A094913 extension

[David W. Wilson]

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).

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