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%I #6 Aug 15 2014 23:06:55
%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,22,24,27,30,33,36,
%T 39,42,45,48,52,56,60,64,68,72,81,90,99,108,117,126,135,144,156,168,
%U 180,192,208,224,243,270,297,324,351,378,405,432,468,504,540,576,624,672,729,810,891,972,1053,1134,1215,1296,1404,1512,1620,1728,1872,2016,2187,2430,2673,2916,3159,3402,3645,3888,4212,4536
%N Paradigm shift sequence for (2,4) 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=4 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-14) for all n >= 51.
%e Paradigm shift sequences with q=4: A029750, A103969, A246074, A246078, A246082, A246086, A246090, A246094, A246098, A246102.
%e Paradigm shift sequences with p=2: A193286, A246088, A246089, A246090, A246091.
%K nonn
%O 1,2
%A _Jonathan T. Rowell_, Aug 13 2014
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