[ December 28, 1999 ] [ Last update [ August 8, 2026 ]
More palindromic_free numbers

This plate is a sequel to WONplate 28

In a attempt to find the largest possible 'free of palindromic substrings' - or just palfree - numbers
- except of course the single digits themselves - like for instance in the following powers :

2 ⁵⁴ = 18014398509481984 - 17 digits
3 ⁶⁷ = 92709463147897837085761925410587 - 32 digits
355 ¹⁵ = 179236021709762370418314530975341796875 - 39 digits
5045 ¹² = 271852947987468014901832813869835078369140625 - 45 digits
Continuing with increasing the basenumber and scanning for increasing
exponents is the way to find more palfree numbers :
Here are some records
38014 ¹⁰ has 46 digits.
58265 ¹⁰ has 48 digits.
65871 ¹⁰ has 49 digits.
81193 ¹⁰ has 50 digits.
117825 ¹² has 61 digits.
1532037 ¹⁰ has 62 digits.
8069502 ⁹ has 63 digits.
80675095 ⁸ has 64 digits.
81568301 ⁹ has 72 digits.
102295816 ¹¹ =
12836203127935236019569704912413948293473049748637093270843840946054091629031657062793216
has 89 digits.
252069167 ¹¹ =
261045967093128764270426159425913502876812085791607246123902536935612014905142152845967251983
has 93 digits.
(You've other record numbers of this kind ! Please submit them to me and I'll display them here also.)

Carlos B. Rivera F. sent a method to produce
infinite large 'free of palindromic substrings' numbers.
First he gives two examples

To produce the palfree number 123123123123
multiply 123 with repunit 111111111111 and divide by 111

To produce the palfree number 1234123412341234
multiply 1234 with repunit 111111111111 and divide by 1111

and then he provides the General Formula

To produce N-N-N-N
multiply N*R(k*n)/R(n)

k = times N appears
n = digits of N
R(n) = (10^n–1)/(10–1) = 11....11 (n times)

Is this the beginning or the end of the palfree numbers story ?

[ December 5, 2021 ]

So many years later I resumed the topic and searched for
palfree Fibonacci numbers.

Here is the largest Fibonacci numbers I could come up with

Fibonacci(61) = 2504730781961 - 13 digits

The search went all the way up to Fibonacci(1200) so it looks
that the above Fibonacci number will be the largest palfree one.

The provisionally complete sequence is :

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 15, 16, 17, 18, 21, 23, 25, 27,
30, 32, 33, 37, 39, 42, 48, 58, 61

Strings of zero's are not allowed, otherwise 109 and 130
would be valid as well.

Let me redo the exercice but now with factorials or ' n! '.

The sequence (n)! starts like this :

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...

Alas from factorial 10 (3628800) onwards you see a string
of zero's appearing that grows larger and larger.

So from here on we will search for palfree factorials with
disregard of these (midway or ending) zero's. Let us see if
we can extend our sequence with that one restriction.
The next one factorial 12 (479001600) is candidate
with 2 x 'two consecutive zero's' in its decimal expansion.

But then I went all the way up to factorial 100 and gave up.
The factorials grow very fast in length and the probability
that a palfree factorial (even with the zero exclusion)
pops up is very low.
So the story ends here with the final provisional sequence

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 12