A094913 extension

[Maximilian Hasler]

PS: sorry for the typo, should read "min" (in the text)

Also, I wanted to comment on your suggestion :
even if I agree that the empty substring should have been included (as
I wrote earlier), I think we should not change this >6 year old
sequence (insofar more as the "new" sequence already exists - I'm
confident that we'll soon have a proof for a(n)+1 = A006697(n));
maybe just add "nonempty" in the definition and if needed, write a comment
%C A094913 It would be more natural to include the empty substring,
which would result in the sequence b(n)=a(n)+1 ; all values computed
so far confirm the conjecture that a(n)+1 = A006697(n)."

Maximilian Hasler

On Dec 17, 2007 3:30 PM, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> It also confirms the formula I gave,
>
> A094913(n) = 2^(m+1)-2 + (n-m)(n-m+1)/2,
>
> with
>
> m = [ n+1-LambertW( 2^(n+1) * log(2) ) / log(2) ]
>   = integer part of the solution to 2^x = n+1-x.
>
> which is NOT based on the conjecture of Jovovic, but on the reasoning
> that you can construct the word of length n such that it contains the
> maximal possible number, equal to max( n-k+1 , 2^k ), of different
> substrings of length k.
> Maybe this was not very clear in my post, and also I shouldn't have
> written it with "b" instead of "2"...
>
> Maximilian Hasler
>
>
> On Dec 17, 2007 11:19 AM, David W. Wilson <wilson.d at anseri.com> wrote:
> >
> >
> >
> >
> > I wrote a brute-force program to compute A094913, which I ran over the
> > weekend. The results are as follows:
> >
> >
> >
> > a(n) = maximum number of distinct substrings of a binary string of length n.
> >
> >
> >
> > b(n) = number of length-n binary strings with a(n) distinct substrings.
> >
> >
> >
> > c(n) = decimal value of d(n) interpreted as a binary number.
> >
> >
> >
> > d(n) = lexically first length-n binary string with a(n) distinct substrings.
> >
> >
> >
> >        n    a(n)    b(n)  c(n)
> >
> >
> >
> >        0       1       1  null
> >
> >        1       2       2  0
> >
> >        2       4       2  01
> >
> >        3       6       6  001
> >
> >        4       9       8  0010
> >
> >        5      13       4  00110
> >
> >        6      17      18  000110
> >
> >        7      22      38  0001011
> >
> >        8      28      48  00010110
> >
> >        9      35      40  000101100
> >
> >       10      43      16  0001011100
> >
> >       11      51      80  00001011100
> >
> >       12      60     210  000010011101
> >
> >       13      70     402  0000100110111
> >
> >       14      81     644  00001001101110
> >
> >       15      93     852  000010011010111
> >
> >       16     106     928  0000100110101110
> >
> >       17     120     912  00001001101011100
> >
> >       18     135     704  000010011010111000
> >
> >       19     151     256  0000100110101111000
> >
> >       20     167    1344  00000100110101111000
> >
> >       21     184    3944  000001000110101111001
> >
> >       22     202    9276  0000010001100101111010
> >
> >       23     221   19448  00000100011001010111101
> >
> >       24     241   37090  000001000110010101111010
> >
> >       25     262   65602  0000010001100101001111011
> >
> >       26     284  107388  00000100011001010011101111
> >
> >       27     307  160760  000001000110010100111011110
> >
> >       28     331  220200  0000010001100101001110101111
> >
> >
> >
> > Where both are defined, a(n) = A094913(n)+1. I would suggest replacing
> > A094913 with a(n), since a(n) treats the empty string as a possible
> > substring.
> >
> >
> >
> > The new elements a(19) through a(28) support Jovovic's conjecture that a(n)
> > = A006697(n).
>



-- 
Maximilian F. Hasler (Maximilian.Hasler(AT)gmail.com)