

A246102


Paradigm shift sequence for (5,4) production scheme with replacement.


10



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, 27, 30, 33, 36, 39, 42, 45, 48, 52, 56, 60, 64, 68, 72, 76, 80, 85, 90, 95, 100, 108, 117, 126, 135, 144, 156, 168, 180, 192, 208, 224, 240, 256, 272, 288, 304, 320, 340, 360, 380, 405, 432, 468, 504, 540, 576, 624, 672, 720, 768, 832, 896, 960, 1024, 1088, 1152, 1216, 1280, 1360, 1440, 1520, 1620, 1728, 1872, 2016, 2160, 2304, 2496, 2688, 2880, 3072, 3328, 3584, 3840, 4096
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OFFSET

1,2


COMMENTS

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=5 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(pq,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 ParadigmShift 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.


LINKS



FORMULA

a(n) = (qd+r) * d^(CR) * (d+1)^R, where r = (nCp) mod q, Q = floor( (RCp)/q ), R = Q mod (C+1), and d = floor ( Q/(C+1) ).
Recursive: a(n) = 4*a(n21) for all n >= 67.


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



