[seqfan] Re: A191476 2^i*3*j and A134583 dimensions of cantor set

[Amiram Eldar]

Let v_3(n) = A007949(n) be the exponent of the highest power of 3 that
divides n.

Using the 2 comments/formulas below:
A033845: Six times the 3-smooth numbers (A003586). - Ralf Stephan, Apr 16
2004
A022329: a(n) = A134583(n) - 1. - Franklin T. Adams-Watters, Mar 19 2009
we have:
A191476(n) = v_3(A033845(n)) = v_3(6*A003586(n)) = v_3(A003586(n)) + 1 =
A022329(n) + 1 = A134583(n).

So A191476 = A134583.


On Sun, May 26, 2024 at 8:03 PM Richard J. Mathar <mathar at mpia-hd.mpg.de>
wrote:

> The comments in A191476 (concerning the ordered sequence of 2^i*3^j) and
> A134583 (Haussdorff dimension of which I don't know anything...) say that
> the sequences differ. The first 1000 terms of the b-files are the same.
> At which index/position start these sequences to differ?
>
> --
> Richard Mathar
>
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