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| A prime factorization problem (Posted on 2006-08-07) |
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Define polynomials Fn(x) such that for all nonnegative integers n:
F0(x)=1, Fn+1(0)=0 and
F'n+1(x) = (n+1)Fn(x+1)
Find the prime factorization of F300(7).
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Submitted by K Sengupta
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Rating: 4.0000 (2 votes)
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Solution:
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(Hide)
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An examination of the low order cases lead one to conjecture that:
F_n(x) = x*(x+n)^(n-1)
Clearly, this satisfies F_0(x) = 1; F_n(0) = 0 for n>=1
Now; F’_n+1(x)
= (x+n+1)^n + n*x*(x+n+1)^(n-1)
= (n+1)(x+1)(x+n+1)^(n-1)
= (n+1)*F_n(x+1)
Hence, F_n(x) = x*(x+n)^(n-1), as conjectured earlier.
Consequently, F_300(7) = 7*(307^299).
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Subject |
Author |
Date |
 | Spoiler | Richard | 2006-08-07 13:45:55 |
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