

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
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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 ParadigmShift 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. [Noninteger 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

Table of n, a(n) for n=1..51.
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^(CR) * (m+1)^R
where C = floor((n+6)/8) [min C=1],
R = n+3 mod C, m = floor((n+33*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(0,3R), and d = max(0, R3)
Recursive: a(n) = 5*a(n8) for all n >= 34


EXAMPLE

For n=20, C = floor(26/8) = 3, R = (23 mod 3) = 2, m = floor (239/3) = floor(14/3)=4; therefore a(20) = 4^(32)*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(2812/4)=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^(52)*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 * A343680 A114440 A334416
Adjacent sequences: A193452 A193453 A193454 * A193456 A193457 A193458


KEYWORD

nonn,easy


AUTHOR

Jonathan T. Rowell, Jul 26 2011


STATUS

approved



