Problems & Puzzles: Puzzles

Puzzle 446. S(p, q)=p.q

Anton Vrba sent the following nice puzzle:

Is the set { (2,5), (3,13), (5,31) and (7,53) } unique?

2+3+5 = 10 = 2*5
3+5+7+11+13 = 39 = 3*13
5+7+11+13+17+19+23+29+31 = 155 =5*31
and similarly
7+11+13+.+53 = 7*53


Q1: Find another two prime numbers p1 and p2 such that the sum of all prime numbers p, p1<=p<=p2 equals the product of p1 and p2.
 
Q2: Find another two prime numbers p1 and p2 such that the sum of all prime numbers p, p1<=p<=p2 is divisible by p1 and p2.

Q3: For p1 =53 find a prime number p2 such that the sum of all prime numbers p, p1<=p<=p2 is divisible by p2. (other values of p1 are 79, 89, 97, 139, 179 )

Q4. For Q1, can you theoretically forecast p2 (perhaps an upper bound) given p1?

 

Farideh Firoozbakht wrote (may 08):

Q2: p1=503 & p2=508213, 503+509+521+...+508213=40*503*508213.

Q3. For p1=97, p2 is equal to 86472413.
Because 97+101+...+86472377+86472413=86472413*2434182.

Q4: According to PNT, sum(prime(k), {k, m, x}) ~ sum(k*log(k), {k, m, x}).
But sum(k*log(k), {k, m, x}) ~ (1/4*x^2(2log(x)-1) - 1/4*m^2(2log(m)-1))
because Integrate(t*log(t), t)=1/4*t^2(2log(t)-1).

So if p1=prime(m) and p2 =prime(x) is the smallest prime p such that
s(p1, p)>=p1*p then x~ solution of the following equation.

x^2*(2*log(x)-1)-4*m*log(m)*x*log(x)-m^2*(2*log(m)-1)=0 (*)

According to the numerical results if p1=prime(m)>190 then p2< prime(m^1.6) .

So in our search for finding solution for Q1 we can use from this result.
For example if p1=541=prime(100) then p2 is less than prime([100^1.6])=
prime(1584)=13337.

It seems that from (*) there exist better upper bounds for p2, but the mentioned
upper bound for p2 that I wrote is useful for our search.

***

Hakan Summakoğlu wrote: (April 2011)

Q2: p1=5 , p2=9557887

5+7+11+13+17+... + 9557887=2935561623745=5*9557887*61427

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