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).

Example:

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

isqrt(r(x))=r(isqrt(x)).......(1.a)

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?

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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)?

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

Question 3:

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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"

Solution:

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:

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:

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:

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