# popular dilutions

Marc LeBrun mlb at well.com
Thu Oct 11 07:47:27 CEST 2001

```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).

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?

```