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A246096
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Paradigm shift sequence for (4,2) production scheme with replacement.
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9
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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, 66, 72, 84, 96, 112, 128, 144, 160, 180, 200, 225, 250, 275, 300, 336, 384, 448, 512, 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
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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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?"
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.
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.
3. The ideal number of implementations per bundle, as measured by the geometric growth rate (p+zq root of z), is z = 4.
4. All solutions will be of the form a(n) = (qm+r) * m^b * (m+1)^d.
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LINKS
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FORMULA
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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) ).
Recursive: a(n) = 4*a(n-12) for all n >= 67.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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