[seqfan] Re: Asymptotic formula

allouche at math.jussieu.fr allouche at math.jussieu.fr
Sat Apr 30 12:43:13 CEST 2011

Here is a hint to obtain an upper bound (I do not write details
as I am on a slow connection today).
Define C(n) = a(n+1) - a(n), get the easy formula giving
C(2n+1) in terms of C(n) and C(2n) in terms of C(n).
Hence a formula giving C(k) in terms of the binary digits
of k, involving the sum of digits and the length of the binary
expansion. Now a(n) = \sum C(k).
I am just afraid that the difficulty will now be in bounding
that last sum in a smart way (i.e., yielding the required bound)...


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