'Super-Factoring' An Integer

[Leroy Quet]

I posted here about some sequence a while back where the sequence's 
closed-form was:

c(m) = 

the number of e()'s where:

e(e(e(...e(m)...))) = 0,

and e(m) is a nonnegative integer such that:
2^e(m) is the highest power of 2 dividing m.

This sequence had a very interesting palindromic structure which was 
generated recursively.
(I called this sequence's symmetric structure "super-palindromic".)


I am wondering now about extending, in a way, the idea somewhat of the 
above 
c-sequence.

If m =

product{p=primes, p|m} p^e(p,m),

where e(p,m) is always a positive integer,
we can factor e(p,m) still as:

e(p,m) = product{q=primes, q|e(p,m)} q^e(q,e(p,m)).

And we can do this, in a tree-like manner (I guess), until we get a 1 at 
the 'leaves' of the tree.

A small example:

m = 


2^(2^(3^1)*3^(2^1*5^(2^1)))*5^(2^(5^(2^1))*7^1)

(Oh, forgive me, please,
if I messed the parentheses..)

But, in any case, if we were to extend (with some modifications) the 
original c-sequence to involving all the primes (not just 2), we might 
ask here for:

C(m) = number of primes (and 1's) in *total* factorization of m.

So, for the m above, whatever that is...,
C(m) = 17.

(I bet this is not a very fast rising sequence...)


  As for this Big-C sequence,...

First, we have the two related sequences depending on whether we want to 
count the one's too or just the primes.

Second, C(m) can be generated recursively.

C(m) = sum{p|m} (1 + C(e(p,m))),

where C(0) = 0 if the 1's are not counted, and is 1 if the one's are 
counted.

(And the sum is over the distinct prime divisors, p, of m.)

3rd, we can generalize the idea so that C(m) = the sum of elements from 
{a(p)}, a sequence with prime indexes.
If we want a(p) to be added every time that the prime p appears in the 
'super-factorization' of m, we might generate the C-sequence by the 
recursion:
(the 'super-factorization' includes the factorization of the exponents 
{and continuing...} as well in the prime-factorization of an integer.)

C(1) = what to add each time a 1 appears at the factorization's "leaves" 
(or = 0 if the 1's are ignored).

C(m) = sum{p|m} (a(p) + C(e(p,m))).


[So,  also, 
sum{m=1 to n} C(m) =

sum{m=1 to n} sum{p=primes <= n/m} (a(p) + C(e(p,m)+1))]




Some examples (figured by hand, so may be wrong):

a(p) = 1, C(1) = 1:
1, 2, 2, 3, 2, 4, 2, 3, 3, 4, 2, 5, 2, 4, 4 ,4

a(p) = 1, C(1) = 0:
0, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 3

a(p) = p, C(1) = 0:
0, 2, 3, 4, 5, 5, 7, 5, 5, 7, 11, 7, 13, 9, 8, 6

a(p) = ln(p), C(0) = 0:

exp(C(m)) -> 1, 2, 3, 4, 5, 6, 7, 6, 6, 10, 11, 12, 13, 14, 15, 8

a(p) = C(pi(p)), C(0) = 1:
(pi(p) = order of prime p)

1, 2, 3, 4, 4, 7, 5, 5, 5, 8, 6, 8


Final questions:


How many m's total can be composed from
(have a super-factorization consisting of)

b(2) 2's , b(3) 3's, b(5) 5's, ...
for a given finite sequence of {b(p)}, p = primes?

(this seems to be a combinatorial problem mostly.)

In each 'branching' of the prime-factorization 'tree', each prime occurs 
at most once. (But any prime can reoccur at the same level of branching, 
but only if it has another parent-branch.)


And, is there anything remotely symmetric about the structure of C for 
any particular nontrivial sequence {a(p)}??

thanks,
Leroy
Quet