About Utterly odd numbers. Consider the following sequence defined here https://oeis.org/A079523 of utterly odd numbers: These are numbers such whose binary representation ends in an odd number of ones.

If $n$ is an utterly odd number, then show that $n+64$ is also an utterly odd number. I did some computer experiments and I think that this is true. But I am not sure how to prove it.
This is not true.

Let $127=1111111_2$. Note that $B(127)=7$. But, since $64=1000000_2$, $$127+64=10111111_2$$ so $B(127+64)=6$.
For a number to be utterly odd, the number of its binary digits is odd.
For example, we have the number 
Converting it into binary:

$$19 = 10011_2$$  where we have total five (odd number) of binary digits

and in the same way, if we see 18,

$$18 = 10010_2$$ It also consists of an odd number of digits in binary. 

So, conclusion:


  A number is called an utterly odd number if it consists of an odd number of digits in its binary representation(omitting zeros on the left side if any )


Now, we check the term n+64 if it is utterly odd or not.
64 has 7 binary digits i.e. 1000000
it will remain a 7 digit binary for $$X&lt;64$$
Because at n=64, 
the expression becomes,
$$ (64+64) =128 $$
Which in binary consists of 8 digits

For n=1 till 200,
the values of n for which $$(n+64)$$ is an utterly odd number are:

$$[0,63] U [192,448] U [960, 1983]$$

let's check the size of the intervals,

length of the first interval is $$63= 2^6-1$$

Second interval length = $$448-192=2^8$$

and Similarly for the third one,
$$1983-960=1023=2^{10} -1$$

So, the length of the interval k in which we get an utterly odd number is :

$$f(k) = {2^2k -1,    {k = Odd}}$$
and $$f(k) =  {2^2k ,  {k = Even} }$$

So, the final conclusion:


  $$n+64$$ is not an utterly odd number for every n, however, it behaves as an utterly odd number in some intervals