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BernardBeduya

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A Pascal's Triangle Hidden Inside the Pascal's Triangle

Hi! I was referred by Rod to this forum.

May I invite you to click the link below to my wordpress blog with the title `A Hiddren Pascal`s Triangle Hidden Inside the Pascal`s Triangle`.

https://bernardbeduya.wordpress.com/2016/02/06/a-pascals-triangle-hidden-inside-the-pascals-triangle/

Please help me make sense of my discovery because I`m not an expert mathematician as you. Thank you and warm regards.

Nehushtan

Member

Registered: 2013-03-09

Posts: 957

Re: A Pascal's Triangle Hidden Inside the Pascal's Triangle

I disagree with this statement from the website:

The same phenomenon appears where the row that bears the numbers 1-4-6-4-1 is the last palindromic sequence.

No, it is not! It is only the last because we are using base 10. If we use higher bases, the palindromic sequence will continue; e.g. in hexadecimal base, the next line would be 1-5-A-A-5-1; and in a still higher base (> 20) the next line would be 1-6-F-K-F-6-1. As you can see, any line in Pascal's triangle can be palindromic, if you choose a high-enough base.

As for the number 1001, it is of the form n³+1 = (n+1)(n²−n+1), where n is the base; hence it is composite regardless of the base chosen. If a number is composite, I'd say it has a good chance of appearing in Pascal's triangle.

Please help me make sense of my discovery because I`m not an expert mathematician as you. Thank you and warm regards.

All you need to know is that mathematics is always a very curious and interesting subject.

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Nehushtan

Member

Registered: 2013-03-09

Posts: 957

Re: A Pascal's Triangle Hidden Inside the Pascal's Triangle

On the subject of palindromic numbers, note the following:

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These are successive powers of n+1, where n is our base; they're palindromic up to some point depending on n. For base 10 (n=10) the palindromicity breaks down at the fifth power, when carrying-over starts to take over:

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[/*]
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To continue the palindromic sequence, all you need to do is choose a higher base (n>10). For example, consider the powers of 17 in hexadecimal:

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[/*]
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Ta da! Whereas (10+1)⁵ fails to be palindromic in base 10, (16+1)⁵ is palindromic in base 16. Choose n > 20 and (n+1)⁶ will be palindromic in base n too.

Last edited by Nehushtan (2016-02-13 23:18:31)

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anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: A Pascal's Triangle Hidden Inside the Pascal's Triangle

Hej, Bernard!

I am failing to see how the "hidden" triangle is different from just the Pascal's triangle multiplied by 1001 and inserted somewhere in-between the numbers of a regular Pascal triangle?

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beduya22

Member

Registered: 2016-06-22

Posts: 2

Re: A Pascal's Triangle Hidden Inside the Pascal's Triangle

Thank you Nehushtan and anonimnystefy for helping me.  Sorry I wasn't able to return to this forum soon because I was busy with work.  I'm trying to edit my wordpress blog now.  I really couldn't have realized that it was wrong for me to think of decimal digits only without your help.  With regards to anonimnystefy's message, it's really no different from just a Pascal's triangle multiplied by 1001 but my goal in inserting this starting at the 13th row of the original Pascal's triangle was because I noticed that there are 4-digit palindromic numbers 1001, 2002 and 3003 in the 14th row, and 3003 and 5005 in the 15th and then 8008 in the 16th row, forming a shape of an inverted triangle.  I then wondered why 4004, 6006, 7007 and 9009 are missing in this region of the original Pascal's triangle.   That's why I conjectured that if I placed the 1st palindromic number 1001 on top of 2002 in the original Pascal's triangle and use that 1001 as the border numbers of my inserted Pascal's triangle, I might be able to locate those missing 4-digit palindromic numbers that I mentioned.  So, when I did see the completed 4-digit palindromic numbers in my inserted Pascal's triangle, I just wondered why this is so.  Anyways, thanks for all your help guys.