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Palindromic Squares
Factorization Distributed Computation All Squares up to length 31 The Sporadic Squares The Subsets Extra Squares Records triangle penta hexa hepta octa nona n(n+1) n(n+2) n(n+x) n^2+1 n^2+x n^2–x n^2+(n+1) n^2+(n+x)

  left to right (forwards) as from the right to left (backwards)

Square numbers are defined and calculated by this extraordinary intricate and excruciatingly complex formula.
So, this line is for experts only

( base ) x ( base )	or	base 2

square 1	square 4	square 9

Your chance to link a record sporadic palindromic square with your name…
Gather all the necessary information from this page at Distributed Computation.

So far this compilation counts 8729 + 4 square palindromic numbers.

All palindromic square numbers can only start or end with digits 0, 1, 4, 5, 6 or 9.
Alas, my palindromes may not have leading 0's! So the zero option must not be investigated.

There exist no palindromic squares of length  2, 4, 8, 10, 14, 18, 20, 24, 30, 38, 40, 46, 54, 56, 62, 64.

A B C →  1 0 0 

A B C →  1 2 1 

Mike Keith made an original study of the palindromic squares.
He was able to classify and enumerate them in a logical manner
and went even further as he tried also to create a general formula for
calculating the number of certain classes of palindromic squares,
except for the 'sporadic' solutions. Wonderful!
[Please refer to the “JRM” source detailed in section 'Sources Revealed'].

Mike had most trouble with the Asymmetric Root Family but thanks to
Dave Wilson's “lemma” he made significant progress.

What are the conditions for creating a pseudopalindrome with a palindromic square ? Open this window and read David W. Wilson's answer.

Unlike Palindromic Triangulars where it is impossible to predict a next higher one, whether its basenumber is palindromic or not,
with the Palindromic Squares (and Cubes) we have an opposite situation. Finding a next higher number is very easy.
Start e.g. with the number 11. Then repeatedly add a zero between the two 'ones' and square them.
A pattern emerges that can go on forever.

There are other numbers with the same properties (e.g. 10101). The repunit numbers like '111' looks also a good candidate
for finding palindromic squares (not for cubes though) but the expansion doesn't go on forever. Nine 'ones' still produce
a palindrome when squared, but not ten 'ones'. It's a near miss as only the number 8 doesn't show up in the left part of the square.
This anomaly stays with longer repunits as well, so we must abandon this candidate.
If we take resort to higher basenumbers only then we can extend the pattern (for a while)!

Here is how Natalie Dickendasher [ November 4, 2005 ] (email) figures them out quickly in her head and impresses friends or boss :
The highest digit of the sequence will always be the quantity of 1's in the number your are squaring.
As an example lets start with my salary for next year (yah right): what is 111111 squared ?
EASY !! This number has a total of 6 digits. Therefore the highest number in the palindrome will be 6
so count to 6 and then back down. The answer is 12345654321 [35].

Finding large palindromic squares is easy if you start with some palindromic basenumbers as explained above.
So, let's concentrate on the difficult ones namely those with a non palindromic basenumber. The first six are
(26 [7]) (264 [13]) (307 [14]) (836 [15]) (2285 [19]) and (2636 [20]) and can be found without too much trouble with a spreadsheet or calculator.

Can we find formulas for all or some of these sequences ?

