[seqfan] Re: A congruence modulo odd prime
Vladimir Shevelev
shevelev at bgu.ac.il
Tue Oct 25 09:36:36 CEST 2011
Using the Leibniz formula modulo p: (f*g)^(n)=sum{i=0, ...,n}f^(i)*g^(n-i) for f=x^(p-1-k)-x^k, g=(x-1)^{-1}, I reduced this congruence to the identity
sum{i=0,...,k}2^(-i)*C(k+i,k)=2^k .
The latter identity is known since it is connected with the Banach problem on match-boxes (see John Riordan, Combinatorial Identities, Wiley&Sons, 1968, Ch.1, problem 7 and W. Feller, An introduction to probability theory and its applications, Wiley, 1958). More exactly, the random variable X in the Banach distribution has probabilities P[X=i]=C(2*n-i,n)*2^(i-2*n), i=0,...,n, such that sum{i=0,...,n}C(2*n-i,n)*2^(i-2*n)=1 or, the same, putting j=n-i, sum{j=0,...,n}C(n+j,n)*2^(-j)=2^n.
Regards,
Vladimir
----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Monday, October 24, 2011 6:36
Subject: [seqfan] A congruence modulo odd prime
To: seqfan at list.seqfan.eu
> Let p be odd prime. I believe that, if 1<=k<=(p-1)/2, then
> the k-th derivative in the point x=2: ((x^(p-1-k)-x^k)/(x-
> 1))^(k) |_(x=2)==0 mod p (generally speaking, one cannot replace
> 2 by another integer).
> E.g., if p=7, k=2, then ((x^4-x^2)/(x-1))^(2)=6*x+2=14 in the
> point x=2.
> Another form of the conjecture: sum{j=1,...,p-2*k-1} 2^{j-
> 1}*C(j+k-1, k)==0 (mod p).
> Questions: 1) Did anyone see similar congruence? 2) Can anyone
> prove (disprove) it?
>
> Regards,
> Vladimir
>
>
>
> Shevelev Vladimir
>
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>
Shevelev Vladimir
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