PALINDROMESEMORDNILAP dp-j 27Dec94 3PM
This subject has created some interest in recent Issues so I would like to
offer the following comment regarding the generation of such numbers.
To simplify presentation I will use a', b' etc to mean a squared, b squared etc
Consider the general palindrome, pal = (abcd.....dcba)
Squaring this algebraically results in a central term 2(a'+b'+c'+....y'+z') if
pal has an even number of terms and 2(a'+b'+c'+.....y') + z' if pal has an odd
number of terms. (NB. z is taken to mean the last alphameric be it b,e,k or
whatever.) All other terms of the expansion are symmetric about this central
term. To take an example pal = abba
pal' = abba * abba
Performing the multiplication column by column as in normal arithmetic and
adding the terms up in each column gives:-
a', 2ab, 2ab+b', 2(a'+b'), 2ab+b', 2ab, a'
which is symmetric about 2(a'+b'). In this example pal has an even no of terms.
If you are feeling ambitious and try squaring abcba you find the central term
is now 2(a'+b')+c'. Throwing caution to the wind if you now try squaring more
and more terms you will find that a pattern emerges such that you can write the
results down without bothering to do the sums at all. But in all cases the
central term will be of the form quoted in the introduction above
If numerical values are now substituted for a,b,c,... it is obvious that if
there are no inter-column carry digits then pal' will also be palindromic.
(Note that this is not to say that there will not be any palindromic squares if
carry digits do occur). Further, since this central term is numerically either
the largest or joint largest of all the terms then provided its magnitude is
less than 10 there will be no inter-column carrys anywhere.
So we can now make some observations about palindromic squares.
1) Any palindrome having an ODD no. of terms will have a palindromic square if
2(a'+b'+c'+...y')+z' is less than 10
2) Any palindrome having an EVEN no. of terms will have a palindromic square if
2(a'+b'+c'+...y'+z') is less than 10
Examining these criteria, the largest pal having a palindromic square will be
infinite. Curiously an infinite pal with an odd no. of terms will be greater
than one which is even ! Strange idea, a sort of alternating infinity ! So the
'biggest' pal with a palindromic square will be:-
2000.......1.......0002 with trillions of 0's in between for the 'odd' infinity
and 2000..............0002 for the biggest 'even' infinity. Their squares will
be 400...4...9...4...004 and 400......8......004 respectively.
WHy?. Well referring to the criteria 1) above and substituting a=2 (a'=4) and
z=1 (z'=1) with all other terms =0 gives 2(4+0+0+.......+0)+1 = 9 and therefore
its square is also palindromic. If you additionally try putting say b=1 (b'=1)
then the sum is now 2(4+1+0+0+....+0)+1 = 11 and therefore its square cannot be
palindromic. Similarly using criteria 2) gives the 'even' infinity result. In
either case there is no way whereby any term greater than 2 will result in a
palindromic square (but see later)
Of further interest, if pal is entirely composed of 1's then the above criteria
indicate that pal=abcdedcba (odd) with a=b=c=d=e=1 ie pal=111111111 (nine 1's)
has a palindomic square whereas pal=abcdeedcba (even) ie pal=1111111111 (ten
1's) does not. So more than nine 1's and pal' will not be palindromic. The
results are not affected by interspersing 0's anywhere you wish, provided
palindromic symmetry is maintained.
Now the Crunch. The above discourse applies to the 'no carry' condition as
stated. Criteria 1) and 2) imply there will be column carry if any pal term is
numerically greater than 2. Now is it possible that although any inter-column
carry will upset the fundamental symmetry there may be an instance in which a
carry beyond the highest place value can still, by what one might say was a
sheer numerical fluke, result in a palindrome? I have, encouraged by Mr Davey's
findings in Issue 39, examined this but will have to await a 'clear thinking'
day. I do not have an inductive proof of non-existance of the sought for
palindromes, and am not yet satisfied that my current programming attempts to
eliminate some of the computer search time are foolproof. For example, if a
carry into the highest place value is not equal to a' MOD 10 then pal' cannot
be palindomic for starters. Testing the carrys may be quicker than squaring the
palindrome and testing the result. Intuitively there must be something amiss if
long computer runs are needed to solve this type of problem, but I don't deny
it may give the clue to a more rational approach. My money is on looking at the
pattern of the carry terms.
Apart from a possible 'rogue' occurrence as yet undiscovered, all the results
produced to date did not require the use of a computer at all. It's slicker in
the head if criteria 1) is stored by the left ear and criteria 2) by the right,
quick march! However someone must surely have already solved this one?
PS. The palindromic cube. Criteria a little different but the biggest odd pal
will be 1000......1......0001 and the even one 1000......0001, rogues excepted.
And the 'paliquads'? The pal whether odd or even will have a maximum of two 1's
and there won't be any others.