Mike Keith gave two formulas in his article 'Classification and enumeration of palindromic squares' published in the “Journal of Recreational Mathematics” Volume 22, Number 2 - 1990, pp. 124-132. One for the Binary Root Family defined for all even N ⩾ 2 and another for the Binary & Ternary Root Family combined, defined for all odd N ⩾ 3. This is a choice that Keith made as it suited more in his scheme of things. Keith_|B| = (N3 – 6*N2 + 32*N) / 48 → for even values of N ⩾ 2 → bnl = N = 30 gives result  470  Keith_|B|+|T| = (N3 – 9*N2 + 59*N – 51) / 24→ for odd values of N ⩾ 3 → bnl = N = 31 gives result  955  ( = 940 + 15) Through David Griffeath we have now also the formula for odd values of N with the |B| separate from the |T|. Griffeath_|B| = (N3 – 9*N2 + 47*N – 39) / 24→ for odd values of N ⩾ 3 → bnl = N = 31 gives result  940  The total number of the Even Root family is simply |E| = 1 if N is even |E| = 2 if N is odd Regarding the Asymmetric family Keith writes “Finding an exact formula for |A| remains an open problem.” David Griffeath [ email from December 1, 2022 ] derives the right formulas for the # bnl binaries → |B|. “Start with a binary palindrome root of size 2n. It needs to have 1s on the ends. There are n–1 remaining spots on the right half where we can choose to put 0,1,2, or 3 additional 1s. Then symmetric spots must be chosen on the left side. So the total number of ways we can do this is C(n-1,0)+C(n-1,1) +C(n-1,2)+C(n-1,3). After some algebra this reduces to n(n^2 - 3n + 8)/6. Now suppose the size is 2n+1. Then any pattern of the type just described works for the digits to the right and left of the central digit, and that can be either 0 or 1. Thus the formula for 2n+1 is simply n(n^2 - 3n + 8)/3.” Griffeath_|B| = n*(n2 – 3*n + 8) / 6 → for even values of N → bnl = 30 (2n) so n=15 gives result  470  Griffeath_|B| = n*(n2 – 3*n + 8) / 3 → for odd values of N → bnl = 31 (2n+1) so n=15 gives result  940  D. Griffeath continues his mail with “As we know, whether there are an even or odd number of digits in a square palindrome root matters substantially. So it is natural to break up the formula this way. It is artificial to force these cases into a single formula. But this can be done by manipulating the coefficients of the two cubics above and adding a multiplier (-1)^n to one of the terms in order to account for parity. This is the essence of what is going on in a more complicated formula. I can go into more details if you like, but nothing illuminating is accomplished.” And yet such a single formula for all even/odd values of N in the |B| family can be found in OEIS A263614 and was added by Colin Barker. I'll give here his Pari/gp function (Barker_|B|) gp > a(n) = (-((-1)^n*(-78+62*n-12*n^2+n^3))+3*(-26+42*n-8*n^2+n^3))/96 After giving this command just type a(30) or a(31) to get  470  and  940  resp. It looks intriguingly complex especially compared with the easyness of programming these Binary Root solutions but less so when we start to dissect it... (   –((–1)n*(n3–12*n2+62*n–78)) + 3*(n3–8*n2+42*n–26)   ) / 96 And again David comes to the rescue and helps us unraveling the Barker formula. See panel on the right ↗   “I've been feeling remiss about not showing you how that messy formula for the number of binaries doesn't involve fancy math, but rather just a bunch of unpleasant, unenlightening algebra. So I've attached a handwritten sketch of the computations. The first step, in (1) and (2) is to change variables and express the formulas for the even and odd cases in terms of the length. Just yucky algebra. Then to get the unified formula, we write the number of binaries b_n as the sum of two cubics, one with the magic multiplier (-1)^n to flip between 1 and -1 in the even and odd cases. The combination needs to flip between (1) and (2) as desired, so we solve for all the c's. I just did the computation of the cubic power coefficients, but the rest are analogous. Just more yucky algebra.” Thanks David for this proper answer, much appreciated!"

Careful observation of some numbers in the list lead me to the discovery of ever greater non-palindromic basenumbers whose squares are palindromic.
Specially the numbers with indexes [48], [85], [90] and [141] attracted my attention because of the CORE number 091 in them.

[48] 1109111
[85] 110091011
[90] 111091111
[141] 11000910011

A second promising CORE number exists. It is 09901 and it shows up from [131], [219], [229], ....

