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*Ref*]**Russian escort Shanghai to get the more**by clain559Sep 30## Latest Messages

Oct 11

Oct 11

Oct 10

Oct 4

Sep 30

Sep 28

Sep 28

Sep 28

Sep 28

Sep 23

Sep 23

Sep 19

Sep 17

Sep 15

I've done a few samples using direct summing and it seems that your:
1/(2^2q) should be 1/(2^q)
Why don't you try it; I could have the formula feeding the summation wrong.
----------------
binomial(10,5)/(2^10);
sum(binomial(2*k,k)*(-1/2)^k*binomial(10,k),k,0,10);
(9/2)!/(5!*sqrt(pi));
----------------------
Answers
63/256
63/256
63/256

Oct 11

Using Maxima's simplify_sum gives for a limit of 2*n
----
$\sum_{p=0}^{2n}\left(\frac{-1}{2}\right)^{p}\binom{2n}{p}\binom{2p}{p}$
$\frac{\left(\frac{2*n-1}{2}\right)!}{\sqrt{(\pi)}\cdot n!}$
Which seems strange until you evaluate
--
$\left(\frac{\left(2\cdot4-1\right)}{2}\right)!=\left(\frac{105}{16}\right)\cdot\sqrt{\pi}$
So the sqrt(pi)'s cancel. Remember $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$
And for odd n
$\sum_{p=0}^{2n+1}\left(\frac{-1}{2}\right)^{p}\binom{2n+1}{p}\binom{2p}{p}$
0
----------------------
Maxima also has the Zeilberger algorithm. I will copy the answer here when I
understand it. If you feel you need it.
--------------------
Incidently simplify_sum doesn't give a proof certificate but Zeilberger does.
----
Ray

Oct 10

Would anybody be interested in discussing a presentation here on using Pascal/Shift matrices to
directly generate various Polynomial sequences via. their generating functions expressed in matrices?
The underlying idea is that most generating functions are still true when indexing variables (t) are replaced with full rank singular matrices (i.e. n-1). In particular the Pascal or Shift matrices. This leads to the generating function expressed in terms of matrices and most familiar generating functions directly stating the polynomials. Also that the series is truncated automatically and exactly.
I do have some theorems rather than just words :)
I have a "blog" where I have stuffed some notes and results.
Ray>

Oct 4

Attention Dr. Puzio: Kindly see my request to Administration.
Would be glad if you would kindly do the needful.

Sep 30

This is to remind administration about my request to either
a) enable " copy and paste operation" on the templates reserved
for articles and messages or b) open a page on facebook which
will automtically enable copying and pasting as well as uploading
snapshots of articles and messages.

Sep 28

Hi parag,
There is no set theory needed for the proof. I think only the
expressions of m and M are strange for you; they simply mean
that m is the least and M the greatest of the given fractions!
Jussi

Sep 28

I am not able to understand that proof at planetmath.org/summednumeratorandsummeddenominator as I am not aware of the set theory. Please give a simple proof for that.

Sep 28

parag,
do you mean such as in
http://planetmath.org/summednumeratorandsummeddenominator?

Sep 28

Many of the Gaussian integers indicated in "Fermat's theorem
in k(i)" and " pseudoprimes in k(i) " are too big to be copied manually.
I, therefore, suggest that the template for articles and
messages be modified to enable "copy and paste" operations. Alternately
planetmath.org open a page on facebook. This will enable us to
upload snapshots of messages/articles.

Sep 23

hi bro, i think this tip will help you :
- Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.
and i found an harshad number digit sum 19:
874

Sep 23

We can construct pseudoprimes in k(i) as follows:
Take a couple of primes in k(1) of the same type i.e. both
being of type 4m+3 or both being of type 4m+1.
Let us take 3*7 = 21 as an example of first type. Choose
as base 21 + i. ((21+i)^20-1)/21 yields a quotient which is
a Gaussian integer; hence 21 is a psedoprime in k(i).
Note a) This is possible only with the aid of software pari
or similar. b) 21 + i, 21 - i, 1 - 21i, -1+21i, all yield
identical Gaussian integers when we apply Fermat's theorem.
b) Second base is chosen as follows: Partition 22 into two parts
such that one is divisible by 3 and the other by 7 i.e.
7 + 15i, 7 - 15i, -7 + 15i and -7 - 15i are four different
points on the complex plane which yield identical Gaussian integers
when Fermat's theorem is applied. Similar remarks apply
to composites in k(1) each of which is of type 4m + 1.

Sep 19

When we apply Fermat's theorem to four different points in the
complex plane we get an invariant result; the four different points
are 21 + i, -21 - i, 1 - 21i and -1 + 21i. i.e. ((21 + i)^20-1)/21=
((-21-i)^20-1)/21 = ((1-21i)^20)/21 = ((-1+21i)^20/21.

Sep 17

Although the above is not functioning I am able to read my older
messages by clicking on "older". However wish the search facility
is restored early.

