## New Articles

[

[

[

[

[

[

[

[

[

[

[

[

[

[

*Ref*]**Numerical verification of the Goldbach conjecture**by Paulo FernandeskySep 28[

*Ref*]**example of contractive sequence**by pahioSep 20[

*Ref*]**contractive sequence**by pahioAug 29[

*Ref*]**Some formulas of partnership**by burgessAug 26[

*Rec*]**Kenosymplirostic numbers**by imaginary.iAug 14[

*Edu*]**How to find whether a given number is prime or not...**by burgessAug 12[

*Edu*]**BODMAS Rule application**by burgessAug 8[

*Edu*]**Tests of Divisibility- Simple tricks**by burgessAug 7[

*Res*]**0/0 is possible and has an answer**by imaginary.iAug 2[

*Ref*]**Sophomore's dream**by pahioJul 9[

*Res*]**examples of growth of perturbations in chemical or...**by rspuzioMay 24[

*Ref*]**proof of Stirling's approximation**by rspuzioMay 8[

*Res*]**Example of stochastic matrix of mapping**by rspuzioApr 23[

*Res*]**6. Discussion**by rspuzioApr 20## Latest Messages

Oct 25

Oct 20

Oct 20

Oct 11

Oct 11

Oct 10

Oct 4

Sep 30

Sep 28

Sep 28

Sep 28

Sep 28

Sep 23

Sep 23

As you can probably see, the front page of the site is a bit broken right now.
I plan to fix it, and improve it a bit in the process, this weekend.
There are some other long-standing bugs that need fixing as well, and I hope to get to those as I have time.
I have a LOT of ideas about ways to improve the site -- as I'm sure many of you do too.
The problem is that we currently have zero budget.
Sincere thanks go to everyone making constructive contributions, and thanks as well for your patience. >

Oct 20

The answer should be in the Picard–Lindelöf theorem.

Oct 20

The search engine of PlanetMath is again functioning.>

Oct 11

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.

## Latest Messages

Oct 25

Oct 20

Oct 20

Oct 11

Oct 11

Oct 10

Oct 4

Sep 30

Sep 28

Sep 28

Sep 28

Sep 28

Sep 23

Sep 23

As you can probably see, the front page of the site is a bit broken right now.
I plan to fix it, and improve it a bit in the process, this weekend.
There are some other long-standing bugs that need fixing as well, and I hope to get to those as I have time.
I have a LOT of ideas about ways to improve the site -- as I'm sure many of you do too.
The problem is that we currently have zero budget.
Sincere thanks go to everyone making constructive contributions, and thanks as well for your patience. >

Oct 20

The answer should be in the Picard–Lindelöf theorem.

Oct 20

The search engine of PlanetMath is again functioning.>

Oct 11

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.