Problems & Puzzles: Puzzles

 

Puzzle 866. R(n^k) is prime for k=2, 3, ...

Abhiram R. Devesh ask for the minimal integers such that the reverse of successive powers 2, 3, 4, ... produce a prime number.

He provides the following examples:

  • For n=4 and k=2; reverse (4^2) = 61 -> prime number.
  • For n=89 and k<=3; reverse (89^2) and reverse (89^3) -> 1297 and 969409 -> both are primes.
  • For n=3244 and k<=4 the primes generated are 63532501, 48705383143 and 692349908447011.
  • For n=3244 and k<=5 the primes generated are 63532501, 48705383143, 692349908447011 and 422250654361652953.
  • For n= 3295489 and k<= 6 the primes generated are 12194774206801, 96151030549962898753, 146272659647782271189449711, 944614969971929473184850883686883, 1658904422328638893877276546926171190821

 Q1. Send your minimal solutions for k=7, 8, ...

Now I will make two small changes: let's make n to be a prime number and let's start k running since k=1.

This is what I have found:

k, n: primes
1, 2: 2
2, 37: 73, 9631
3, 3121: 1213, 1460479, 16504500403
4,10429: 92401, 140467801, 9853810034311, 18694641661692811
5,10282339: 93328201, 129013594627501, 9124210088606665117801, 14286816400200203701819087111, 996592728588610999557150173929639411

 Q2. redo Q1 for this new sequence, for k=6, 7, 8, ...

 


Contribution came from Jan van Delden and Emmanuel Vantieghem

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Jan wrote:

Q2:

 

6, 10859529881

I used search bounds as specified below (only for starting digit 1, until the first hit).

 

Q1,Q2:

 

Since the ending digits of R(n^m) should be in {1,3,7,9} the starting digit of n^m should be in the same set.
If this is true for values m, with  all m<=k, this will restrict the intervals where one should search n.

If the starting digit of n is equal to 1 and k>=6 we have: n should be in [10^a,10^a*f] with f=2^(1/k).
If k<6 there is a second interval where n might be.


If the starting digit of n is equal to 3 we have: n should be in [10^a*f,10^a*g] and for k=2s+1 or k=2s+2:
f=(3*10^s)^(1/(2s+1))
g=(4*10^s)^(1/(2s+1))

If k=5 or 6 we have f=3.1291 ,g=3.3145
If k=7 or 8 we have f=3.1386, g=3.2702

 

If the starting digit of n is equal to 7 we canít have m=2, since the numbers 49..64 either start with an even number or 5.

If the starting digit of n is equal to 9 we have: n should be in [f*10^a,10^a] with f=0.9^(1/k).

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Emmanuel wrote:

Q1 :
m  generates the six primes  R(m^i)  for  i = 2 to 7 : 
6131656137244631701, 4687780371448722592434398011, 6581567870392263594933439116371287411, 4209013642499904869343202306994378627082708811, 6942025800052486133299979790007600729941009796735379221, 4855198802235397910707696704196786012811051145957489672498582721

 
Q2 :
The prime  p = 9835884797  generates  6  primes  R[p^i]  for  i = 1 to 6 :
7974885389, 90213755893792644769, 375925129175150146948230965159, 1861066504084295533606008771053833259539, 75734523610423559008595363570321983115945739195029, 923265319911534838659059685817760318004507052104872426384509

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