[131] 10109901101
[219] 1010099010101
[229] 1011099011101
[346] 100110990111001
[354] 101000990100101
[364] 101010990110101
[536] 10010109901101001
[553] 10100009901000101
[563] 10100109901100101
[570] 10101009901010101

I think a new third CORE number is emerging... It is 0999001
Watch out from

[340] 100109990011001
[531] 10010099900101001
[545] 10011099900111001

Quite interesting. Let's put these three CORE numbers in a row :

091
09901
0999001

David W. Wilson took up the thread where I left off.
Read what he posted to me about these core 'pseudopalindromes'
[ February 16, 1998 ].

If we allow a digit "n" with value "–1", then 091 is equal to the palindrome 1n1,
while 09901 = 10n01, 0999001 = 100n001, etc. Thus

          10110n01101
        x 10110n01101
          -----------
          10110n01101
        10110n01101..
       10110n01101...
     n0nn010nn0n.....
   10110n01101.......
  10110n01101........
10110n01101..........
---------------------
102210100272001012201

Click here to display the full listing up to length 31.   Click here to display the sporadics listing of all such square palindromes (SSP).   Click here to display  the subsets of palindromic squares. Three categories.

And what about this repeat pattern of the item 3069
and that is finalized alternatively with a chopped off and rounded version of it namely a 3 or 307...

[4] 3
[14] 307
[31] 3069 3
[56] 3069 307
[95] 3069 3069 3
[161] 3069 3069 307
[263] 3069 3069 3069 3

3069 occurs as displacements to powers of ten so that this number forms the nearest (probable) prime to those axes :

10¹⁰⁰¹ – 3069 is the largest prime of 1001 digits

NEVER ODD OR EVEN

Read title backwards please !

It looks as though the decent sufficiently-random nonpatternlike unpredictable nonpalindromic basenumbers become very rare.
In the table (see Full Listing or the Subsets) I highlighted them using a lightblue background in the side-cells.
And even then a further distinction can be made. Consider the even and odd length values for a minute.
Sporadic Palindromic Squares with an even length are just a plain minority. Until now I can only list twenty of them.
These are highlighted in the table by means of a aquamarine background color in the side-cells.

Sporadic Palindromic Squares of EVEN length

Score 'subset E' = 22

I 'almost' solved a very ancient problem namely : squaring the circle (pi = 3,1415....)
MORELAND PI welcomes PALINDROMES

[415]    314155324482867

This basenumber starts with the the number 2201 which is (to me) the only known nonpalindromic basenumber
of a palindromic cube ! The cube of 2201 equalling 10662526601.

[481]    2201019508986478

Here is a good place to talk about a famous conjecture that nobody
could prove or disprove for more than 80 years.
I must thank Mike Keith for letting me know this fact through an email message.
Mike has heard that this conjecture originates from
L. E. Dickson's famous “History of the Theory of Numbers”.
Can someone with access to this source confirm this ?

History :
My table lists exhaustively all palindromic squares up to lengths 31.
I also listed all the palindromic squares of length 32.
Palindromic squares of  'even' nature seem to be very popular among the record
breaking submitters. They told me there are no gaps for the 'even' length 32 !
So, in fact I had to finish checking only length 33...
Of which I checked all squares beginning with 1, 5 and 9 !
( and 50 % of those beginning with digits 4 and 6 ).
I collected a total of 151 palindromic squares beginning with digit 1.
(Un)lucky for me Dickson's number 102030405060708090807060504030201
( 10101010101010101² ) [572] is the 86^th one. Anyway after checking these 86 squares
I could proudly announce that the conjecture was finally proven [ May 3, 1998 ].
The second smallest number [633] that answers to these rules is this one (also of length 33) :
11843191515764821² = 140261185279083838380972581162041.

Sloane's A001110 gives the first numbers that are both Square and Triangular.

Sloane's A036353 gives the first numbers that are both Square and Pentagonal.

Sloane's A036354 gives the first numbers that are both Square and Heptagonal.

