Square numbers are defined and calculated by this extraordinary intricate and excruciatingly complex formula. So, this line is for experts only _{}
( base ) x ( base )orbase ^{2}
An alternative way to get a 'structural' insight as how to imagine squares is to follow these links :
All palindromic square numbers can only end with digit 1, 4, 5, 6 or 9.
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)!
[ November 4, 2005 ] Here is how Natalie Dickendasher (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.
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) (264) (307) (836) (2285) and (2636) and can be found without too much trouble with a spreadsheet or calculator.
Mike Bennett made an attempt in finding out what his chances are in discovering palindromic squares and palindromic cubes. Here is his mathematical analysis and the probability of findingthose interesting palindromic polygonal numbers.
The second highest one (3069306930693) 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.
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.
In the mean time I was making a table of palindromic squares from scratch myself, hoping to find larger numbers then the ones mentioned above. And in [ October 1996 ], I succeeded beginning from [307] ! (Un)fortunately soon afterwards this record was broken by submitters like Graham Lyons, Pete Leadbetter and Mike Bennett who sent in numbers of even greater calibre ! Observe that the top three palindromic square submitters were all British ! Was I detecting a new pattern here ?
" Pete wrote that this square was found using an Athlon 1200 MHZ PC programmed in Powerbasic about a third of the way into a 23 day search. The algorithm is basically the same as the one used to find the previous 44 digit number. A faster machine plus the more efficient compilation that Powerbasic affords allowed the 52 digit range to be looked at within a 'not so mind boggling' time. By the way, the find reported to you was the result of a 'burn in' for my new PC, as was my previous 44 digit result! Also I'd recommend Powerbasic for number crunching, it's seriously quick! Pete remarks that of the set of EVEN length palindromic squares none of themseems to end in digit 5. Is there a proof for this ? If the base ends in 5 then the square must end in 25. Perhaps this reduces the freedom of one of the digits thereby reducing the likelyhood of such a number. Or are there other number theory reasons ? "
Pete remarks that of the set of EVEN length palindromic squares none of themseems to end in digit 5. Is there a proof for this ? If the base ends in 5 then the square must end in 25. Perhaps this reduces the freedom of one of the digits thereby reducing the likelyhood of such a number. Or are there other number theory reasons ? "
" The palindromic square number searching algorithm starts by assuming a generic quadratic formula F(n) = An^2 + Bn + C, where A, B, and C are non-negative integer numbers. Now the code tries to find palindrome number when n is within a certain range 'from' to 'to'. It divides the range into subranges such that for the first and last n within the range F(n) have the same length and same first digit. For that range, it searches all the last number of n within 0..9 such that the latest digit of F(n) matches the first digit of F(n). Now we know when n is within that range, and when n is of the form 10 * m + f, the first and last digit of F(n) are always the same (f). After that the code substitute n = 10 * m + f into F(n), remove the highest digit, divide by 10 to remove the last digit to get F(m). The routine is called recursively to work on the new formula until the range is small enough to work with using an iterative algorithm. I'm using a 1.4Ghz Pentium 4 machine, running Windows XP. Program is written in Microsoft Visual C++ 6.0. The time the algorithm needs to complete search for n-digit palindrome squares is around O(10^(n/4)), that is to say time increases 10 times for every 4 digits. Using the current code, I estimate that it will take around 200 days to complete search all the 52-digit numbers. I hope I can reduce that to something like 50 days. "
I'm using a 1.4Ghz Pentium 4 machine, running Windows XP. Program is written in Microsoft Visual C++ 6.0. The time the algorithm needs to complete search for n-digit palindrome squares is around O(10^(n/4)), that is to say time increases 10 times for every 4 digits. Using the current code, I estimate that it will take around 200 days to complete search all the 52-digit numbers. I hope I can reduce that to something like 50 days. "
" The complete search for 45-digit palindromic numbers can be completed in around 30 hours. How is that done? I doubled the speed of my code and got a 2.0 Ghz dual-CPU machine. "
