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| thyristor |
 Posted: Thu Apr 17, 2008 1:22 pm Post subject: Binomial coefficients |
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Ever thought of that the binomial koefficients equal 11n?
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| serpicojr |
| Posted: Thu Apr 17, 2008 2:51 pm Post subject: |
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| JaneBennet |
| Posted: Thu Apr 17, 2008 3:10 pm Post subject: |
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The sum of binomial coefficients is equal to 2n. For example, 5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 1 + 5 + 10 + 10 + 5 + 1 = 32 = 25.
Proof: Take x = 1 in the binomial expansion of (1+x)n.
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| serpicojr |
| Posted: Thu Apr 17, 2008 3:59 pm Post subject: |
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| Yeah, I thought that's perhaps what he meant. |
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| MagiMaster |
| Posted: Thu Apr 17, 2008 9:42 pm Post subject: |
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Joined: 16 Jul 2006
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I think he meant that the digits of 11^n are binomial coefficients.
1
11
121
1331
14641
161051 <- It breaks here though
101^n lasts a while longer. In general, any 10^k+1 works for a while and the reason is pretty simple. (a + b)^n = sum over i of (n choose i)*(a^i)*(b^(n-i)), so for a=10, b=1 you're just left with powers of 10 times the binomial coefficients. |
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| thyristor |
| Posted: Sat Apr 19, 2008 3:56 am Post subject: |
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You are right about what I meant Magi Master but the only reason that the binomial coefficients aren't palindroms all the time even if they equal 11n is that ten symbols, as we use, aren't enough.
There will be overflow.
If you want to you can write a computer program with for example 15 symbols and there you'll see that 115 is a palindrom.
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| MagiMaster |
| Posted: Sun Apr 20, 2008 1:28 pm Post subject: |
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| Yes, in a higher base, 11n will stay a palindrome for longer, but never indefinitely (101n is about the same as 11n in base-100). I suppose you could define something like a base-infinity (based on tuples) that would stay a palindrome but I don't think there's much point in that (correct me if I'm wrong). |
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| thyristor |
| Posted: Mon Apr 21, 2008 4:08 am Post subject: |
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The only point is that palindromes are beautiful.
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