a too-short sequence related to Latin squares
Hugo Pfoertner
all at abouthugo.de
Wed Sep 8 21:09:34 CEST 2004
Brendan McKay wrote:
>
> * Brendan McKay <bdm at cs.anu.edu.au> [040908 14:18]:
> > 96260050927125657231057045653340232713369826309730222706933915414681058441
>
> Interesting factorization:
> (408476406083) (1658631171224660115434923) (681082357) (29409077641)
> (1089775321) (3413191) (1907)
> Took Maple about a minute.
>
> B.
The factorization is what one expects for a randomly chosen number C of
this size. As a rule of thumb the expected number k of prime factors is
k=ln(ln(C)) +- sqrt(ln(ln(C))). For C~=9.62*10^73 we get k=5.1+-2.3. The
largest prime factor is expected to be < C^0.6065 in 50% of all cases,
the second largest PF < C^0.2117, so we expect 39 and 16 digits for the
top 2 factors. (From Knuth TAOCP vol. 2, pp. 383-384).
Maple seems to be rather slow doing factorizations. Dario Alpern's ECM
http://www.alpertron.com.ar/ECM.HTM (one of the better sequence
producing tools in my opinion) took 6 seconds for the complete
factorization + sum of 3 squares on an 800 Mhz Athlon.
Hugo
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