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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