Progress on Sum of Prime Factors Function
For any positive integer N let s(N) denote the sum of the prime
factors of N. For example, s(12)=2+2+3=7. A previous note
described some investigations into this function, which led to a
consideration of the set of integers {m1,m2,..,mk} for any given
N such that mj + s(mj) = N. The article defined the function
tau(N) in terms of this set of numbers as follows
k
tau(N) = N - SUM s(m_j)
j=1
It was noted that tau(N) is positive for nearly all N, and the
average value of tau(N) increases as N increases. In fact, for
positive values of tau(N) it seems that for any x there is an
index k such that tau(N) > x for all N > k.
However, for a small number of integers N the value of tau(N) is
*negative*. The first such value is 239, for which tau(239)= -21.
The article asked if tau(N)=0 for any integer N. This is Problem
#9 on the MathPages Most Wanted List, and the answer is still
not known. However, some progress has been made, thanks to some
further numerical results computed by Jud Mccranie.
Jud checked integers up to 10^9 and found no cases with tau=0, but
he did continue to find occassional integers with relatively small
negative values of tau. (He found about 2000 in all.) Interestingly,
he noted that only a specific set of values occur in the negative
range. In particular he found the following 15 integers with the
tau(N) = -29:
N tau(N) terms factorization of N
97547759 -29 5 97 1005647
204393239 -29 5 2663 76753
223052759 -29 5 223052759
236437559 -29 5 5449 43391
341332559 -29 5 341332559
372043799 -29 5 41 113 131 613
476690039 -29 5 7 68098577
488336759 -29 5 23 21232033
492404039 -29 5 492404039
552351239 -29 5 11 311 161459
563508359 -29 5 31 18177689
633501359 -29 5 317 1998427
646689239 -29 5 7 92384177
677462759 -29 5 71 9541729
693707399 -29 5 7 11^2 819017
From this it's not hard to see what's happenning, and it's kind
of neat. It depends on the existence of certain Egyptian fraction
partitions of 1.
Suppose we can find an integer N such that each of the following
five numbers is a prime:
a = (N-2)/3
b = (N-3)/4
c = (N-4)/5
d = (N-5)/6
e = (N-19)/20
It follows that m + s(m) = N where m is any of the five numbers
2a 3b 4c 5d 19e
If no other "m value" exists for this N, then tau(N)=-29.
What's so special about the set of numbers {2,3,4,5,19}? Well, to
see why this works, remember that according to the definition of
tau(N) we have
tau(N) = N - (2+a) - (3+b) - (4+c) - (5+d) - (19+e)
Substituting the above expressions for a,b,c,d,e and separating terms
we have
/ 1 1 1 1 1 \
tau(N) = N - 33 - N( - + - + - + - + -- )
\ 3 4 5 6 20 /
/ 2 3 4 5 19 \
+ ( - + - + - + - + -- )
\ 3 4 5 6 20 /
The sum of the unit fractions in the first parentheses is 1, and
the sum of the complement fractions in the second parentheses is
4, giving the result tau(N) = -29 independent of N. So it turns
out the ancient Egyptian fractions play an important part in this
phenomenon. To make this work for some number besides -29 we would
need to find another sum of distinct unit fractions that sum to 1,
and whose denominators are all 1 greater than a prime (counting 4
as a prime because s(4)=4).
Let's say we have a set of k distinct primes {p1,p2,...pk} (again
allowing 4 to be considered as a prime) such that
p1 + p2 + ... + pk = M
and
1 1 1
---- + ---- + ... + ---- = 1
p1+1 p2+1 pk+1
It follows that if N is any integer such that each of the numbers
a1 = (N-p1)/(p1+1)
a2 = (N-p2)/(p2+1)
etc.
ak = (N-pk)/(pk+1)
is a prime integer then tau(N) = k-1-M. This pretty much rules
out finding an N *of this form* with tau(N)=0, because the sum of
the k primes will clearly always be greater than k-1. Still, I
find this to be a very interesting result.
This raises some interesting questions. For any fixed set of
primes {p1,p2,..,pk} are there infinitely many N that give all
prime values of a1,a2,..ak? Also, are all negative values
of tau members of families of this type? The answer to this
second question is no, because most of the integers with negative
tau are demonstrateably not of this form. For example, the very
first occurrance of a negative tau is tau(239)=-21, which has the
set of "m values"
158 = (2)(79)
177 = (3)(59)
188 = (4)(47)
203 = (7)(29)
209 = (11)(19)
In general, for any N such that (N-2)/3, (N-3)/4, etc are all primes
(and no other m-values exist) we have we have
/ N-2\ / N-3\ / N-4\
tau(N) = N - ( 2 + --- ) - ( 3 + --- ) - ( 4 + --- )
\ 3 / \ 4 / \ 5 /
/ N-7\ / N-11\
- ( 7 + --- ) - (11 + ---- )
\ 8 / \ 12 /
which reduces to
N + 1
tau(N) = ------- - 23
120
Notice that if N=2759 were such that (N-2)/3 etc were primes, then
it would give tau = 0. It actually happens that (2759-2)/3 and
(2759-11)/12 are both primes, but the rest of the conditions are
not satisfied, so tau(2759) does not equal zero. However, there
doesn't seem to be any a_priori reason why something like this
couldn't eventually happen for some N, giving a tau of zero.
Anyway, returning to the families of equal tau values, I've been
able to identify the family for most of the persistent small
negative values of tau that appeared in Jud's results. These
correspond to the following sets of "primes U 4"
pk tau(N)
---------------------- ---------
2 3 4 5 19 -29
2 3 4 5 29 59 -97
2 3 4 7 19 23 -53
2 3 5 7 11 23 -46
2 3 5 7 13 23 83 -130
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