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%I #7 Sep 11 2014 08:33:37
%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,18,21,24,28,32,36,40,45,50,55,60,
%T 66,72,84,96,112,128,144,160,180,200,225,250,275,300,336,384,448,512,
%U 576,640,720,800,900,1000,1125,1250,1375,1536,1792,2048,2304,2560,2880,3200,3600,4000,4500,5000,5625,6250,7168,8192,9216,10240,11520,12800,14400,16000,18000,20000,22500,25000
%N Paradigm shift sequence for (4,2) 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=4 steps), or implement the current bundled action (which requires q=2 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-12) for all n >= 67.
%Y Paradigm shift sequences with q=2: A029744, A029747, A246080, A246084, A246088, A246092, A246096, A246100.
%Y Paradigm shift sequences with p=4: A193456, A246096, A246097, A246098, A246099.
%K nonn
%O 1,2
%A _Jonathan T. Rowell_, Aug 13 2014
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