by **sopadeajo** » Fri Jun 09, 2006 5:37 pm

Here is the set of distint number of permutations for the 1-off subset: S-1, of your 29-permutable set with 17 cells and 5 variables :

S-1={5,6,8,9,11,12,14,15,16,18}.

I am not certain if S-1 is complete. Aren´t there more elements missing from elements of the 29-permutable set in different positions???

Note that we get 50% of them that are multiple of 3.

So a 3 permutation structure appears too often.

For example: removing 6 in r8c2-->18 permutations, do not lead (by removing another cell ) to 17 or 13 permutations, because this 18 is mostly a 6*3 permutation.

Removing 6 in r8c3 (in a different 29-permutation), with 11 permutations ,is a 11=2+3+2*3. So you can obtain 5, by removing another number in the appropriate cell, but not 7=3+4=2+5 or 7=2+2+3.

I do believe that there is a different 29-permutable set, probably with more cells (or maybe not), which will lead to much more prime permutations.

In fact your 29-permutable set is not unique, neither your 11-permutable; and so your disproof is invalid.

Note too, that the minimal prime permutation of S-1, is 5=17% of 29.

So ,for the first 4 digits prime: 1009, we could probably reach a minimal 30% prime permutation with just 1 element off, and go on removing to find the lowest primes.

Nethertheless, i restate my conjecture to a nice question:

What is the maximal n , such that a p(n)-permutable set, will lead to p(k) permutable sets for all 1<=k<n, by removing 1 or more of the elements of the set, in successive operations ?(readd the element, if needed).

We´ll call them :n- Fully Prime Permutable Sets (n-FPPS)

Your 5-permutable (probably minimal) set, leaded to 3 and 2.

So, n>=3.

.

Could you try to build a 4-FPPS from a 3-FPPS, and then a 5,6,..,10?

I am not sure if a n-FPPS must be built from a n-1 one, or if this condition is not necessary.