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A193455 Paradigm shift sequence with procedure length p=3.

%I A193455 #31 Mar 24 2021 09:52:27
%S A193455 1,2,3,4,5,6,7,8,9,12,16,20,25,30,36,42,49,64,80,100,125,150,180,216,
%T A193455 256,320,400,500,625,750,900,1080,1296,1600,2000,2500,3125,3750,4500,
%U A193455 5400,6480,8000,10000,12500,15625,18750,22500,27000,32400,40000,50000
%N A193455 Paradigm shift sequence with procedure length p=3.
%C A193455 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?"
%C A193455 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.
%C A193455 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.]
%C A193455 3. The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n+ 3 - 3*i_max
%C A193455 4. All solutions will be of the form a(n) = m^b * (m+1)^d
%H A193455 Jonathan T. Rowell, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Rowell/rowell3.html">Solution Sequences for the Keyboard Problem and its Generalizations</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.10.7.
%H A193455 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,5).
%F A193455 a(n) =
%F A193455 a(25)   = 256  [C = 4 below]
%F A193455 a(1:24) = m^(C-R) * (m+1)^R
%F A193455                 where C = floor((n+6)/8) [min C=1],
%F A193455                   R = n+3 mod C, m = floor((n+3-3*C)/C)
%F A193455   a(n>=26) = 4^b * 5^(C-(b+d)) * 6^d
%F A193455                   where C = floor((n+6)/8), R = n+6 mod 8,
%F A193455                   b = max(0,3-R), and d = max(0, R-3)
%F A193455 Recursive:  a(n) = 5*a(n-8) for all n >= 34
%e A193455 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.
%e A193455 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.
%e A193455 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.
%o A193455 (Python)
%o A193455 def a(n):
%o A193455     c=(n + 6)//8
%o A193455     if n<25:
%o A193455         if n<10: return n
%o A193455         r=(n + 3)%c
%o A193455         m=(n + 3 - 3*c)//c
%o A193455         return m**(c - r)*(m + 1)**r
%o A193455     elif n==25: return 256
%o A193455     else:
%o A193455         r=(n + 6)%8
%o A193455         b=max(0, 3 - r)
%o A193455         d=max(0, r - 3)
%o A193455         return 4**b*5**(c - (b + d))*6**d
%o A193455 print([a(n) for n in range(1, 101)]) # _Indranil Ghosh_, Jun 27 2017
%Y A193455 Paradigm shift sequences: A000792 (p=0), A178715 (p=1), A193286 (p=2), A193455 (p=3), A193456 (p=4), A193457 (p=5).
%K A193455 nonn,easy
%O A193455 1,2
%A A193455 _Jonathan T. Rowell_, Jul 26 2011