popular dilutions

[karttu at megabaud.fi]

Marc wrote:

> Start with 11 (just for example) which is prime.  We'll "dilute" it by
> inserting one 0 to the left of each digit, giving 0101 = 101, which is
> still prime.  But if we dilute it twice by inserting two 0s we get 001001 =
> 1001, which is composite.  Call n's "potency" the dilution at which n's
> primality dissolves in this way.
> 
> Thus all non-primes are impotent, while 11 has potency of 2, as do 13 and
> 17.  Then 19 jumps to potency 4 (since 19, 109, 1009 and 10009 are all
> prime while 100009 isn't).  But the potency of 23, the next prime, is just
> 1 (since 203 is composite).  And so on...
> 
> The above examples present decimal potency for readability, but I'm actually
> interested in binary potency:
>    2 = 10 --> 0100 = 4 so its (binary) potency is 1
>    3 = 11 --> 0101 = 5 --> 001001 = 9 so its potency is 2
>    5 = 101 --> 010001 = 17 --> 001000001 = 65 giving 2 again
>    etc
> (note that the 5 that appears as the first dilution of 3 has a different
> "successor" than undiluted 5 does).

A dumb question: Why? (it has a different successor in that case)


Yours,

Antti


> 
> The binary potency of n is given by the new sequence A064891:
>    0 1 2 0 2 0 1 0 0 0 1 0 1 0 0 0 2 0 1 0 0 0 2 0 0 0 0 0 3 0 1 0 0 0 0...
> 
> The binary potency of the n-th prime is now A064892:
>    1 2 2 1 1 1 2 1 2 3 1 1 1 3 3 3 1 3 1 1 1 1 2 1 1 1 1 3 1 1 1 1 1 3 4 ...
> 
> The smallest value with each binary potency is A064893, to the extent that
> I have been able to compute it:
>    1 2 3 29 149 4079...
> The next term, if it exists, is a prime greater than a million.
> 
> Need I say "more"?  Or is sixual potency somehow taboo?
> 
> Questions:
>    What is the next term?
>    Do all potencies appear?
>    Is the sequence monotonic?
>    Are there integers with infinite potency?
>    What can we say about the distribution of primes in the sequence of
> dilutions of n?
> 
>