[ November 10, 2022 ] [ Last update November 1, 2023 ]
Smallest multipliers to make a number palindromic
in bases 2 to 12 (b ≠ 10)
by Xinyao Chen (email)

First of all if this topic is new to you familiarise
yourself with it by studying WONplate 36 and WONplate 96
which deals only with the base 10 variant.

Xinyao Chen went further and delved into
the other bases as well. This gives cause to a lot of new material to list
and lots of opportunity to extend the tables.

Of course, in base b the numbers which are multiples of b will have no
palindromic multiples, since all multiples of these numbers end with 0
but not begin with 0, thus not palindrome.
Only consider the X not divisible by the base, since all multiples of X
divisible by the base has trailing zeros, thus cannot be palindromic
since a number cannot have leading zeros.

For bases 2 to 7, the smallest multipliers of the numbers ⩽ 2^12
(which are not multiples of b) are in the worksheets, and for base 8,
I found the smallest palindromic multiples of all numbers ⩽ 10000
(which is 2^12 in decimal) which are not multiples of 8, except three numbers:
 6161 ,  6771 ,  7601 , they are like the numbers 8181, 8991, 9801 in base 10,
and seems to be hard.
You can also consider the concatenations of 61 in base 8 and the sequence
677...771 in base 8 (they should be like the concatenations of 81 in base 10
and the sequence 899...991 in base 10, respectively).

Also, in base b, a palindromic number is a multiple of b^2-b+1 (i.e. 81 in base 10)
if and only if its digits sum is a multiple of b^2-b+1, this is why in general,
the least multiplier is larger if the number is a multiple of b^2-b+1

Also, it appears that the three largest term below b^4: b^4-2*b^3+2*b^2-2*b+1
(i.e. 8181 in base 10), b^4-b^3-b+1 (i.e. 8991 in base 10), b^4-2*b^2+1
(i.e. 9801 in base 10) have three largest multipliers, and it appears that
the concatenation of b^2-b+1 (i.e. 81 in base 10) in base b and numbers of
the form b^n-b^(n-1)-b+1 (i.e. 8999...9991 in base 10) have larger multipliers.

Instead of showing long lists that might be rather boring I asked Xinyao Chen
for his program that recreates these lists on demand. The pari/gp code is
very intricate and uses functions that call other functions. Here it is:

\\ This is the PARI/GP program code of WONplate 223: (just type f(b)
\\ to print the smallest palindromic (in base b) multiple of the
\\ numbers <= 2^12 not divisible by b) e.g. for b = 8 use (01:54) gp > f(8) 

a(n,b)={my(d, i, r);
r=vector(#digits(n-b^(#digits(n\(b+1),b)),b)+#digits(n\(b+1),b));
n=n-b^(#digits(n\(b+1),b)); d=digits(n,b); for(i=1, #d, r[i]=d[i];
r[#r+1-i]=d[i]); sum(i=1, #r, b^(#r-i)*r[i])}

c(k,b)=for(n=2,2^64,if(a(n,b)%k==0,return(a(n,b))))

d(n)=if(n<10,n+48,n+55)

e(n,b)=for(k=1,length(digits(n,b)),print1(Strchr(d(digits(n,b)[k]))))

f(b)=for(k=1,2^12,if(k%b,e(k,b);print1(",");e(c(k,b),b);print()))

For bases {6}, {8}, {10}, {12} X. Chen found the smallest palindromic multiples
of the harder numbers (all written in corresponding bases).

Xinyao Chen has not confirmed that the numbers for A1A1 (base 12),
6771 (base 8), ABB1 (base 12), BA01 (base 12) are the smallest solutions, but
they are genuine solutions, and these four numbers have no small solutions, the
solutions for these four numbers are only conjectured, found by the law of the
equivalent numbers in base 6 and base 10.

[ October 30, 2001 ]

Xinyao Chen made two webpages

https://sites.google.com/view/wonplate-223-base-6/
and
https://sites.google.com/view/wonplate-223-base-8/

You write that b ≠ 10, but b = 10 is also on my research,
currently I am computing b = 10 and b = 12,
but multiples for 81 (for b = 10) and 121 (for b = 12) takes longer times.