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A193456 Paradigm shift sequence with procedure length p=4.

%I A193456 #17 Oct 14 2015 12:37:59
%S A193456 1,2,3,4,5,6,7,8,9,10,12,16,20,25,30,36,42,49,56,64,80,100,125,150,
%T A193456 180,216,252,294,343,400,500,625,750,900,1080,1296,1512,1764,2058,
%U A193456 2500,3125,3750,4500,5400,6480,7776,9072,10584,12500,15625,18750,22500,27000
%N A193456 Paradigm shift sequence with procedure length p=4.
%C A193456 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?"
%C A193456 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 A193456 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.]
%C A193456 3. The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n+ 4 - 4*i_max
%C A193456 4. All solutions will be of the form a(n) = m^b * (m+1)^d
%H A193456 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.
%F A193456 a(n) =
%F A193456       a(8:10) = 8; 9; 10     [C=1 below]
%F A193456       a(18:20) = 49; 56; 88  [C=2 below]
%F A193456       a(28:29) = 294; 343    [C=3 below]
%F A193456       a(38:39) = 1764; 2058  [C=4 below]
%F A193456       a(48) = 10584          [C=5 below]
%F A193456       a(58) = 63504          [C=6 below]
%F A193456        a(1:67) = m^(C-R) * (m+1)^R
%F A193456                 where C = floor((n+2)/10) +1 [min C=1]
%F A193456                 m = floor ((n+4)/C)-4, and R = n+4 mod C
%F A193456       a(n>=68) = 5^b * 6^(C-b-d) * 7^d
%F A193456                 where C = floor((n+2)/10) +1
%F A193456                 R = n+2 mod 10
%F A193456                 b = max(0, 8-R); d = max(0, R-8)
%F A193456 Recursive: for n>=69, a(n)=6*a(n-10)
%e A193456 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
%e A193456 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.
%Y A193456 Paradigm shift sequences: A000792, A178715, A193286, A193455, A193456, and A193457 for p=0,1,...,5.
%K A193456 nonn
%O A193456 1,2
%A A193456 _Jonathan T. Rowell_, Jul 26 2011