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%I #6 Aug 15 2014 23:07:09
%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,21,24,27,30,33,36,40,44,
%T 48,52,56,60,65,72,81,90,99,108,120,132,144,160,176,192,208,224,243,
%U 270,297,324,360,396,432,480,528,576,640,704,768,832,896,972,1080,1188,1296,1440,1584,1728,1920,2112,2304,2560,2816,3072,3328,3584,3888,4320,4752,5184
%N Paradigm shift sequence for (3,3) 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=3 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 = 4.
%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) = 4*a(n-15) for all n >= 48.
%Y Paradigm shift sequences with q=3: A029747, A029750, A246077, A246081, A246085, A246089, A246093, A246097, A246101.
%Y Paradigm shift sequences with p=3: A193455, A246092, A246093, A246094, A246095.
%K nonn
%O 1,2
%A _Jonathan T. Rowell_, Aug 13 2014