Re: [math-fun] Fracfac expansions (fractions in factorial base)
Using mere long doubles in C, here are the fracfac coefficients that I believe, for 1/p, p prime <= 19. Apparently, we always have: a) (p-1)/2 is the coefficient of 1/(p-2)! (for p > 3), b) 0 is the coefficient of 1/(p-1)!, and c) p-1 is the coefficient of 1/p!. Can someone give a nice proof of these apparent patterns? --Dan coefficient of 1/factorial of 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ------------------------------------------------------------------------------------------------------------- prime p ------- 2: 1 3: 0 2 5 0 1 0 3 7: 0 0 3 2 0 6 11: 0 0 2 0 5 3 1 4 0 10 13: 0 0 1 4 1 2 5 4 8 5 0 12 17: 0 0 1 2 0 2 3 6 8 9 0 9 2 7 0 16 19: 0 0 1 1 1 6 2 0 9 5 2 6 11 11 13 8 0 18 ------------------------------------------------------------------------------------------------------------- coefficient of 1/factorial of 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 I wrote: << . . . [E]very x in (0,1) has a unique representation of the form x = Sum_{n=1..oo} c_n/(n+1)! if the integers c_n satisfy 0 <= c_n <= n. . . .
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On Sunday 01 January 2012 23:33:19 Dan Asimov wrote:
Using mere long doubles in C, here are the fracfac coefficients that I believe, for 1/p, p prime <= 19.
Apparently, we always have:
a) (p-1)/2 is the coefficient of 1/(p-2)! (for p > 3),
b) 0 is the coefficient of 1/(p-1)!, and
c) p-1 is the coefficient of 1/p!.
Can someone give a nice proof of these apparent patterns?
I don't know about "nice", but the following is very straightforward. We have 1/p = a2/2! + a3/3! + ... + ap/p! (p-1)! = a2 (p!/2!) + a3 (p!/3!) + ... + ap Working mod p and applying Wilson's theorem gives your (c). Call the last few coefficients ..., x, y, z. Then we have (p-1)! = p-1 + p(y + (p-1)(x + (p-2)T)) for some T. Now everything except py is a multiple of p-1, so py is too, so y is; since 0 <= y < p-1, we must in fact have y=0. And now we have (p-1)! = p-1 + p(p-1)(x + (p-2)T) (p-2)! = 1 + p(x + (p-2)T) so working mod p-2 we have 0 = 1 + px = 1 + 2x Since p-2 is odd, there is a unique x satisfying this, namely (p-3)/2. (Not (p-1)/2, which presumably was a typo.) -- g
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Dan Asimov -
Gareth McCaughan