Sequences Involving Binary & Prime-Factorization

[Leroy Quet]

[posted also to sci.math]

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