" My name is Takahashi, and I live in Osaka. I am untrained and doesn't know any English at all, so ask a Japanese to translate this message. ( ps. done so! ) Since I love mathematics, I was very interested when I saw these numbers. In Japanese we call these palindromic squares 'kaibunheihousuu', and my program is an improvement of Assembler code made 15 years ago. Its efficiency is still bad, but because I investigate very thoroughly, there are no unexpected slovenly errors. Palindromic squares will arise 66 days after the program starts when it is ready of course. I expect to finish it the second half of July. "
" I found a new one last night using my 8-core machine. Happy New Year. " " The machine I'm using is Dell Precision PWS690, 2.33GHz. It's dual QuadCore, so 8 cores in total. I'm trying to use all 8 cores at the same time to boost performance. The complete search for 47 digits took 35,540 seconds (clock time), around 10 hours. It's still 32-bit programs. 64-bit would be definitely faster, but I do not have it installed on my machine. Searching 48 digits came back empty. Now I'm in the middle of searching for 49 digits. "
" The machine I'm using is Dell Precision PWS690, 2.33GHz. It's dual QuadCore, so 8 cores in total. I'm trying to use all 8 cores at the same time to boost performance. The complete search for 47 digits took 35,540 seconds (clock time), around 10 hours. It's still 32-bit programs. 64-bit would be definitely faster, but I do not have it installed on my machine. Searching 48 digits came back empty. Now I'm in the middle of searching for 49 digits. "
I found Feng Yuan's record palindromic square of 55 digits in the following archived blog http://blogs.msdn.com/fyuan/archive/2008/01/31/sts-a-palindromic-word-i-will-remember.aspx
Click here to display the full listing upto length 31. Click here to display the full listing from length 32 onwards.
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 [47], [84] and [140] attracted my attention because of the CORE number 091 in them.
[47] 1109111 [84] 110091011 [89] 111091111 [140] 11000910011
A second promising CORE number exists. It is 09901 but it shows up only from [130].
[130] 10109901101 [218] 1010099010101 [228] 1011099011101 [345] 100110990111001 [353] 101000990100101 [363] 101010990110101 [535] 10010109901101001 [552] 10100009901000101 [562] 10100109901100101 [569] 10101009901010101
I think a new third CORE number is emerging... It is 0999001 Watch out from
[339] 100109990011001 [530] 10010099900101001 [544] 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
What are the conditions for creating a pseudopalindrome with a palindromic square ? Open this window and read David W. Wilson's answer.
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 at the end of this page].
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...
[3] 3 [13] 307 [30] 3069 3 [55] 3069 307 [94] 3069 3069 3 [160] 3069 3069 307 [262] 3069 3069 3069 3
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. Palindromic squares with an even length are just a plain minority. Until now I can only list sixteen of them. These are highlighted in the table by means of a aquamarine background color in the side-cells. See Sporadic Palindromic Squares of EVEN length
I 'almost' solved a very ancient problem namely : squaring the circle (pi = 3,1415....) MORELAND PI welcomes PALINDROMES
[414] 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.
[480] 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 upto 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 ( 10101010101010101^{2} ) 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 (?) that answers to these rules may be this one (also of length 33) : 11843191515764821^{2} = 140261185279083838380972581162041.
Click here to display the subsets of palindromic squares.
Feng Yuan (email) from Washington State, USA, sent a record sporadic palindromic square 53 of digits [ January 16, 2008 ].
Feng Yuan (email) from Washington State, USA, sent a palindromic square with even numbers 50 of digits [ January 8, 2008 ].
Pete Leadbetter informs me that he checked for 40 digit palindromes and found none. He's convinced that there aren't any 38's since his 'old' program searched that far. There are no more 44 digit squares that are palindromic other than the above one. In his letter he says also that for the case of 42 digits it will take nearly as long (12 days on his PC !) and leave the search for someone else with enough free computer timecycles.
Alain Bex (email) gave also some palindromic squares in base 12. Link to Palindromic Numbers other than Base 10
David W. Wilson's (email) contribution regarding the 'Pseudopalindromes with a palindromic square' is very remarkable and therefore much appreciated.
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".
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