Sequences Involving Binary & Prime-Factorization

[Leroy Quet]

I asked for the asymptotics of {a(k)} and {b(k)} in another email.

We can rewrite the b-recursion (which the original version I give below 
in copied message), so as to get:

b(m) =
 b(floor(sqrt(m))) +  sum{k=1 to floor(sqrt(m))} a(k) pi(m/k^2),

where pi(x) is the number of  primes <= x, for x = positive real.

So, very roughly (and shunning rigor),

I assume, using this new recursion, that

b(m) = O(pi(m)) = O(m/ln(m)),

using the prime-number theorem (of course).  

More precisely (and still unrigorously), I get:

limit{m -> oo}

(b(m) - b(floor(sqrt(m)))) *ln(m)/(m - sqrt(m))

=  sum{k=1 to oo}  a(k)/k^2 = C,

which of course converges, since the a's are each either 0 or 1.

So, on average, each a(k) has a
C/ln(k) 
probability of being 1.

Am I right in each of my assumptions? 

And, what is the constant C =
sum{k=1 to oo}  a(k)/k^2
numerically?

thanks, 
Leroy
Quet


I wrote:
>First, we have the sequence formed by simply letting
>
>a(m) = 1, if the sum of all exponents of the prime-factorization of m has 
>no carries when summed in in base-2.
>
>a(m) = 0, if there are any carries in the summing of the exponents of the 
>prime-factorization of m.
>
>So, for example, 
>
>a(12) = 1 because 12 = 2^2 *3^1,
>and, in base-2,  2 = '10', 1 = '1',
>and '10' and '1' have their ones in different positions.
>
>But, 
>a(24) = 0, because 24 = 2^3 *3^1,
>and, in base-2, 3 = '11', 1 = '1',
>which both share a rightmost one.
>
>Unless I made an error, this sequence is not yet in the Encyclopedia Of 
>Integer Sequences.
>
>a(k) -> 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1,...
>
>(Defining a(1) = 1 makes b()-recursion below work.)
>
>
>Now, let b(m) = 
>
>sum{k=1 to m} a(k).
>
>b(m) gives the number of positive integers  <= m and with an a() of 1, 
>obviously.
>
>My main math result related to these sequences:
>
>(a somewhat interesting method of calculating {b()} using recursion):
>
>b(1) = 1;
>
>
>b(m)  =
>
>
>b(floor(sqrt(m))   +  sum{p=primes<=m} b(floor(sqrt(m/p))),
>
>
>
>(or, if we choose to define 1 as a prime:
>
>b(m)  =
>
>sum{p=primes<=m} b(floor(sqrt(m/p)))  )
> 
>
>And, unless I made an error, this sequence is not yet in the Encyclopedia 
>Of Integer Sequences either.
>
>b(k) -> 1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 11, 11, 11, 12, 13,...
>
>thanks,
>Leroy Quet