[seqfan] Re: A combinatorial problem
heinz at hs-heilbronn.de
Tue Aug 3 16:11:55 CEST 2010
with the help of a computer and a little program it is a matter
of a few seconds to compute several hundred values of this
sequence. I will not do it by hand, because this is error prone
and wastes my time.
So I computed a(p*q*r*s) = 5712 for distinct primes p,q,r,s.
Now we have:
a(p), a(p*q), a(p*q*r), a(p*q*r*s) = 1, 2, 18, 5712
which looks like the beginning of sequence
Can you prove that there is any connection?
Vladimir Shevelev schrieb:
> 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), but I absolutely do not agree that a(n) is calculated "so easily". Try, please, find a recursion for it. Try, please, calculate by handy a(p*q*r*s). So I think that a(n) is very hard calculated (at least, till now).
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