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%I #6 Aug 15 2014 23:07:55
%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,
%T 28,30,33,36,39,42,45,48,51,54,57,60,64,68,72,76,80,84,88,92,99,108,
%U 117,126,135,144,153,162,171,180,192,204,216,228,240,256,272,288,304,324,351,378,405,432,459,486,513,540,576,612,648,684,720,768,816,864,912,972
%N Paradigm shift sequence for (3,5) 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=5 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-18) for all n >= 66.
%Y Paradigm shift sequences with q=5: A103969, A246074, A246075, A246076, A246079, A246083, A246087, A246091, A246095, A246099, A246103.
%Y Paradigm shift sequences with p=3: A193455, A246092, A246093, A246094, A246095.
%K nonn
%O 1,2
%A _Jonathan T. Rowell_, Aug 13 2014
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