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More Bernoulli proof

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• Message 1 of 1 , May 27, 2002
<<<
Here's the outline of another proof valid for even m.
One can easily show that S(p^r,m)/p^r - S(p^{r+1},m)/p^{r+1}
is an integer for prime p and r >=1. As S(p^r,m)/p^r -> B_m
p-adically as r -> infinity then we get
S(p^r,m)/p^r is an integer for some r >= 1
iff
S(p^r,m)/p^r is an integer for all r >= 1
iff
B_m is p-integral.
>>>

Robin Chapman's on form it appears!
That one looks far simpler for its half of the cases.

Phil

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