Problems & Puzzles:
Puzzles
Puzzle 29. P_{i}
= P_{ i1}&nxtprm(P
_{i1}), P_{i}
= prime for i => 1
Patrick De Geest sent this puzzle
at September 10, 1998. In its own words:
"Step 1.
=======
Concatenation of prime p and nextprime(p) is a prime. (See A030459)
E.g. I.
smallest prime p = 2
nextprime(p) = 3
concatenate 2 and 3 and we get prime 23 [length : 2]
E.g. II. prime p = 5297
nextprime(p) = 5303
prime concatenation = 52975303 [length : 8]
Not so difficult I hear you utter
and indeed larger examples are easy to find. So we
continue with the next step. After all, we are warmed up now.
Step 2.
=======
Take a prime concatenation from step 1 and repeat the
procedure exactly as described previously.
E.g. I. (step 1)
smallest prime p1 = 467
nextprime(p1) = 479
prime concatenation = 467479 > becomes prime p2
!
(step 2)
prime p2 = 467479
nextprime(p2) = 467491
prime concatenation = 467479467491 [length : 12]
E.g. II. (step 1)
prime p1 = 25457
nextprime(p1) = 25463
prime concatenation = 2545725463 = prime p2
(step 2)
prime p2 = 2545725463
nextprime(p2) = 2545725499
prime concatenation = 25457254632545725499 [length :
20]
467 is the first one. The full
series goes like :
467, 941, 13681, 14461, 21787, etc.
The prime concatenations quickly reach considerable
lengths. But finding and testing them is still possible.
That's why I propose to move on to step 3. As a
coincidence our prime 467 is also the smallest prime that
allows three steps before reaching a prime concatenation.
Step 3.
=======
E.g. I. see {Step 2 > E.g. I} for the first two
steps.
(step 3)
prime p3 = 467479467491
nextprime(p3) = 467479467527
prime concatenation = 467479467491467479467527 [length :
24]
E.g. II. (step 1)
prime p1 = 959941
nextprime(p1) = 959947
prime concatenation = 959941959947 = prime p2
(step 2)
prime p2 = 959941959947
nextprime(p2) = 959941959953
prime concatenation = 959941959947959941959953
(step 3)
prime p3 = 959941959947959941959953
nextprime(p3) = 959941959947959941959989
prime concatenation =
959941959947959941959953959941959947959941959989
[length : 40]
The first three primes that make a
cycle of three steps possible are :
467, 941 and 959941
Now the level of difficulty reaches major proportions. Is
there a prime that gives us a cycle of 4 steps ? The
answer is yes.
Step 4.
=======
The smallest prime number that gives a cycle of four
steps is '1936227'. The final prime concatenation has
length 112 !
1936227 is up to now the only prime I found and tested OK
! The composite '7335450' is the second one in this
series but then again, it is a composite number that we
start the steps with, so not suitable for the puzzle.
(step 1)
smallest prime p1 = 1936227
nextprime(p1) = 1936237
prime concatenation = 19362271936237 = prime p2
(step 2)
nextprime(p2) = 19362271936243
prime concatenation = 1936227193623719362271936243 =
prime p3
(step 3)
nextprime(p3) =1936227193623719362271936273
prime concatenation =
19362271936237193622719362431936227193623719362271936273
= prime p4
(step 4)
nextprime(p4 =
19362271936237193622719362431936227193623719362271936447
prime concatenation =
19362271936237193622719362431936227193623719362271936273/
19362271936237193622719362431936227193623719362271936447
[length : 112]
After this adventure till step 4, I
need help from my puzzle comrades, to boldly go where
no...
Is there a prime that gives us a cycle of 5 steps ? I
don't know yet.
So, it's time to pose some final puzzle questions :
 P U Z Z L E Q U E S T I O N S .
 A.

 Find six more prime numbers that give cycles of three
steps.

 1. 467

 2. 941

 3. 959941

 4. ?

 ...

 9. ?



 B.

 Find two following prime numbers that give cycles of
four steps.

 1. 1936227

 2. ?

 3. ?



 C.

 Find the smallest prime numbers that gives cycles of '5
and more' steps".
Solution
Question A " Find six more
prime numbers that give cycles of three steps"
Jo
Yeong Uk has sent us an email (16/11/98) that
says: "I've tested prime numbers that give cycles of
any steps up to 1.8*10^7,and I found 15 numbers that give
cycles of 3 steps bigger than 959941. The number is 3396199, 4858943, 5696101
,6475643, 7566133, 7584253, 7592261 ,9305281, 9463877,
11430491, 13442243, 14374837, 15941473, 17414497,
17691997"
Question B
"Find a prime number that gives a cycle of four
steps"
Again Jo
Yeong Uk discovered at 17/11/98 that the first
prime number that gives a cycle of four steps is
127787377 (the first reported by Patrick De Geest  1936227  was not a prime number...)... and
today 22/11/98 he reports the second one:
1510818931.Finnally today, 29/11/98 he sent the third,
fouth and fifth ones: 3147482977, 3307903909 and
3621408103
***
I have
made my own program to find and/or verify the results for
this puzzle and this is the concatenated prime resulting
from the 127707377 found by Jo:
127787377 (initial number)
127787377127787431
(Step 1)
127787377127787431127787377127787497
(step 2)
127787377127787431127787377127787497127787377127787431127787377127787503
(step 3)
127787377127787431127787377127787497127787377127787431127787377127787503
127787377127787431127787377127787497127787377127787431127787377127787641
(step 4)
