[seqfan] Re: A000108(n) ≡ 1 (mod 6)

[L. Edson Jeffery]

>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