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%I #12 Mar 30 2024 23:08:45
%S 1,1,0,1,2,2,3,5,7,9,13,20,28,39,57,82,116,166,239,342,488,699,1002,
%T 1433,2050,2936,4203,6014,8608,12323,17638,25244,36134,51722,74030,
%U 105961,151669,217091,310729,444760,636607,911202,1304240,1866817,2672058,3824629
%N Number of ways are there to score a break of n points at snooker. Assuming an infinite number of reds are available, along with the usual six colors, and a break alternates red-color-red-...
%C Equivalently, a(n) is the number of compositions of n with odd-indexed parts being 1 and even-indexed parts being between 2 and 7. - _Andrew Howroyd_, Jan 14 2020
%H Andrew Howroyd, <a href="/A180158/b180158.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,1,1,1,1,1,1).
%F a(n) = a(n-3) + a(n-4) + a(n-5) + a(n-6) + a(n-7) + a(n-8) for n >= 8.
%F G.f.: (1 + x)/(1 - x^3 - x^4 - x^5 - x^6 - x^7 - x^8). - _Andrew Howroyd_, Jan 14 2020
%e For n=0 there is one way to score zero points - pot nothing, potting a single red is the one way to have a break of one, a break of two is impossible (as defined here), a break of three can only be achieved with a red followed by a yellow, ...
%o (PARI) Vec((1 + x)/(1 - x^3 - x^4 - x^5 - x^6 - x^7 - x^8) + O(x^50)) \\ _Andrew Howroyd_, Jan 14 2020
%K nonn,easy
%O 0,5
%A Rob Cummings (robc(AT)rocoto.demon.co.uk), Aug 14 2010
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