|
| |
|
|
A246089
|
|
Paradigm shift sequence for (2,3) production scheme with replacement.
|
|
9
|
|
|
|
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 21, 24, 27, 30, 33, 36, 40, 44, 48, 52, 56, 63, 72, 81, 90, 99, 108, 120, 132, 144, 160, 176, 192, 216, 243, 270, 297, 324, 360, 396, 432, 480, 528, 576, 648, 729, 810, 891, 972, 1080, 1188, 1296, 1440, 1584, 1728, 1944, 2187, 2430, 2673, 2916, 3240, 3564, 3888, 4320, 4752, 5184, 5832, 6561, 7290
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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=2 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.
|
|
|
LINKS
|
|
|
|
FORMULA
|
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-11) for all n >= 41.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
