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A193456 Paradigm shift sequence with procedure length p=4. 10
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 80, 100, 125, 150, 180, 216, 252, 294, 343, 400, 500, 625, 750, 900, 1080, 1296, 1512, 1764, 2058, 2500, 3125, 3750, 4500, 5400, 6480, 7776, 9072, 10584, 12500, 15625, 18750, 22500, 27000 (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 action, bundle all existing actions (which requires p =4 steps), or apply the current bundled action. The first use of a novel bundle erases (or makes obsolete) all prior actions.  How many total actions (simple) can be applied in n time steps?"

1. 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 Procedural Lengths p=1 and 2, respectively.

2. The optimal number of pastes per copy, as measured by the geometric growth rate (p+z root of z), is z = 6. [Non-integer maximum between 5 and 6.]

3. The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n+ 4 - 4*i_max

4. All solutions will be of the form a(n) = m^b * (m+1)^d

LINKS

Table of n, a(n) for n=1..53.

Jonathan T. Rowell, Solution Sequences for the Keyboard Problem and its Generalizations, Journal of Integer Sequences, Vol. 18 (2015), Article 15.10.7.

FORMULA

a(n) =

      a(8:10) = 8; 9; 10     [C=1 below]

      a(18:20) = 49; 56; 88  [C=2 below]

      a(28:29) = 294; 343    [C=3 below]

      a(38:39) = 1764; 2058  [C=4 below]

      a(48) = 10584          [C=5 below]

      a(58) = 63504          [C=6 below]

       a(1:67) = m^(C-R) * (m+1)^R

                where C = floor((n+2)/10) +1 [min C=1]

                m = floor ((n+4)/C)-4, and R = n+4 mod C

      a(n>=68) = 5^b * 6^(C-b-d) * 7^d

                where C = floor((n+2)/10) +1

                R = n+2 mod 10

                b = max(0, 8-R); d = max(0, R-8)

Recursive: for n>=69, a(n)=6*a(n-10)

EXAMPLE

For n = 18, a(18) uses the general formula given for n in [1:67], but uses C=2 (rather than C=3).  m = floor(22/2)-4 = 7; R = 22 mod 2 = 0; therefore a(18) = 7^(2-0)*8^0 = 49

For n=37, a(37) has: C = floor(39/10) +1 = 3+1=4.  m = floor(41/4)-4 = 10-4=6, R = 41 mod 4 = 1; therefore, a(37) = 6^(4-1)*7^(1) = 6^3 *7 = 1512.

CROSSREFS

Paradigm shift sequences: A000792, A178715, A193286, A193455, A193456, and A193457 for p=0,1,...,5.

Sequence in context: A280863 A059765 A180479 * A143289 A064807 A235591

Adjacent sequences:  A193453 A193454 A193455 * A193457 A193458 A193459

KEYWORD

nonn

AUTHOR

Jonathan T. Rowell, Jul 26 2011

STATUS

approved

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Last modified May 26 15:50 EDT 2020. Contains 334626 sequences. (Running on oeis4.)