[seqfan] Re: A000108(n) ≡ 1 (mod 6)

[Emmanuel Vantieghem]

The conjecture is true up to  m = 20000.
Moreover, C(2^n -1)  is divisible by 3 for all m > 8 (<= 20000).

2015-11-15 7:35 GMT+01:00 L. Edson Jeffery <lejeffery2 at gmail.com>:

> >I have tested the conjecture up to 250000 and found no other terms except
> for 0, 1 & 31. Bob
>
>
> Hello Bob,
>
> Are you saying that you tested the conjecture by checking each C(n) modulo
> 6, for n=0..250000? If so, then there are only 17 odd C(n) within that
> range (unless I miscounted), so the sample size is too small to draw any
> conclusions.
>
> We know that C(n) is odd if and only if n = 2^m - 1, where m >= 0 is an
> integer. Using that, we could test the conjecture only for those C(2^m -
> 1), m=0,1,... . But the number of digits of C(2^(m+1) - 1) appears to be
> roughly twice that of C(2^m - 1) which causes obvious problems with
> computations. The number of digits of C(2^m - 1), for m >= 0 (as counted by
> Mathematica), is the sequence starting
>
> {1, 1, 1, 3, 7, 17, 35, 74, 150, 304, 612, 1228, 2460, 4926, 9857, 19721,
> 39449, 78905, 157818, 315644, 631296, 1262601, 2525212, 5050435, 10100879,
> 20201769, 40403550, 80807112, ...}
>
> I tested the conjecture for C(2^m - 1) up to the still very small m = 31,
> or n = 2147483647, using Mathematica, but got an overflow error (or
> something like that) because C(2147483647) is already so large: I guess
> that C(2^31 - 1) has a little more than 1292913800 digits. So is there a
> better way to test these numbers?
>
> Finally, note that if C(n) is odd, then C(n) == 1 (mod 3) implies that C(n)
> == 1 (mod 6) (or 1 (mod 12)) because also C(n) == 1 (mod 4).
>
> Ed Jeffery
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>