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Can anybody be kind enough to explain what exactly is a super palindrome?

Also consider the following example :

$923456781-123456789=799999992:9=88888888$

The largest prime factor of $88888888$ is $137$

$88888888:137=648824:101$ ($101$ being largest prime factor of $648824$) = $6424:73$ ($73$ being largest prime factor of $6424$) this in turn equals to $88$ (a palindrome)

Finally, $88:11$ ($11$ being largest prime factor of 88 and a palindrome) equals to $8$ which is also a palindrome.

Now, given this post and its comments, will it be correct to call $88888888$ a super palindrome?

Can it be possible to call a number a super palindrome if it generates non palindromes along with palindromes, provided the non-palindromes finally generate using the method of division by largest prime factor, a palindrome other than 1?

Vepir
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Pradip Saha
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  • @celtschk Please tell me : "Can it be possible to call a number a super palindrome if it generates non palindromes along with palindromes, provided the non-palindromes finally generate using the method of division by largest prime factor, a palindrome other than 1?" – Pradip Saha Nov 03 '12 at 17:28
  • @amWhy Thanks a lot for the edit.. – Pradip Saha Nov 03 '12 at 17:35
  • no problem...it's easy to do, for future reference. Enclose a phrase to use as the active link in square brackets (in your post I used [this post and its comments]) followed immediately with the hyperlink enclosed in regular parentheses. – amWhy Nov 03 '12 at 17:38
  • @amWhy Do you think that the argument that I have given for calling 88888888 a super palindrome is logically sound? Please refer to 9 hidden in every number regardless of number of digits?, you will see I am trying to understand whether my approach of generating palindromes, can in turn generate a family of super palindromes.. Please feel free to correct me if i am wrong.. – Pradip Saha Nov 03 '12 at 17:47

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Based on the linked post's definition of super-palindrome:

"I define a super-palindrome as a palindrome which is either 1, or which gives another super-palindrome if divided by its largest prime factor."

88888888 is not a super palindrome since division by its largest prime factor does not yield another super-palindrome. Also, you ended your algorithm early by stopping at 8 since you can say the largest prime factor of 8 is 2 etc. Your number still seems to exhibit interesting properties however and would certainly warrant more investigation!

For example, any $n$-digit number with all of its digits a single number has the same prime divisors as the other 8 $n$-digit numbers with all their digits the same. Also note that your numbers are nearly symmetric upon division of their largest prime factor. If you provided a more rigorous definition of this, that could be interesting to study.

Just some thoughts.

Jemmy
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  • if you refer to my earlier question, [9 hidden in every number regardless of number of digits?] (http://math.stackexchange.com/questions/227692/9-hidden-in-every-number-regardless-of-number-of-digits). – Pradip Saha Nov 03 '12 at 18:44
  • I am trying to generate palindromes by using the following approach : Take an n-digit number, interchange the position of the terminal digits, take the difference between the 2 resulting numbers and then divide the difference by 9, so I am getting a symmetrical palindrome, so will symmetrical palindromes not classify as a special class of palindromes, since when we use the 'divide by its largest prime factor' method, we can generate palindromes and in some cases non-palindromes, which by using the 'divide by its largest prime factor' method, we can eventually generate palindromes.. – Pradip Saha Nov 03 '12 at 18:48
  • Anyways, I thank you for your thoughts on this, would really love and appreciate if we can further discuss my above comments.. – Pradip Saha Nov 03 '12 at 18:50