Problems & Puzzles: Puzzles

Puzzle 18.- Some special sums of consecutive primes :

1. S(p1->pK) = S(pK+1->pL )

I have found only two sum of this types :

2 + 3 = 5 ; S(p1->p2) = S(p3->p3 ) and

2 + 3 + … 3833 = 3847 + … + 5557 ; S(p1->p532) = S(p533->p733 )

Find 3 more examples

1. I have found that S(pk)@ pk = 0 for :

pk = 5, 71, 369119, 415074643

Find three k values more.

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At 31/08/98 Jud McCranie informs the following:

part 1:  "no more solutions for L < 57,442,974 (PL = 1,137,118,693)"
part 2: no more solutions for primes < 2^32.".

Later (7/7/2000) he added to part 2: "No others p < 29,505,444,491 (sum < 2^64)... but see A007506". According to that sequence it results that the part 2 was established time ago (?) by Robert G. Wilson.

***

By obliviousness I (C.R.) tackled (24/2/2001) again the question 1, but this time I recorded all the solutions such that S(p1->pK) - S(pK+1->pL ) <10. Thanks to that missing I discover the following two almost solution:

1+S(2 ->23117) =  S(23131 -> 33359)
S(2 -> 8358529) = S( 8358563 -> 11956103) +1

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Giovanni Resta found (May 2003) the following surprising solution for the question 1:

I would like to submit a new solution for Puzzle 18, "Some special sums of consecutive primes" In particular about the problem 18.1: p_1 + ...+ p_k = p_(k+1) + .... + p_h whose largest solution was

2+3+...3833 = 3847+...+5557 (k=532, h=733)

I found this new solution:

for k = 18151265107 and h = 25492021989, that is:

2+3+... + 468872968241 = 468872968243+ ...+ 667515565537 =  4169395490114624428834

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