[seqfan] Re: Exponent of highest power of 2 dividing n is equal to exponent of highest power of 3 dividing n

[israel at math.ubc.ca]

The number of members of the sequence <= N is the sum over k>=0 of the 
number of positive integers <= N/6^k congruent to 1 or 5 mod 6, thus

Sum_{k=0 .. log_6(N)} floor((N/6^k+5)/6) 
+ Sum_{k=0 .. log_6(N/5)} floor((N/6^k+1)/6)

This is within O(log N) of 2 Sum_{k>=0} N/6^(k+1) = 2*N/5.

So you should indeed get an error term O(log n).

Cheers,
Robert

On Sep 1 2016, Charles Greathouse wrote:

>Juri-Stepan Gerasimov submitted a nice sequence which turned out to be a
>duplicate of
>https://oeis.org/A064615
>
>It's not hard to prove that a(n) ~ 5n/2. But I can't find a reasonable
>error term. Can anyone help?
>
>For example, a(174142594) = 435356467 is at a record distance of 18 from
>the asymptotic at 435356485, and this is suggestive of a logarithmic error
>(which is what I would naively expect).
>
>Charles Greathouse
>Case Western Reserve University
>
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>Seqfan Mailing list - http://list.seqfan.eu/
>
>