n such that there is one sequence of length n having equal sum and product
hv at crypt.org
hv at crypt.org
Fri Apr 8 04:13:28 CEST 2005
"David Wilson" <davidwwilson at comcast.net> wrote:
:Hugo van der Sanden wrote:
:
:> It doesn't appear to be being treated as a "trivial" solution so I guess
:> it is somehow considered invalid: are the a_i constrained to be positive
:> integers? I think the book and the EIS sequence would both benefit from
:> clarification.
:
:Your observation is correct. In the title, "sequence" would be more
:accurately replaced with "multiset of positive integers".
:
:> I think it is also worth mentioning the minimal constraint that terms > 2
:> in A033179 must be of the form p+1 (since [ a+1, b+1, <1> x (ab - 1) ]
:> is always a solution for k = ab+1, and non-trivial when a >= b >= 2).
:>
:> Hugo van der Sanden
:
:Pretty observation. I will submit the corrections.
Thanks, though I'd be surprised if it hadn't been noted before.
Similarly when there are precisely three elements > 1, we will have
k >= 5 and odd (and therefore already catered for) except when there
are two even elements and one odd one, giving:
k = 2(4abc + 2bc - a - b - c + 1)
for the multiset [ 2a+1, 2b, 2c, <1> x (k - 3) ], with a, b, c all >= 1.
Setting b = c = 1 gives k = 6a + 2, which caters for all primes of the
form 3a + 1, so only k = p + 1 : p == 5 (mod 6) are still candidates.
Similarly setting b = 2, 3, 4 etc rules out primes of the form 7a+3,
11a+5, 15a+7 etc; when c = 2 we rule out 15a+9, 23a+15, 31a+21 etc,
but the returns quickly diminish, and I don't see a simple way to
characterise the primes that remain.
With four elements, we get (a,2,2,2) taking out primes of the form
7a+4, (a,3,2,2) taking out 11a+7, and so on.
I suspected a proof that the set is finite (and complete as listed)
would come from finding a prime modulus that is covered by an
appropriate selection of partial multisets. However I note that
the contributions to a prime modulus p come from the various possible
factorisations of (p+1), so we need a p+1 that has at least p
distinct factorisations - that is we need A1055(p+1) >= p.
On inspection it seems unlikely that the peaks of A1055 increase fast
enough for any value to satisfy - while ln(A1055(n))/ln(n) does increase
with successive maxima, A1055(n)/n seems to be decreasing: calculating
the values for each n listed in A25487, the best was A1055(1440) = 171
(with ln(1440)/ln(171) =~= 0.707, strangely close to 1/sqrt(2)).
I will try extending my calculations to check A1055(n) for higher
values, but the evidence so far suggests it won't help - is there
an upper bound on A1055(n)/n someone can find that would save me
the effort?
And if there is no p > 3 such that A1055(p+1) >= p-1, I don't think
a covering set is possible. Which suggests that I won't be able to
establish the completeness of A33179 this evening. :(
Hugo
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