[seqfan] Constant with 2-Adic Valuation Continued Fraction
Paul D Hanna
pauldhanna at juno.com
Thu Nov 18 20:04:19 CET 2004
Thanks Bob -
It is clear that the large partial quotients of x^2
are doubly exponential, likely of the form ~ c*2^(2^k).
Incidentally, there are actually many constants with
this '2-adic valuation power' property:
Let x(m) be defined by the continued fraction:
x(m) = [1;m,1,m^2,1,m,1,m^3,1,... m^A007814(n),...]
where A007814(n) = valuation(n,2);
then the contfrac(m*x(m)) equals contfrac(x(m)) but
interleaved with m's:
m*x(m) = [m;1,m,m,m,1,m,m^2,m,1,... m,m^A007814(n),...]
So there are an infinite number of constants with
that property.
But I believe that there is only one x(m) with the
property that
x(m)^k has doubly-exponential partial quotients
for some integer k, and that is x(2).
This makes one want to find the function:
F(z) = [1;z,1,z^2,1,z,1,z^3,1,... z^A007814(n),...]
but I was unable to find the coefficients for F(z) -
does F(z) exist?
At any rate, does this investigation imply that
there is some intimate connection between
2-adic valuation and the continued fraction?
Thanks,
Paul
-- "Robert G. Wilson v" <rgwv at rgwv.com> wrote:
Et al,
In Mathematica:
f[n_] := Block[{k = 0}, While[ Mod[n, 2^k] == 0, k++ ]; 2^(k - 1)]; RealDigits[
FromContinuedFraction[ Table[ f[n], {n, 133}]], 10, 105][[1]]
1, 3, 5, 3, 8, 7, 1, 1, 2, 8, 4, 2, 9, 8, 8, 2, 3, 7, 4, 3, 8, 8, 8, 9, 4, 0, 8,
4, 0, 1, 6, 6, 0, 8, 1, 2, 4, 2, 2, 7, 3, 3, 3, 4, 1, 6, 8, 1, 2, 1, 1, 8, 5, 5,
6, 9, 2, 3, 6, 7, 2, 6, 4, 9, 7, 8, 7, 0, 0, 1, 8, 4, 2, 0, 6, 4, 8, 2, 6, 0, 5,
4, 8, 4, 3, 1, 9, 6, 9, 7, 6, 0, 1, 7, 4, 6, 5, 6, 9, 7, 9, 6, 6, 8, 5
And the continued fraction of x^2 by ContinuedFraction[ FromContinuedFraction[
Table[ f[n], {n, 1000}]]^2, 50] are:1, 1, 4, 1, 74, 1, 8457, 1, 186282390, 1, 1,
1, 2, 1, 430917181166219, 11, 37, 1, 4, 2, 41151315877490090952542206046, 11, 5,
3, 12, 2, 34, 2, 9, 8, 1, 1, 2, 7,
13991468824374967392702752173757116934238293984253807017, 3, 4, 1, 3, 100, 4, 1,
59, 1, 2, 3, 3, 1, 3, 3, ..., .
The increasing PQ's are:
1, 4, 74, 8457, 186282390, 430917181166219, 41151315877490090952542206046,
13991468824374967392702752173757116934238293984253807017,
36598324105860613463778234224980053029230366701976106078497358565722574084911463505856982751401419117789561,
..., . But the positions of the above PQ's are: 1, 3, 5, 7, 9, 15, 21, 35, 71,
143, 291, 635, 1407, 2979, 6101, 12339, 25019, 50413, 101340, 202792, ..., out of
the 812695 that I generated. Also the last one is a 170781 decimal digit number.
Bob.
Paul D. Hanna wrote:
> Consider the constant (newly added to OEIS as A100338):
>
> x=1.353871128429882374388894084016608124227333416812118556923672649787...
>
> The continued fraction of this constant is A006519 (greatest power of 2
> dividing n):
> contfrac(x) = [1;2,1,4,1,2,1,8,1,2,1,4,1,2,1,16,...A006519(n),... ]
>
> This constant x has the special property that the
> continued fraction expansion of 2*x is equal to the
> continued fraction expansion of x interleaved with 2's:
> contfrac(2*x) = [2;1, 2,2, 2,1, 2,4, 2,1, 2,2, 2,1, 2,8,...
> 2,A006519(n),...].
>
> PARI code to get 1000 digits:
> \p1000
> CF=vector(1500,n,2^valuation(n,2));
> PQ=contfracpnqn(CF);
> x=PQ[1,1]/PQ[2,1]*1.0
>
> The continued fraction of x^2 is interesting: contfrac(x^2) =
> [1,1,4, 1,74, 1,8457, 1,186282390, 1,1,1,2,1,430917181166219, 11,37,1,4,2,
> 41151315877490090952542206046, 11,5,3,12,2,34,2,9,8,1,1,2,7,
> 13991468824374967392702752173757116934238293984253807017, ...]
> and some of the partial quotients of x^2 seem to grow exponentially.
>
> Has anyone seen this constant before?
> I wonder if it has some nice series representations as well ...
>
> Thanks,
> Paul
More information about the SeqFan
mailing list