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A246081
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Paradigm shift sequence for (0,3) 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, 14, 16, 18, 21, 24, 27, 30, 33, 36, 42, 48, 54, 63, 72, 81, 90, 99, 108, 126, 144, 162, 189, 216, 243, 270, 297, 324, 378, 432, 486, 567, 648, 729, 810, 891, 972, 1134, 1296, 1458, 1701, 1944, 2187, 2430, 2673, 2916, 3402, 3888, 4374, 5103, 5832, 6561, 7290, 8019, 8748
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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=0 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?"
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.
5. For large n, the sequence is recursively defined.
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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) ).
a(n) = 3*a(n-9) for all n >= 15.
G.f.: x*(1+x+x^2)^2 * (1-x+x^3) * (1+x+x^2+2*x^3+x^4+x^6) / (1-3*x^9). - Colin Barker, Nov 19 2016
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PROG
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(PARI) Vec(x*(1+x+x^2)^2 * (1-x+x^3) * (1+x+x^2+2*x^3+x^4+x^6) / (1-3*x^9) + O(x^100)) \\ Colin Barker, Nov 19 2016
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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