Lets define n as the product of 2 and k distinct primes. We can split n in 2^k-1 distinct ways as a product of an even number and a product of primes. And in 2^k ways if we allow the split (1,n), which by the way would lead to another solution 1+210=211 in the example given. Not all these partitions will lead to a prime sum. But if they do the maximum number would be 2^k (for simplicity).

 

If n is a general even number, products of 2 and other primes (not all distinct), the number of splits are less compared to the situation above, due to the primes with multiplicity greater than 1.

 

So maybe a good way to measure the number of splits would be:

 

log(# distinct sums prime)/ (log(2)* #prime factors (besides 2), counted with multiplicity)

 

The maximum is 1, and is achieved by 2, 6, 30, 210 (if one allows the split (1,n)) and probably a whole bunch of other numbers (for instance 10, 22, 42)

***

Torbjörn Alm & Seiji Tomita wrote more or less the samethe same:

The first solution > 7 pairs is:

Number: 66378
Factor count: 32
Factors
1 2 3 6 13 23 26 37 39 46 69 74 78 111 138 222 299 481 598 851 897 962 1443 1702 1794 2553 2886 5106 11063 22126 33189 66378
Combinations: 15
2+33189=33191
3+22126=22129
6+11063=11069
13+5106=5119
23+2886=2909
26+2553=2579
37+1794=1831
39+1702=1741
46+1443=1489
69+962=1031
74+897=971
78+851=929
111+598=709
138+481=619
222+299=521

There are 4 more such solutions < 5000000, all with 32 factors ant 15 combinations. N+1 is prime.

Number: 186162
Number: 899970
Number: 1624462
Number: 3047410

***

Luis wrote:

Any primorial serves to construct a great deal of primes
with the sum of its two factors. This is a list:

Primorial         Number of primes
7# = 210              7
11# = 2310           11
13# = 30030          21
17# = 510510         40
19# = 9699690        70

***

Farideh wrote:

Note that for 210 we also have 1+210 is prime so for each divisor d of 210 we have d+n/d is prime.

 

All such numbers are squarefree because if p^2 divides such number n then p divides p+n/p so it can't be prime. Hence if n is a such number then number of divisors of n is of the form 2^k.

 

Smallest such numbers n for k = 1, 2, 3, 4 & 5 are respectively  2, 6, 30, 210 & 186162. So for 186162 we get 16 primes, namely if d divides 186162 then d+186162/d is prime.

 

Next term of this sequence is greater than 2*10^10

...

I didn't find a number with 32 corresponding primes yet.

But Next term of the sequence, 186162, 899970, 3047410, 603843982, 3162524682 , ... is 14581858930.

Tony D Noe wrote:

Instead of nontrivial divisors, it seems more natural to include all the
divisors of a number.  Then for 210 we obtain one additional prime: 1 + 210
= 211.

Let a(n) be the number of distinct primes of the form d + n/d, where d is a
divisor of n.  Then it is easy to see that for n > 1, we have a(n) > 0 only
if n is even.  This function a(n) has been studied by several authors.

See sequence http://oeis.org/classic/A088627 for a(2n).  Mattias Engelhardt
has calculated this function for n up to 10^6.  In sequence
http://oeis.org/classic/A091350 he tabulates the smallest k such that a(k)
= n for n up to 29.  I have extended the sequence up to n=90.

Let p# denote the primorial of p: p# = 2*3*5*7*11*...*p.  Then a(p#)
appears to be an increasing function.  The sequence a(prime(n)#) is
tabulated by Lei Zhou in http://oeis.org/classic/A103787.

In sequence http://oeis.org/classic/A080715 Matthew Vandermast is
interested in finding n such that d + n/d is prime for every divisor d of
n.  In this case, n must be an even number that is the product of distinct
primes.  It is easy to find n that are the product of 2, 3, or 4 primes.
In fact, 210 is the smallest example consisting of 4 primes.  The first n
consisting of 5 primes is 186162 = 2*3*19*23*71, which produces 16 primes
of the form d + n/d.  An interesting question is whether there are numbers
n with a greater number of prime factors.