Sloane's A036428 gives the first numbers that are both Square and Octagonal.

Sloane's A036411 gives the first numbers that are both Square and Nonagonal.

Alain Bex (email) gave also some palindromic squares in base 12.
Link to Palindromic Numbers other than Base 10

A very interesting webpage by William Rex Marshall (1930-2010) is titled “Palindromic Squares”
and gives an overview of the first 67 palindromic squares known up to May 28, 2001.
The classification of these squares as set out by Michael Keith in the JRM from 1990 is also revealed.
http://www.geocities.com/williamrexmarshall/math/palsq.html (archived)

The second highest one (3069306930693 [263]) is copied from Martin Gardner's book “The Ambidextrous Universe”
see page 40.

Another source “Curious and Interesting Numbers” by David Wells, page 185, provided me the basenumber 798644 [37].

While surfing the Internet I came across the work of Keith Devlin namely All the Math that's Fit to Print.
In chapters 17, 75 and 110 he printed some palindromic squares he got from other readers.
The highest one that originates from his publication is 6360832925898 [264].

Gustavus J. Simmons published an article about palindromic squares in the “Journal of Recreational Mathematics”,
J. Rec. Math., 3 (No. 2, 1970), 93-98, titled “Palindromic Powers”.
OEIS Source (annotated scanned copy) → https://oeis.org/A002778/a002778_2.pdf

Gustavus J. Simmons published an article about palindromic squares in the “Journal of Recreational Mathematics”,
Volume 5, No. 1, 1972, pp. 11-19, titled “On Palindromic Squares of Non-Palindromic Numbers”.
OEIS Source (annotated scanned copy) → https://oeis.org/A002778/a002778.pdf

Mike Keith (email) (website) published an article about palindromic squares in the “Journal of Recreational Mathematics”,
Volume 22, Number 2 - 1990, pp. 124-132, titled “Classification and enumeration of palindromic squares”,
and was followed by an article by Charles Ashbacher, pp. 133-135, “More on palindromic squares”.
OEIS Source (annotated scanned copy) → https://oeis.org/A002778/a002778_1.pdf

I found Feng Yuan's record palindromic square of 55 digits [3992] in the following archived blog
http://blogs.msdn.com/fyuan/archive/2008/01/31/sts-a-palindromic-word-i-will-remember.aspx

Henry Ernest Dudeney, author of “536 Puzzles and Curious Problems” (originally published in 1967), dedicated one short
puzzle to Palindromic Square Numbers, namely Puzzle 112 from the chapter 'Arithmetic and Algebraic Problems' :

  This is a curious subject for investigation - the search for square numbers
the figures of which read backwards and forwards alike. Some of them are
very easily found. For example, the squares of 1, 11, 111, and 1111 are respec-
tively 1, 121, 12321, and 1234321, all palindromes, and the rule applies for any
number of 1's provided the number does not contain more than nine. But there
are other cases that we may call irregular, such as the square of 264 = 69696
and the square of 2285 = 5221225.

  Now, all the examples I have given contain an odd number of digits. Can
the reader find a case where the square palindrome contains an even number
of figures?
Answer
  The square of 836 is 698896, which contains an even number of digits and
reads backwards and forwards alike. There is no smaller square number con-
taining an even number of figures that is a palindrome.

Jean-Marie De Koninck, author of “Those Fascinating Numbers” (originally published in french in 2008 as
“Ces nombres qui nous fascinent”), exhibits interesting properties for the number 26 :

The smallest number which is not a palindrome, but whose square is a palin-
drome; a palindrome is a number which reads the same way from the left as
from the right; the first ten numbers n satisfying this property are listed below:

n
n2
 
n
n2
26
264
307
836
2285
676
69696
94249
698896
5221225
 
2636
22865
24846
30693
798644
6948496
522808225
617323716
942060249
637832238736

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Patrick De Geest - Belgium - Short Bio - Some Pictures
E-mail address : pdg@worldofnumbers.com