[seqfan] Re: 4 interesting constants and 3 new sequences.

[Hugo van der Sanden]

Ah, the upper and lower bounds won't match up in the same way: e.g. 125 /
5^3 != 625 / 5^3 (mod 5), so the next digit won't be forced to the same
choice of {d, d+5}.

I'm not sure how n! works - we have n! divisible by 2^(n - g(n)) where g(n)
is (?) the index of the last zero in the binary representation of n. But you
are (presumably) stripping off the trailing zeros when you reverse, so
that'll be reducing the powers of 2 by O(n/4). Off the top of my head, I
don't know how that'll affect the limiting constants.

Hugo

2009/3/23 Simon Plouffe <simon.plouffe at gmail.com>

> Hello, this is interesting in the case of 2^n,
> it should then apply to n! (?) apparently the limiting constants
> are the same. In the case of 5^n, the 2 limiting constants
> do not add up to an integer, the sum is something like
> 1.0478702883108308... do you have an explanation for that ?
> me not.
> good afternoon,
> Simon Plouffe
>
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