Problems & Puzzles: Puzzles

Puzzle 52.- Sum of K consecutive primes equal to the sum of their reversible ones.

Let’s define pi as the reverse of Pi, being both prime numbers, such that pi ? Pi (this last condition prohibits Pi to be a palindrome). Some couples (Pi, pi’) are : (13, 31), (17,71), (113, 311), (157, 751), etcetera.

Now let’s ask for those consecutive primes P1, P2, … ,,Pk such that :

P1 + P2 + … + Pk = N =  p1 + p2 + … + pk for k=>2

(Of course, p1 + p2 + … + pk not necessarily are consecutive primes)

For K=2, the smallest two examples are :

32213 + 32233 = 64446 = 31223 + 33223

36353 + 36373 = 72726 = 35363 + 37363

But I haven’t found any example for K=3.

Maybe you can find the least case for some K=>3, or show a reason why these sequences can not exist.


Jud McCranie has obtained (5/5/99) other 29 solutions for the case K=2, for primes < 2^32, and no one solution for K=>3.This is his largest example for K=2:

 3738668363 + 3738668383 = 7477336746 = 3638668373 + 3838668373

***

Felice Russo wrote (10-11/05/99) "I didn't find any solution for k=4 and p<10^8...also for k=5 and p<10^8  I didn't find any solution. My feeling is that for k>2 there isn't any solution. But how to prove this...?"

***

J. C. Rosa found the next solution for k=2:

7326556207+7326556267=14653112474=7026556237+7626556237

And one more again the 20/11/01:

Looking at the first and/or last digits of the sums P1+P2 and p1+p2 starting from the first and /or last digits of P1 and P2, I got the following result :

The equality P1+P2=p1+p2 implies: P2=P1+20*d , d is an integer not equal at 5;50;500;...Thus P2=P1+20,40,60,80,120,140,....

Using this result I have found to-day a next solution for k=2: 7736776367+7736776387=15473552754=7636776377+7836776377

***

Later (Aug. / 2003) Rosa wrote again:

I have found the second and the third solution of this kind where the sum is a palindromic number :
 
132040201 + 132040261 = 264080462 = 162040231 + 102040231
 
31002320003 + 31002320023 = 62004640026 = 32002320013 + 30002320013
 
Now I am searching for the fourth solution....(12 digits)

...

Here are the 4 th ,the 5 th , the 6 th and the 7 th solution where the sum is a palindromic number :

1134202024301 + 1134202024321 = 2268404048622 = 1234202024311 + 1034202024311

3400334330033 + 3400334330053 = 6800668660086 = 3500334330043 + 3300334330043

3414420244133 + 3414420244153 = 6828840488286 = 3514420244143 + 3314420244143

34410133101413 + 34410133101473 = 68820266202886 = 37410133101443 + 31410133101443

...

I have found 15 new solutions where the sum is palindromic (that is to say :from the 8-th to 22-th).

Here is the 8-th :

114422212224401+114422212224421 = 228844424448822 =124422212224411+104422212224411

and the 22-th:

343110030011303+343110030011383 = 686220060022686 =383110030011343+303110030011343

(No other solution up to 10^15). Now , I stop the search.....but maybe a new challenge: Find a semi-titanic solution .

***

J. C. Rosa also sent (22 August, 2003) some solutions for k=3:

For k=3 and up to 10^11, I have only found two solutions :

33799199711+33799199741+33799199747 =101397599199 =74799199733+14799199733+11799199733

35285258231+35285258237+35285258291 = 105855774759 =19285258253+73285258253+13285258253

But maybe there are others solutions because I have not tested all possibilities up to 10^11....Now I stop the search my PC being really too slow...

But the most surprising thing is that four days before J. K. Andersen sent one solution for k=3 and two k=4, but remained unpublished by personal mine reasons and then unknown to J. C. Rosa. This is what Andersen wrote (18 August 2003):

Some solutions can be found with certain patterns.

For K=3, let A be any palindrome and consider the concatenations 14A01, 14A31, 14A91. Since 4+4+4 = 0+3+9, these 3 numbers have the same sum as their reversals. If the 3 numbers are consecutive primes and the reversals are also primes then it is a solution. One such solution is for A=0040831941491380400, giving consecutive reversible primes 14004083194149138040001 + 0, 30, 90

A possible pattern for K=4 is palindrome A in 13A11, 13A21, 13A41, 13A51. This requires 8 simultaneous primes of which 4 must be consecutive. Two solutions are: 13289766156665166798211 + 0, 10, 30, 40 and 13715818054445081851711 + 0, 10, 30, 40

The above are unlikely to be the least solutions for K=3 and K=4.

***

Then - regarding the J. C. Rosa aim- I asked immediately to Andersen for titanic or semi-titanic solutions for k=2. This is what he responded:

K=2 with my approach requires 4 simultaneous primes including the 2 reversals. The complexity at n digits is O(n^6). Titanic seems too hard. Semi-titanic is also hard, considering there is no apparent way to sieve without resorting to individual trial factoring (August 19) ...A pattern for K=2 is palindrome A in 11A01 and 11A21. A 301-digit solution: 11*10^299 + 1570220307030220751*10^141 + 01, 21. The prp solution was found with use of Michael Scott's Miracl bigint library. It was proved with Marcel Martin's Primo. No digits above 4 in A would give a palindromic sum. (August 23)

So the road has been opened, in order to get a titanic solution...

***

Giovanni Resta wrote (Nov. 2004)

I can confirm that the smallest solutions with K=3 are the two solutions reported by J. C. Rosa

The next two smaller solutions for K=3 are
769681186933 + 769681186969 + 769681186999 =
339681186967 + 969681186967 + 999681186967
956654456599 + 956654456629 + 956654456749 =
995654456659 + 926654456659 + 947654456659

while the 3 smallest solutions for K=4 are

775878563 + 775878569 + 775878577 + 775878599 = = 365878577 + 965878577 + 775878577 + 995878577

74858485829 + 74858485847 + 74858485853 + 74858485859 = = 92858485847 + 74858485847 + 35858485847 + 95858485847

744509905409 + 744509905439 + 744509905441 + 744509905499 = = 904509905447 + 934509905447 + 144509905447 + 994509905447

***

 

 

 

 

 

 


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