Problems & Puzzles: Puzzles

 Puzzle 369. Approximations to unit adding prime reciprocals. Anton Vrba asks for approximation to the unit (1) by the algebraic sum of prime reciprocals. An approximation to unit means here that the algebraic sum of the prime reciprocals is a rational number a/b, such that a= b+1 or b-1. Example (by Vrba): 1/2 + 1/3 + 1/5 - 1/31  = 931/930 The asked solutions assumes that all primes used are distinct. Anton has sent examples using up to nine (9) distinct primes. BTW, all his solutions use the prime "2" but  they are not minimal in the sense stated below. A solution for a given quantity k of primes used is said to be minimal if b(k) is the smallest possible for this k value. Questions. 1) Please send your minimal solutions for 1<=k<=10, and then for k=20, 30, & 50. 2) If possible redo your solutions for 1, using only odd primes.

Solutions came from J. Wroblewski & D. G. Degiorgi.

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Wroblewski wrote, for the Q1:

Somehow I feel you are being very optimistic asking for a solution
to Puzzle 369 with k=50 :-))))))

I didn't work out any efficient search method, since I wasn't able
to find any solution with k=9, but I doubt significant improvement
of Anton's 9 is realistic, since numbers involved seem to get large
pretty quickly with growth of k.

I can't wait to see the Puzzle solution ;-)

Yours,
Jarek

1 -> 1/2 = 1/2 = 1 - 1/2
2 -> 1/2 + 1/3 = 5/6 = 1 - 1/6
3 -> 1/2 + 1/3 + 1/5 = 31/30 = 1 + 1/30
4 -> 1/2 + 1/3 + 1/11 + 1/13 = 859/858 = 1 + 1/858
5 -> 1/2 + 1/3 + 1/5 - 1/13 + 1/23 = 8969/8970 = 1 - 1/8970
6 -> 1/2 + 1/3 + 1/5 - 1/37 - 1/43 + 1/59 = 2816069/2816070 = 1 - 1/2816070
7 -> 1/2 + 1/3 + 1/11 + 1/13 + 1/37 - 1/41 - 1/263 = 342317119/342317118 = 1 + 1/342317118
8 -> 1/2 + 1/3 + 1/5 - 1/7 + 1/17 + 1/29 + 1/73 + 1/397 = 3000402931/3000402930 = 1 + 1/3000402930

***

Daniele wrote:

Regarding Q1:

I tried a brute force approach, but it worked only up to k=8 and found no solution for k=9.
The idea was to try all possible primes < 282 (the first 60) in the first k-1 position.
This finds for k from 3 to 8 respectively 2,5,11,8,11,1 solutions and the minimal are

K=3: 31/30 = 1/2 + 1/3 + 1/5
K=4: 859/858 = 1/2 + 1/3 + 1/11 + 1/13
K=5: 8969/8970 = 1/2 + 1/3 + 1/5 - 1/13 + 1/23
K=6: 2816069/2816070 = 1/2 + 1/3 + 1/5 - 1/37 - 1/43 + 1/59
K=7: 342317119/342317118 = 1/2 + 1/3 + 1/11 + 1/13 + 1/37 - 1/41 - 1/263
k=8: 3000402931/3000402930 = 1/2 + 1/3 + 1/5 - 1/7 + 1/17 + 1/29 + 1/73 + 1/397

The solutions for k=1,2 are trivially 1/2,5/6 = 1/2 + 1/3.

Regarding Q2:
The sum of the reciprocal of the first 8 odd primes (111429982/111546435) is  such that a < b-1.
Replacing one of them with a larger one gives a larger difference between a and b (as the value decreases and the denominator increases).

It follows that there are no solutions without 2 if k is 8 or less.
On the other side, if k is odd and only odd primes are used, then both a and b are also odd and thus cannot differ by 1.

It follows that the smallest k that could allow a solution in odd primes is 10, and it is not casual that all solutions you get from Anton contains 1/2.