Sep 15

Regretable that search facility is still not functioning.

## Latest Messages

Oct 11

Oct 11

Oct 10

Oct 4

Sep 30

Sep 28

Sep 28

Sep 28

Sep 28

Sep 23

Sep 23

Sep 19

Sep 17

Sep 15

I've done a few samples using direct summing and it seems that your:
1/(2^2q) should be 1/(2^q)
Why don't you try it; I could have the formula feeding the summation wrong.
----------------
binomial(10,5)/(2^10);
sum(binomial(2*k,k)*(-1/2)^k*binomial(10,k),k,0,10);
(9/2)!/(5!*sqrt(pi));
----------------------
Answers
63/256
63/256
63/256

Oct 11

Using Maxima's simplify_sum gives for a limit of 2*n
----
$\sum_{p=0}^{2n}\left(\frac{-1}{2}\right)^{p}\binom{2n}{p}\binom{2p}{p}$
$\frac{\left(\frac{2*n-1}{2}\right)!}{\sqrt{(\pi)}\cdot n!}$
Which seems strange until you evaluate
--
$\left(\frac{\left(2\cdot4-1\right)}{2}\right)!=\left(\frac{105}{16}\right)\cdot\sqrt{\pi}$
So the sqrt(pi)'s cancel. Remember $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$
And for odd n
$\sum_{p=0}^{2n+1}\left(\frac{-1}{2}\right)^{p}\binom{2n+1}{p}\binom{2p}{p}$
0
----------------------
Maxima also has the Zeilberger algorithm. I will copy the answer here when I
understand it. If you feel you need it.
--------------------
Incidently simplify_sum doesn't give a proof certificate but Zeilberger does.
----
Ray

Oct 10

Would anybody be interested in discussing a presentation here on using Pascal/Shift matrices to
directly generate various Polynomial sequences via. their generating functions expressed in matrices?
The underlying idea is that most generating functions are still true when indexing variables (t) are replaced with full rank singular matrices (i.e. n-1). In particular the Pascal or Shift matrices. This leads to the generating function expressed in terms of matrices and most familiar generating functions directly stating the polynomials. Also that the series is truncated automatically and exactly.
I do have some theorems rather than just words :)
I have a "blog" where I have stuffed some notes and results.
Ray>

Oct 4

Attention Dr. Puzio: Kindly see my request to Administration.
Would be glad if you would kindly do the needful.

Sep 30

This is to remind administration about my request to either
a) enable " copy and paste operation" on the templates reserved
for articles and messages or b) open a page on facebook which
will automtically enable copying and pasting as well as uploading
snapshots of articles and messages.

Sep 28

Hi parag,
There is no set theory needed for the proof. I think only the
expressions of m and M are strange for you; they simply mean
that m is the least and M the greatest of the given fractions!
Jussi

Sep 28

I am not able to understand that proof at planetmath.org/summednumeratorandsummeddenominator as I am not aware of the set theory. Please give a simple proof for that.

Sep 28

parag,
do you mean such as in
http://planetmath.org/summednumeratorandsummeddenominator?

Sep 28

Many of the Gaussian integers indicated in "Fermat's theorem
in k(i)" and " pseudoprimes in k(i) " are too big to be copied manually.
I, therefore, suggest that the template for articles and
messages be modified to enable "copy and paste" operations. Alternately
planetmath.org open a page on facebook. This will enable us to
upload snapshots of messages/articles.

Sep 23

hi bro, i think this tip will help you :
- Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.
and i found an harshad number digit sum 19:
874

Sep 23

We can construct pseudoprimes in k(i) as follows:
Take a couple of primes in k(1) of the same type i.e. both
being of type 4m+3 or both being of type 4m+1.
Let us take 3*7 = 21 as an example of first type. Choose
as base 21 + i. ((21+i)^20-1)/21 yields a quotient which is
a Gaussian integer; hence 21 is a psedoprime in k(i).
Note a) This is possible only with the aid of software pari
or similar. b) 21 + i, 21 - i, 1 - 21i, -1+21i, all yield
identical Gaussian integers when we apply Fermat's theorem.
b) Second base is chosen as follows: Partition 22 into two parts
such that one is divisible by 3 and the other by 7 i.e.
7 + 15i, 7 - 15i, -7 + 15i and -7 - 15i are four different
points on the complex plane which yield identical Gaussian integers
when Fermat's theorem is applied. Similar remarks apply
to composites in k(1) each of which is of type 4m + 1.

Sep 19

When we apply Fermat's theorem to four different points in the
complex plane we get an invariant result; the four different points
are 21 + i, -21 - i, 1 - 21i and -1 + 21i. i.e. ((21 + i)^20-1)/21=
((-21-i)^20-1)/21 = ((1-21i)^20)/21 = ((-1+21i)^20/21.

Sep 17

Although the above is not functioning I am able to read my older
messages by clicking on "older". However wish the search facility
is restored early.

Sep 15

Regretable that search facility is still not functioning.