[seqfan] Re: Binary Expansion of Constant A303340

[Paul Hanna]

Maximilian,
      Yes, I agree, finding a closed form expression for the binary digits
may be unreachable.
And I appreciate your insightful comments.

Yet a method of constructing the binary expansion may still be feasible,
perhaps by rewriting the sums in terms of powers of 1/2, like so

B = 1/2 + 1/2^2 + 1/2^3 + 0/2^4 + 1/2^5 - 1/2^6 + 1/2^7 - 2/2^8 + 2/2^9 -
3/2^10 + 1/2^11 - 1/2^12 + 1/2^13 - 5/2^14 + 7/2^15 - 7/2^16 + 1/2^17 +
3/2^18 + 1/2^19 - 12/2^20 + 16/2^21 - 9/2^22 + ... + b(n)/2^n + ...,
where
b(n) = Sum_{d|n} binomial(n/d-1, d-1) * (-1)^(d-1) for n>=1.

But now you have those messy negative coefficients to deal with.  Oh, well.

Thanks for your thoughts,
     Paul


On Wed, Apr 25, 2018 at 9:54 AM, M. F. Hasler <seqfan at hasler.fr> wrote:

> On Tue, Apr 24, 2018 at 6:21 PM, Paul Hanna <pauldhanna.math at gmail.com>
> wrote:
> > SeqFans,
> >       Can someone find a nice construction for the binary expansion of
> the
> > constant defined by https://oeis.org/A303340 ?
> > (2) B = 1/2 + 3/2^4 + 7^2/2^9 + 15^3/2^16 + 31^4/2^25 + 63^5/2^36 +
> > + ... +  (2^n - 1)^(n-1) / 2^(n^2) + ...
> > In base  2: B = 0.11100100010010111111111100111101001100001001001100...
> >
> > From the above series, it seems like the binary digits of this constant
> > should have some pattern or method of construction.
>
> In B = Sum_n  (2^n - 1)^(n-1) / 2^(n^2)
> the length of the numerator is  1, 2, 6, 12, 20, 30, 42, 56, = n*(n-1) for
> n>1.
> Since these numerators are shifted n^2 bits to the right,
> there is a large overlap between successive numerators:
> n=1: 0.1  (length 1 ; positions 1..1
> n=2: 0.0011 (length 2, positions 3..4 : no overlap with preceding term
> n=3: 0.000110001 (length 6, positions 4..9 : overlap of the leftmost
> digit with the preceding term)
> n=4: 0.0000110100101111 (length 12, positions 5..16 : overlap of 5
> digits with the preceding term)
> n=5: 0.0000011100001011110000001 (length 20, positions 6..20 : overlap
> of 11 digits with the preceding term)
> ...
> Although nonzero bits of numerator n only start at position n+1, they
> will influence earlier digits of several preceding terms:
> e.g., the second nonzero bit in B comes from terms n=2 and n=3, but
> the third nonzero bit comes from much later:
> the sum from n=3 and n=4 yields a 1 in position 4 (result of the sum
> of two bits 1 in position 5 of both),
> and only the sum of terms n=5 and n=6 will result in adding 1 to that
> bit, which yields the "final" 1 in position 3.
>
> I think it is highly nontrivial to get a closed expression for the
> nonzero bits, and one could add the list of indices of these as a
> separate OEIS sequence.
> Note that the indices of this "cons" sequence are, by current
> conventions, 0,-1,-2,-3, ...
> In the hope of maybe getting one day a simple formula for this, I
> think it would be much more sensible to index the terms 1,2,3,...;
>  in that case, the index would be the power of 2 which corresponds to
> the given digit.
> That sequence would start
> 1,2,3,6,10,13,15,16,17,18,19,20,21,22,23,24,27,28,29,30,32,
> 35,36,41,44,47,48,51,52,53,54,55,57,62,65,66,68,69,72,74,81,
> 82,86,87,88,...
>
> [As a side note, irrelevant for this problem, [[Warning: politically
> incorrect "free speech" following - please ignore!]]
> I've been suggesting for many years that "cons" sequences should
> generally be indexed in this very logical way, as to have:
> * const = Sum_{i >= offset} a(i)*b^(-i),
> * increasing indices for all sequences,
> * at most a finite number of "ugly" negative terms
> * no contradiction between indices in the "list" and in the b-file,
> * none of all the other problems arising from the IMHO weird current
> index convention for "cons" sequences, which will certainly change
> sooner or (possibly much) later.]
>
> One could also add the sequence of run lengths of 0's and 1's
> (independent of the indexing scheme):
> 3,2,1,3,1,2,1,1,10,2,4,1,1,2,2,4,1,2,1,2,2,2,5,1,1,4,1,2,2,
> 1,2,2,1,1,1,6,2,3,3,1,1,3,1,2,4,1,1,1,1,1,1,1,1,3,1,6,6,4,1,
> 1,2,1,4,1,1,4,....
> Neither for this nor for the run lengths of 1's or of 0's,
> 3,1,1,1,10,4,1,2,1,1,2,5,1,1,2,2,1,1,2,3,1,1,4,1,1,1,1,1,6,
> 1,2,4,1,6,3,1,2,1,1,1,1,3,2,2,1,1,1,1,1,2,2,1,1,1,2,1,3,2,2,
> 1,1,3,6,3,5,1...
> respectively for 0's:
> 2,3,2,1,2,1,2,4,2,2,2,1,4,2,1,2,1,6,3,1,3,2,1,1,1,1,3,6,4,1,
> 1,1,4,4,2,1,1,3,1,1,1,1,4,2,4,1,1,4,2,1,2,1,1,1,3,2,1,3,1,1,
> 4,1,1,1,3,1,...
> I or OEIS could find something.
>
> Cheers,
> Maximilian
>
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