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### Primes that are Sums of Primes

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Date: 06/22/99 at 03:38:59
From: Anand Anil Tikotekar

My question is about certain prime numbers. The number 5 is the sum
of two prime numbers (2+3). The number 5 is the third prime number and
5 = 3+2 (summing all primes up to 3, because 5 is the 3rd prime).
17 is the 7th prime, and 17 = 7+5+3+2 (the summation of all prime
numbers up to 7). Similarly, 41 is the 13th prime and 41 = 13+11+7+5+
3+2.

My problem is to find the next number(s) in the series 5, 17, 41, ...
Can you find the next number(s)? If there is no next number in the
series, can you prove this?
```

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Date: 06/23/99 at 16:03:21
From: Doctor Nick

Hi Anand -

This is a nice problem.

Let's start by defining some notation.

Let p(n) be the n-th prime, so p(1) = 2, p(2) = 3, etc.
Let s(n) be the sum of the first n primes,
so s(n) = p(1) + p(2) + p(3) + ... + p(n).

You have noted that

5 = s(2) = p(p(2)) = p(3)  =  5
17 = s(4) = p(p(4)) = p(7)  = 17
41 = s(6) = p(p(6)) = p(13) = 41

Your question is then: are there any other solutions to s(m) =
p(p(m))?

The answer is no. The way you can prove this is to show that

s(m) > p(p(m))

for all m greater than some number M. Then, check that for all m less
than M, the only solutions are m = 2, m = 4, and m = 6.

Now, how do we show s(m) > p(p(m))? This is the interesting part of
the problem.

What we need is some idea of how big p(m) is, since everything stems
from this. It's a very useful fact that p(m) is about m log m, where
log is the natural logarithm. In fact, you can find positive real
numbers A and B such that

A m log m < p(m) < B m log m

for all m greater than some bound.

In particular, it's true that for all m,

p(m) < 3 m log m.

From this, it follows that

p(p(m)) <  3 p(m) log p(m) < 20 m (log m)^2.

for all m.

Now, s(m) is pretty big. Since p(m) > m, it follows that

s(m) > 1 + 2+ 3 + 4 + ... + m = m*(m+1)/2.

So s(m) grows like m^2, while p(p(m)) grows like m(log m)^2, which is
smaller. Hence, for m sufficiently large, s(m) will be much bigger
than p(p(m)).

How big does m have to be?  Well, we need

m*(m+1)/2 > 20 m (log m)^2

which is

m+1 > 40 (log m)^2

which is true if m > 2429, by direct calculation.

Thus, if m > 2429, s(m) > p(p(m)). Hence, if s(m) = p(p(m)), then m
must be less than 2430. By checking s(m) against p(p(m)) for all m
less than 2430, we can find that m = 2, m = 4, and m = 6 are the only
solutions.

These estimates can be improved with some work, so that the range you
have to check can be reduced.

With this problem, you are venturing into the area known as analytic
number theory. I recommend you consider getting and reading a book on
the subject. One especially good one is by Tom Apostol, and should be
available in most college/university libraries, or from the publisher,
or bookstores (Amazon.com, for instance).

Feel free to write back if you have further questions.

Have fun,

- Doctor Nick, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Discrete Mathematics
High School Number Theory

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