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Here is appropriate to exhibit the solutions (not always minimal, as said above) sent by Anton when he proposed this puzzle:

K=1: 1/2
K=2: 1/2+1/3 = 5/6
K=3: 1/2 + 1/3 + 1/5  = 31/30
K=4: {2, 3, 5, -29} = 31/30 -1/29 =  869/870
K=5: {2, 3, 7, 41, -1721} = 1723/1722 - 1/1721 = 2963561/ 2963562
K=6, 7 & 8: {2,  3,  11,  13, -863, –148091, -109654573313, +12024125448565746369281}
K=9: {2, 3, 5, -13, 23, 8971, 80469869, -6475399897347037, -5990114832937424569918790178587}

Regarding the case K=9, Anton added:

(This series also is valid for the first 5,6 and 7 terms and for 8 terms the numerator and denominator differ by  7)

***

As a matter of fact, I can see now - after receiving the solutions sent by Wrobleswki & Daniele - that the 'minimal condition' was never stated by him, but by me at the moment of editing the puzzle.

Another thing never stated explicitly by Anton, but inducted by me studying his examples, was that when you have a solution a/b for K primes, then you can get a new solution for K+1 primes if at least one of the following is prime: b+1 or b-1 (some times you have the luck that both b+1 & b-1 are primes -twins - and this time you can get two new solutions each using K+1 primes).

The four (4) algebraic laws that I have obtained behind this fact are:

(b+1)/b -1/(b+1)= [b(b+1)+1]/[b(b+1)]..........(1)

(b+1)/b -1/(b-1)= [b(b-1)-1]/[b(b-1)]...........(2)

(b-1)/b +1/(b+1)= [b(b+1)-1]/[b(b+1)]..........(3)

(b-1)/b +1/(b-1)= [b(b-1)+1]/[b(b-1)]...........(4)

and you will use the appropriate formula to the particular situation you face.

Example:

From the solution for K=2, a/b=5/6 = 1/2+1/3, it happen that:

a=b-1, and both 6-1 & 6+1 are primes so you can use 3 & 4 in order to obtain two new solutions for K=3:

5/6+1/7 = 42/43, by (3)

5/6+1/5 = 31/30, by (4)

Almost for sure Anton obtained this way the chain of solutions for K=6, 7 & 8. Perhaps this was the way he obtained his solution for K=9?

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One more thing, using the formulas shown in the above paragraph we can obtain two solutions K=9 from the solution K=8 obtained by both Wroblewski & Daniele, because... 3000402930 -1 & 3000402930 +1 are both prime numbers! Nevertheless, no one of these two solutions are expected to be the minimal one for K=9, but both are certainly smaller than the K=9 solution obtained by Anton.

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Wroblewski continued hunting more solutions (I suppose) with the aid of the four equations shown above, disregarding the asked minimal condition. His results are so many (281 at this very moment, Aug, 18, 06) that I prefer just to link to his own page: http://www.math.uni.wroc.pl/~jwr/pp369.txt . He has gotten three examples with the k value 12.

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Fred Schneider sent two solutions.

A solution for n=10 (this is also the only known Giuga number with 10 factors):

(x+1)/x =

1/2 + 1/3 + 1/11 + 1/23 + 1/31 + 1/47059 + 1/2217342227+1/1729101023519 +
1/8491659218261819498490029296021 + 1/58254480569119734123541298976556403

where x=4200017949707747062038711509670656632404195753751630609228764416142557211582098432545190323474818

A solution for n=11:

This is based on the above solution, as x+1 is a prime number.

(x^2+x+1)/(x^2+x) =
1+1/17640150777867267329330115650723717627898516596703613717910645022270320079
74907238552397132826470002218371504833064952368953752992545025252572554603713540
2846075397352168769479761231988203607942 =

1/2 + 1/3 + 1/11 + 1/23 + 1/31 + 1/47059 + 1/2217342227+1/1729101023519 +
1/8491659218261819498490029296021 + 1/58254480569119734123541298976556403 -
1/4200017949707747062038711509670656632404195753751630609228764416142557211582098432545190323474819

Sorry, I don't know if these are minimal or not. I suspect not. :)

¿Giuga?... See this paper: http://www.ams.org/mcom/2000-69-229/S0025-5718-99-01088-1/S0025-5718-99-01088-1.pdf

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