A193457 - OEIS

%I #13 Oct 14 2015 12:37:48

%S 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,

%T 150,180,216,252,294,343,392,448,512,625,750,900,1080,1296,1512,1764,

%U 2058,2401,2744,3136,3750,4500,5400,6480,7776,9072,10584,12348,14406

%N Paradigm shift sequence with procedure length p=5.

%C 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?"

%C 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 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.]

%C 3. The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n+ 5 - 5*i_max

%C 4. All solutions will be of the form a(n) = m^b * (m+1)^d

%H 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 a(n) =

%F       a(13) = 16              [C=2 below]

%F       a(24) = 100             [C=3 below]

%F       a(46) = 3750            [C=5 below]

%F       a(57) = 22500           [C=6 below]

%F       a(68) = 135000          [C=7 below]

%F       a(1:68) = m^(C-R) * (m+1)^R

%F                 where C = floor((n+8)/11) [min C=1]

%F                 m = floor ((n+5)/C)-5, and R = n+5 mod C

%F       a(n>=69) = 5^b * 6^(C-b-d) * 7^d

%F                 where C = floor((n+8)/11)

%F                 R = n+8 mod 11

%F                 b = max(0, 3-R); d = max(0, R-3)

%F Recursive: for n>=80, a(n)=6*a(n-11)

%e 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.

%e 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.

%e 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

%Y A000792 (n>=1), A178715, A193286, A193455, A193456, and A193457 are paradigm shift sequences for p=0,1,...5 respectively.

%K nonn

%O 1,2

%A _Jonathan T. Rowell_, Jul 27 2011