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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.
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
KEYWORD
nonn,easy
AUTHOR
Jonathan T. Rowell, Jul 26 2011
STATUS
approved

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Last modified June 17 13:58 EDT 2024. Contains 373445 sequences. (Running on oeis4.)