number of octagons in the unbranched catapolyoctagon. We get the sequence 0, 0, 0, 2, 2, 12, 12, 62, 62, 312, 312, ... whose bijection (apart for the case r = 1) is the current sequence.

%C In addition, the number of unbranched catapolyoctagons of the symmetry C_{2v} is given by m_r = (1/2) * (3 - 2*(-1)^r) * 5^(floor(r/2) - 1) - (1/2), where again r is the number of octagons. We get the sequence 0, 0, 2, 2, 12, 12, 62, 62, 312, 312, 1562, 1562, ... whose bijection is the current sequence.

%C The total number of unbranched catapolygons (with respect to all the symmetry point groups D_{8h}, D_{2h}, C_{2h}, and C_{2v}) is given by i_r = A121101(r).

%C (End)

%D S. J. Cyvin, B. N. Cyvin, and J. Brunvoll. Enumeration of tree-like octagonal systems: catapolyoctagons, ACH Models in Chem. 134 (1997), 55-70, eq. (6).

%H P. E. Weidmann, <a href="https://worldwidemann.com/the-sequencer-oeis-survey/#a121177">The OEIS Sequencer survey</a>, Apr 11 2015.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Point_groups_in_three_dimensions">Point groups in three dimensions</a>.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-5).

%F a(n) = (5^n-5)/10 = 2*A003463(n-1) for n >= 1. - _Philipp Emanuel Weidmann_, cf. link.

%F G.f.: 2*x^2 / ( (5*x-1)*(x-1) ). - _R. J. Mathar_, Jul 31 2019

%Y Cf. A121101, A125831.

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_, Aug 15 2006

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