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A246090
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Paradigm shift sequence for (2,4) 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, 16, 17, 18, 19, 20, 22, 24, 27, 30, 33, 36, 39, 42, 45, 48, 52, 56, 60, 64, 68, 72, 81, 90, 99, 108, 117, 126, 135, 144, 156, 168, 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
(list;
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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=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?"
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 = 3.
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) = 3*a(n-14) for all n >= 51.
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EXAMPLE
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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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