A193286 a(n) is the maximal number of a's that can be produced in a blank document with n "keystrokes".

1, 2, 3, 4, 5, 6, 7, 9, 12, 16, 20, 25, 30, 36, 48, 64, 80, 100, 125, 150, 192, 256, 320, 400, 500, 625, 768, 1024, 1280, 1600, 2000, 2500, 3125, 4096, 5120, 6400, 8000, 10000, 12500, 16384, 20480, 25600, 32000, 40000, 50000, 65536, 81920, 102400, 128000, 160000, 200000, 262144, 327680

Offset: 1

Author

N. J. A. Sloane, Jul 20 2011

Keywords

• nonn
• nice

Comments

A "keystroke" means one of the following:

a (i.e., hit the letter "a" on the keyboard)

Ctrl-a ("select all")

Ctrl-c (copy selected text to clipboard)

Ctrl-v (paste from clipboard to cursor location)

Alternatively, a(n-2) = maximal value of Product (k_i-2) for any way of writing n = Sum k_i

1. Note that the copy command does not deselect the text.

2. This sequence is a "paradigm-shift" sequence with procedure length p =2 (in the sense of A193455).

3. The optimal number of pastes per copy, as measured by the geometric growth rate (p+z root of z), is z = 4. [Non-integer maximum between 4 and 5.]

4. The function a(n) = maximum value of the product of the terms k_i, where Sum (k_i) = n + 2 - 2*i_max.

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

Examples

			For n=25, C = floor(28/6) = 4, R = (27 mod 4) = 3, and Q = floor(27/4)-2 = 4; therefore, a(25) = 4^(4-3)*5^(3) = 4*5^3 = 500.
For n=9, we use the general solution, but with C=2 (rather than C=1). R=(11 mod 2)=1, Q=3, and a(9)=3^(2-1)*4^1 = 12.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• John Derbyshire, Solutions to puzzles in my National Review Online Diary: June 2011
• John Derbyshire, Solutions to puzzles in my National Review Online Diary: June 2011 [Cached copy, pdf version only, with permission]
• Jonathan T. Rowell, Solution Sequences for the Keyboard Problem and its Generalizations, Journal of Integer Sequences, Vol. 18 (2015), Article 15.10.7.

Crossrefs

See A178715 for another version. Cf. A000792.

A000792, A178715, A193286, A193455, A193456, and A193457 are paradigm shift sequences of procedure lengths p=0,1,...,5, respectively.

Programs

Haskell

-- See Theorem 5 in John Derbyshire link.
a193286 n = p n [] where
   p 0 ks       = product ks
   p n []       = p (n-1) [1]
   p n (k:ks)
    | n < 0     = 0
    | otherwise = max (p (n-1) ((k+1):ks)) (p (n-3) (1:k:ks))
-- Reinhard Zumkeller, Jul 22 2011, Jul 21 2011

Mathematica

a[n_ /; 1 <= n <= 7] := n; a[8] = 9; a[n_ /; 9 <= n <= 27] := (c = Max[1, Floor[(n+3)/6]]; r = Mod[n+2, c]; q = Floor[(n+2)/c]-2;q^(c-r)*(q+1)^r);a[n_ /; n >= 28] := ({q, r} = QuotientRemainder[n+2, 6]; 4^(q-r)*5^r);Table[a[n], {n, 1, 60}] (* Jean-François Alcover, May 28 2015 *)

Python

def a(n):
    if n<8: return n
    elif n==8: return 9
    elif n>8 and n<=27:
        c=max(1, ((n + 3)//6))
        r=(n + 2)%c
        q=((n + 2)//c) - 2
        return q**(c - r)*(q + 1)**r
    else:
        q=((n + 2)//6)
        r=(n + 2)%6
        return 4**(q - r)*5**r
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 27 2017, after Mathematica code

Formula

a(n) = 4*a(n-6) for n >= 34. [corrected by Georg Fischer, Jun 09 2022]

a(n) = a(8;9;15;21;27) = 9; 12; 48; 192; 768 - corresponding to [C=2;2;3;4;5 below].

a(n=9..27) = Q^(C-R) * (Q+1)^R where C = floor((n+3)/6) [minimum value 1], R = (n+2) mod C, and Q = floor((n+2)/C)-2.

a(n>=28) = 4^(C-R) * 5^R, where C = floor((n+2)/6), R = (n+2) mod 6.

Extensions

Additional comment and formula from David Applegate, Jul 21 2011

More terms from Reinhard Zumkeller, Jul 22 2011, Jul 21 2011

Additional comments, formulas, examples and CrossRefs from Jonathan T. Rowell, Jul 30 2011