Problems & Puzzles: Puzzles

Puzzle 194. The Palinpoints

Joseph L. Pe defined (see his paper) an entertaining functional equation:

f(r(x)) = r(f(x))      ....................... (1)

  • x is an integer
  • f(x) is a given arithmetic function of x, and an integer itself
  • r(x) means the digital-reverse of the integer x

The palinpoints are all these x values that solve the equation (1) for a given specific function f(x).


Let f(x)=isqrt(x); isqrt(x) is the integer square root of x. Thus (1) becomes:


Palinpoints =  1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 67 76 77 88 89 98 99 100 121 131 141 144 154 164 169 179 189 400 441 451 461 484 494 505 515 525 900 961 971 981 ...and so on

Example: x=154 then isqrt(r(154))=isqrt(451)=21 & r(isqrt(154)=r(12)=21

Pe has obtained the earliest palinpoints for several well known arithmetical functions f(x) such as: x2, φ(x), σ(x), p(x).

Maybe Pe named palinpoints the solutions of (1) because usually these are:

  1. palindromic numbers or
  2. pairs of reversal numbers.

But not always happens b.

Question 1.

Can you establish when b. is not followed?


If f = p(x), being p(x) the x-th prime number, the equation (1) becomes:

p(r(x)) = r(p(x))...........(1.b)

For (1.b) Pe has found only seven palinpoints:

x = 1, 2, 3, 4, 5, 12 & 21 ( see EIS A069469 )

Question 2.

Can you find one more palinpoint to (1.b)?


Some minutes before the final edition of this page, Joseph sent two new puzzling questions about his equation (1):

Question 3:

a) What is the minimum number of palinpoints an arithmetical function can have? b) Are there such functions with no palinpoints at all?


For tackling the question 2 maybe you'll find useful to know the very efficient strategy applied by the authors (?) of the well known "The Nth Prime Page"


Question 1 was solved by J. K. Andersen:

I interpret the question as: Assume x is a palinpoint. Determine when r(x) is not a palinpoint.

The assumption is f(r(x)) = r(f(x))
r(x) is by definition a palinpoint if and only if: f(r( r(x) )) = r(f( r(x) ))
Let r2(x) = r(r(x)), insert the assumption and the requirement is:
f(r2(x)) = r( r(f(x)) ) = r2(f(x))
r2(x) is x with any ending 0's removed.
r(x) is a palinpoint if and only if:
f(x with ending 0's removed) = f(x) with ending 0's removed.

If neither x nor f(x) has ending 0's then it is trivially satisfied.
It can be satisfied in other cases, e.g.. x=10 is a palinpoint for f when: f(10)=20 and f(1)=2. Here r(10)=1 is also a palinpoint.

Question 2 was solved also by J. K. Andersen!!!

The next palinpoint is x = 8114118 with p(x) = 143787341.

Later he added:

I did not know the problem in advance. I modified a prime-finding C program for the search and got the result in a few seconds running. I continued searching for bigger palindromic solutions (and only that) with p(i)<2^32 (meaning i<=203280221) but found none. I then realized a number with an even number of digits could never be a palprime (except 11). I could have stopped at 10^9 but it only took a few minutes anyway.

Wondering if others had noted p(8114118) = 143787341, I just did a Google search on "143787341". I was surprised to see that you had noted it, even in one of your own puzzles: puzzle 51 with no solutions yet. I saw you also submitted it to Prime Curios and EIS. It is also at Mathworld.

It appears that the search done by J.K.A. was simplified using the following lemma pointed out by him in his email: "x is always a palinpoint of f if x and f(x) are both palindromes".

In any case I wonder if this is the claimed "next" 8th palinpoint or just - as asked - "another palinpoint". BTW, as a curio note, 143787341 is palprime!

J.K.A. replied this to my worries:

Maybe I was a little unclear in my earlier post. I made an exhaustive search to x = 10 million and found the first new solution p(8114118) = 143787341. This should be guaranteed to be the 8th solution if my program is correct. It was after this I switched to only looking for palindromic solutions. There may still be non-palindromic solutions with x above 10 million.


Question 3 was solved by Frank Rubin and J. K. Andersen:

F. R.:An answer to Part 3 occurred to me right away. Let f(x) be 10^(abs(x)+2)-2.
So f(0)=98, f(1)=998, f(2)=9998, etc. Then there are no integers x and y for which f(x)=r(f(y)) so there cannot be any integer n for which f(r(n))=r(f(n)). f has no palinpoints.

J. K. A.: I don't know whether any limitation on "arithmetical function" is intended. It is easy to construct functions with no palinpoints at all. f(x) = c for a non-palindromic constant c has no palinpoints since: f(r(x)) = c, and r(f(x)) = r(c). I think f(x) = 10 is the simplest function without palinpoints.

Asked by me if he could produce a non-constant f(x) function with the asked property, he replied a solution similar to that one sent by Rubin:

It is still easy to construct "artificial" functions without palinpoints e.g.. f(x) = 10^(x+1)+2.

r(f(x)) = f(r(x)) will never be satisfied if the set of all function values has no reversible pairs, i.e.. if there is not even a solution to r(f(x)) =f(y) for any combination of x and y.

I have not looked for "normal" functions. I don't know what that would mean anyway. Maybe a small challenge would be a function which can take any value.


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