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%I #8 Aug 15 2014 23:06:41
%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,18,21,24,27,30,33,36,40,44,48,
%T 52,56,63,72,81,90,99,108,120,132,144,160,176,192,216,243,270,297,324,
%U 360,396,432,480,528,576,648,729,810,891,972,1080,1188,1296,1440,1584,1728,1944,2187,2430,2673,2916,3240,3564,3888,4320,4752,5184,5832,6561,7290
%N Paradigm shift sequence for (2,3) production scheme with replacement.
%C This sequence is the solution to the following problem: "Suppose you have the choice of using one of three production options: apply a simple incremental action, bundle existing output as an integrated product (which requires p=2 steps), or implement the current bundled action (which requires q=3 steps). The first use of a novel bundle erases (or makes obsolete) all prior actions. How large an output can be generated in n time steps?"
%C 1. A production scheme with replacement R(p,q) eliminates existing output following a bundling action, while an additive scheme A(p,q) retains the output. The schemes correspond according to A(p,q) = R(p-q,q), with the replacement scheme serving as the default presentation.
%C 2. This problem is structurally similar to the Copy and Paste Keyboard problem: Existing sequences (A178715 and A193286) should be regarded as Paradigm-Shift sequences with production schemes R(1,1) and R(2,1) with replacement, respectively.
%C 3. The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 3.
%C 4. All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.
%F a(n) = (qd+r) * d^(C-R) * (d+1)^R, where r = (n-Cp) mod q, Q = floor( (R-Cp)/q ), R = Q mod (C+1), and d = floor ( Q/(C+1) ).
%F Recursive: a(n) = 3*a(n-11) for all n >= 41.
%Y Cf. A178715, A193286.
%K nonn
%O 1,2
%A _Jonathan T. Rowell_, Aug 13 2014
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