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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 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 April 17 11:50 EDT 2021. Contains 343064 sequences. (Running on oeis4.)