Kok Seng Chua has studied this question and posted his results as sequences
http://oeis.org/classic/A128276, http://oeis.org/classic/A128277,
http://oeis.org/classic/A128278, and http://oeis.org/classic/A128279.  He
correctly states that the prime k-tuple conjecture implies that r primes
can be found to produce 2^(r-1) primes of the form d + n/d.  Let me explain
how this works for finding an n that is the product of r=6 primes.

The first 5 primes must be 2 and any four distinct odd primes.  Let's use
2,3,5,7,11.  Let the sixth prime be p.  So n=2*3*5*7*11*p.  We form all
possible values of d + n/d for divisors d of n, obtaining the following 32
linear polynomials:

 1 + 2310p, 2 + 1155p, 3 + 770p, 5 + 462p, 7 + 330p, 11 + 210p,
 6 + 385p, 10 + 231p, 14 + 165p, 22 + 105p, 15 + 154p,
 21 + 110p, 33 + 70p, 35 + 66p, 55 + 42p, 77 + 30p, 30 + 77p,
 42 + 55p, 66 + 35p, 70 + 33p, 110 + 21p, 154 + 15p, 105 + 22p,
 165 + 14p, 231 + 10p, 385 + 6p, 210 + 11p, 330 + 7p, 462 + 5p,
 770 + 3p, 1155 + 2p, 2310 + p

(Because we also require p to be prime, there are actually 33 polynomials.)
This is exactly the form that the prime k-tuple conjecture requires.  If
the conjecture is true, then a prime p exists.  The conjecture also gives
an estimate of the number of primes p < x that make the 33 linear
polynomials prime: C*x/(log x)^33 for some constant C.  This implies that
it could take a long time to find a solution to the 6-prime case.

Carlos Rivera wrote:

In the search for the smallest numbers having 32 prime sums, my best shot was 95881258 = 2*11*17*19*103*131 which has 27/32 prime sums.

***

Fred Schalekamp wrote:

After 5 minutes I found for 2310 (=2*3*5*7*11): 11 prime sums...By testing 30030 = (2*3*5*7*11*13) I found 21 prime sums...  for 9699690 = 2*3*5*7*11*13*17*19 gives 70 prime sums.

***

Fred Schneider wrote:

I picked primorials (r=primorial(p)) because d + r/d is clearly always
p unsmooth.  You could I suppose use any squarefree number as d and
n/d are still always relatively prime.

At pIndex=4, primorial=210, primeCount=8.
At pIndex=5, primorial=2310, primeCount=11.
At pIndex=6, primorial=30030, primeCount=21.
At pIndex=7, primorial=510510, primeCount=41.
At pIndex=8, primorial=9699690, primeCount=71.
At pIndex=9, primorial=223092870, primeCount=118.
At pIndex=10, primorial=6469693230, primeCount=263.
At pIndex=11, primorial=200560490130, primeCount=449.
At pIndex=12, primorial=7420738134810, primeCount=703.
At pIndex=13, primorial=304250263527210, primeCount=1385.
At pIndex=14, primorial=13082761331670030, primeCount=2423.
At pIndex=15, primorial=614889782588491410, primeCount=5502.
At pIndex=16, primorial=32589158477190044730, primeCount=8617.
At pIndex=17, primorial=1922760350154212639070, primeCount=18250.
At pIndex=18, primorial=117288381359406970983270, primeCount=29353.
At pIndex=19, primorial=7858321551080267055879090, primeCount=61970.
At pIndex=20, primorial=557940830126698960967415390, primeCount=103568.
At pIndex=21, primorial=40729680599249024150621323470, primeCount=209309.
At pIndex=22, primorial=3217644767340672907899084554130, primeCount=404978.
At pIndex=23, primorial=267064515689275851355624017992790, primeCount=853279.
At pIndex=24, primorial=23768741896345550770650537601358310, primeCount=1609502.
At pIndex=25, primorial=2305567963945518424753102147331756070,
primeCount=3008915.

Anton Vrba wrote:

A quick search yields 331170 has 29/32 prime.

Farideh reported more solutions with 16 prime sums, totalizing 11 solutions up today: