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 A193457 Paradigm shift sequence with procedure length p=5. 10
 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, 81, 100, 125, 150, 180, 216, 252, 294, 343, 392, 448, 512, 625, 750, 900, 1080, 1296, 1512, 1764, 2058, 2401, 2744, 3136, 3750, 4500, 5400, 6480, 7776, 9072, 10584, 12348, 14406 (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=5 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 6 and 7.] 3. The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n+ 5 - 5*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..54. 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(13) = 16 [C=2 below] a(24) = 100 [C=3 below] a(46) = 3750 [C=5 below] a(57) = 22500 [C=6 below] a(68) = 135000 [C=7 below] a(1:68) = m^(C-R) * (m+1)^R where C = floor((n+8)/11) [min C=1] m = floor ((n+5)/C)-5, and R = n+5 mod C a(n>=69) = 5^b * 6^(C-b-d) * 7^d where C = floor((n+8)/11) R = n+8 mod 11 b = max(0, 3-R); d = max(0, R-3) Recursive: for n>=80, a(n)=6*a(n-11) EXAMPLE For n=30, C=floor(38/11)=3, m=floor(35/3)-5 = 11-5 = 6, and R= (35 mod 3) = 2; therefore a(30) = 6^(3-2)*7^2 = 6*7^2 =294. For n=13, use the general formula with C=2 (rather than C=1), with R = (18 mod 2) = 0, m=floor(18/2)-5=9-5=4; therefore a(13)=4^2*5^0=16. For n=80, C = floor(88/11)=8, R=(88 mod 11) = 0, b = max(0,3)=3, and d=max(0,-3)=0; therefore a(80) = 5^3*6^(8-3)*7^0 = 5^3*6^5 = 972000 CROSSREFS A000792 (n>=1), A178715, A193286, A193455, A193456, and A193457 are paradigm shift sequences for p=0,1,...5 respectively. Sequence in context: A004841 A293715 A276393 * A161950 A111470 A227508 Adjacent sequences: A193454 A193455 A193456 * A193458 A193459 A193460 KEYWORD nonn AUTHOR Jonathan T. Rowell, Jul 27 2011 STATUS approved

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Last modified June 17 13:47 EDT 2024. Contains 373445 sequences. (Running on oeis4.)