Problems & Puzzles: Puzzles

Puzzle 422. 3N consecutive prime JM Bergot sent another nice puzzle: I notice that 7+11=18 and 13+17+19+23=72. The pattern is for 3N consecutive primes: the sum of the first N divides the sum of the second 2N.   Q1. Find a moderately large example for N=2 Q2. How large N examples can you find?

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Contributions came from Farideh Firoozbakht & Enoch Haga.

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Farideh wrote:

Answer to Q1:

I found two smaller solutions for N=2
3+5=8 ;7+11+13+17=48 and 5+7=12 ;11+13+17+19=60
But I haven't found larger solution for N=2.

Answer to Q2.

I found the following two solutions for N=4.


19+23+29+31=102 ; 37+41+43+47+53+59+61+67=408
47+53+59+61=220 ; 67+71+73+79+83+89+97+101=660

And for N=19 the solution is 37+41+...+113=1433 , 127+131+...+337=8598.

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Enoch wrote:

If the ratio of 1 to 2 primes is to be maintained no matter the value of N,
if N=2 then 3,5,7 are the only starting primes producing an integral
solution for a total of 6 primes.


If the value of N=3, a total of 9 primes, maintaining the 1 to 2 ratio, then

a starting value of 2 produces an integral solution, namely 2+3+5=10 and
7+11+13+17+19+23=90, so that 90/10 = 9.

If the value of N=4, a total of 12 primes, maintaining the 1 to 2 ratio, we

find primes 19+23+29+31=102 and 37+41+43+47+53+59+61+67=408 so that

408/102=4. Also 47-61=220 and 67-101=660, so that 660/220=3.

Since the problem assumes the ratio is maintained, there will be an infinite
number of solutions greatly outnumbered by those non-solutions converging on
2.

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