[seqfan] Re: A combinatorial problem
heinz at hs-heilbronn.de
Wed Aug 4 23:09:36 CEST 2010
if you consider a computer program equivalent to a (closed) mathematical
formula, then we have solutions to A003043 and to A179926. But programs
tend to take more time to run as the input (or argument) gets larger.
A179926(n) is fast, if n is less than a few hundred. So it is easy to
a (p_1*p_2*p_3*p_4) = 5712 (and not 1080 or 1368). For example, see
all the 5712 permutations of divisors of 2*3*5*7=210 that are counted by
I do not have time to improve existing programs or to compute more values
of these sequences. But a submitter of a new sequence should make sure
that the first few submitted values and the formulas are correct and not
Vladimir Shevelev schrieb:
> Unfortunately, maybe, I don't know all possibilities of seqfan list, but I see at the first time that you "told me" in http://list.seqfan.eu/pipermail/seqfan/2010-August/005517.html. Maybe, it is a good idea, and, maybe, I did not take into account anything in my combinatorial construction. If you are right, then, in any case, as I anderstand, there not exist any general formula for A003043(n).
> You wonder how do my last results fit to my own submission? I have already written the following:
> "Alois, when I by handy calculated a(12) etc., this was concordant with a general model of computation of a(n) which turnes out to be only a lower estimate. I agree that a(12) is, indeed, 3 (not 2)", moreover, I sent Doug's and your corrections into A179926. In its turn, I can wonder: " how do your last conclusions fit to your own opinion that A179926(n) "is so easily calculated""? You state that A179926 contains A003043; so, maybe, you can easily calculate of A003043? In such case,
> please, add the sixth term after the Max's fifth one.
> Summarizing , we see that the problem is very difficult, but I believe that either you or I or anyone from our colleagues will find an unerring full combinatorial solution of it.
> Best regards,
More information about the SeqFan