The OEIS is supported by the many generous donors to the OEIS Foundation.

 Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 59th year, we have over 358,000 sequences, and we’ve crossed 10,300 citations (which often say “discovered thanks to the OEIS”). Other ways to Give
 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A193455 Paradigm shift sequence with procedure length p=3. 10
 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 16, 20, 25, 30, 36, 42, 49, 64, 80, 100, 125, 150, 180, 216, 256, 320, 400, 500, 625, 750, 900, 1080, 1296, 1600, 2000, 2500, 3125, 3750, 4500, 5400, 6480, 8000, 10000, 12500, 15625, 18750, 22500, 27000, 32400, 40000, 50000 (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=3 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 = 5. [Non-integer maximum between 4 and 5.] 3. The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n+ 3 - 3*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. Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,5). FORMULA a(n) = a(25) = 256 [C = 4 below] a(1:24) = m^(C-R) * (m+1)^R where C = floor((n+6)/8) [min C=1], R = n+3 mod C, m = floor((n+3-3*C)/C) a(n>=26) = 4^b * 5^(C-(b+d)) * 6^d where C = floor((n+6)/8), R = n+6 mod 8, b = max(0,3-R), and d = max(0, R-3) Recursive: a(n) = 5*a(n-8) for all n >= 34 EXAMPLE For n=20, C = floor(26/8) = 3, R = (23 mod 3) = 2, m = floor (23-9/3) = floor(14/3)=4; therefore a(20) = 4^(3-2)*5^(2) = 4*5^2 = 100. For n=25, the same general formula is used, but C=4 (instead of 3). R=28 mod 4 =0, m = floor(28-12/4)=4; therefore a(25) = 4^4 = 256. For n=35, C = floor(41/8)=5, R = 1, b = max(0,2)=2, d=max(0,-2)=0; therefore a(35) = 4^2*5^(5-2)*6^0 = 2000. PROG (Python) def a(n): c=(n + 6)//8 if n<25: if n<10: return n r=(n + 3)%c m=(n + 3 - 3*c)//c return m**(c - r)*(m + 1)**r elif n==25: return 256 else: r=(n + 6)%8 b=max(0, 3 - r) d=max(0, r - 3) return 4**b*5**(c - (b + d))*6**d print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 27 2017 CROSSREFS Paradigm shift sequences: A000792 (p=0), A178715 (p=1), A193286 (p=2), A193455 (p=3), A193456 (p=4), A193457 (p=5). Sequence in context: A130588 A079238 A079042 * A356349 A343680 A114440 Adjacent sequences: A193452 A193453 A193454 * A193456 A193457 A193458 KEYWORD nonn,easy AUTHOR Jonathan T. Rowell, Jul 26 2011 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified December 5 20:39 EST 2022. Contains 358589 sequences. (Running on oeis4.)