If n is a positive integer and p is a prime factor of 2n,
then p divides 2(2^(2n)-1) ?? Presumably true, but is it easy?

The primes p in question make up the denominator of Bernoulli number
B_{2n}, so an alternate formulation is: 2(2^n-1)B_n is an integer,
where B_n is a Bernoulli number.

Hi

Any clue as to how to (bound/give an asymptotic expression for/
evaluate exactly° the following series?

\sum_m \sum_n 1/sqrt((a + m^2) (a + n^2) (a + (m+n)^2))

Here "a" is any positive real. The variables m and n go through all
integers.

Yours
Harald

Call for papers (extended): International Conference on Systems
Engineering and Engineering Management (ICSEEM 2008)
From: International Association of Engineers (IAENG)
San Francisco, USA, 22-24 October, 2008
http://www.iaeng.org/WCECS2008/ICSEEM2008.html

Important Dates:
Draft Paper Submission Deadline (extended): 22 July, 2008
Camera-Ready Papers Due & Registration Deadline: 30 July, 2008
WCECS 2008: 22-24 October, 2008

I assume you mean (n-k)! rather than (n-1)!. But even so, this formula
does not
even work for n=0, where the left hand side yields 0. Are there other
typos?

On Jul 17, 1:12 pm, "G. A. Edgar" math.ohio-state.edu.invalid>
wrote:

I would like a reference for the following.
I know how to prove it, for example using the generating function.
But better would be to provide a reference.

sum_{k=0}^n binomial(n,k) (2^k-1) B_k = 1

binomial(n,k) is a binomial coefficient n!/k!/(n-1)!

Repeat: I already know how to prove it, so I am not asking for your
proof. Do you know a reference for this...

Final CFP: Workshop on Workflow and Process Management (WfPM'08),
September 2008, Timisoara, Romania
EXTENDED DEADLINES
___________________________________________________________________________­
___________________

WfPM'08 -- Workshop on Workflow and Process Management
in conjunction with
September 29, 2008

SYNASC-2008
10th International Symposium on
Symbolic and Numeric Algorithms for Scientific Computing
Timisoara, Romania
September 26 - 29, 2008

Workshop Program Chairs
=======================
Laura Maruster, Groningen University of Groningen, The Netherlands
Teodor-Florin Fortis, West University of Timisoara, Romania

My question is mostly a "research" question in the librarians' sense
of the term. It is about the origins of various proofs of the
Chevalley-Warning theorem. Just about the only thing I know is that
Chevalley showed that if
P_1(t),...,P_r(t) are polynomials in F_q[t_1,...,t_n] with zero
constant term and such that sum_i deg(P_i) < n,
then there exists 0 =/= x in (F_q)^n such that P_1(x) = ... = P_r(x) =
0; whereas Warning improved this by removing the assumption that they
have zero constant term and concluding that the number of simultaneous
solutions is divisible by p (the characteristic of F_q).

Does anyone know what the original proofs of Chevalley and Warning
were? In Ireland and Rosen it says that Warning's proof actually
showed that the number of simultaneous solutions is at least q^{n-d}
(assuming, I suppose, that there is at least one solution). Nowadays
the proof that you almost invariably see is Ax's incredibly short
proof, which begins with the easy observation that the number of
simultaneous solutions is equal, as an element of F_q to
(equivalently, is congruent mod p to)

sum_{x in F_q^d} prod_{i=1}^n (1-P_i(x)^{q-1})

The basic idea here is that proofs which use only elementary algebra
(high-school type) and induction can be considered more elementary (I
use the phrase “exceedingly elementary”) than proofs which use
arbitrary polynomials, the binomial theorem, the arithmetic-geometric
mean inequality, or other concepts that involve sets of numbers where
the number of items in the set can vary.

This was inspired by two things:

1. my work on “exceedingly elementary” proofs that n^{1/n} -> 1 for
large integral n;
2. the recent proof, based on the arithmetic-geometric mean
inequality, that (1+1/n)^n is increasing and (1+1/n)^{n+1} is
decreasing as integral n increases. The proof is ingenious: using n
values of 1+1/n and 1 of n we get (1+1/(n+1))^{n+1} > (1+1/n)^n (i.e.
(1+1/n)^n is increasing); using n values of 1-1/n and 1 value of 1 we
get (1+1/n)^{n+1} < (1+1/(n 1))^n (i.e. (1+1/n)^{n+1} is decreasing).
Since 0 < (1+1/n)^{n+1} (1+1/n)^n = (1/n)(1+1/n)^n < 4/n > 0, the
limit of the two sequences exists – I propose that we call it “e”.

For (1), part of it involved the following “contra-Bernoulli”
inequality (CBI):

Hi everyone,

I've come up with a simple way to "derive" Gauss's prime density
function
using probability heuristics, alone. The "derivation" involves
setting up a
nonlinear differential equation whose solution happens to agree
exactly
with Gauss's estimate. I don't know if I've discovered
something interesting, or [more likely] old. At any rate, here's my
argument:

========================================================

I have found a simple way to derive Gauss's estimate of the prime
density function using probability heuristics.

Background

Excerpt From MathWorld: http://mathworld.wolfram.com/PrimeNumberTheorem.html

..
In 1792, when only 15 years old, Gauss proposed that pi(n), the prime
density function

I'm studying the semidefinite programming problem:
Given real symmetric m x m matrices (well, rational matrices in
practice) F_0, F_1, ..., F_n, obtain x_1, ..., x_n real coefficients
such that
F = F_0 + \sum_i x_i F_i >= 0 (I mean, is semidefinite positive).

We suppose the F_i to be linearly independent.

The set S of acceptable (x_1, ...., x_n) vectors is:
* closed and convex (it is the intersection of the halfspaces defined by
x^t F_0 x + \sum_i (x^t F_i x) x_i >= 0 for all vector x)
* semialgebraic: F is semidefinite positive if and only if the
coefficients of its characteristic polynomial are alternatively
nonpositive and nonnegative (the determinant is nonnegative, the next
one is nonpositive, etc.), so it is defined by m polynomial inequalities.

I would be tremendously helped if I could compute the linear algebraic
variety generated by S (I mean, the least set of the form v_0 +
Vect(v_1, ..., v_d) such that S is included in it.)

By "computing" I mean actually obtaining v_0, v_1, ..., v_d. My
intuition is that if the F_i are rational, then these should also be
rational if there is at least one rational point inside S.