## Search found 431 matches

- Sun Aug 14, 2016 9:23 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Primes, sequences and infinite

It seems you are looking for a closed formula for this ? Exactly! But I want to reach it using other sequence easy to find. I still do not know how to simplify the second sequence the negative. All the results will be the first step to have another formula for the prime counting function. To comput...

- Sun Aug 14, 2016 9:09 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Primes, sequences and infinite

I start working on the details. A(p) could be divided in 2 sequences : one positive A`(p) and the second negative A``(p). A`(p)=p*S(p) (this one is solved A``(p) depends mainly on the gaps between 2 consecutive p(n+1)-p(n) (it is more complicated to simplify. I`m not looking for an approximation of ...

- Sun Aug 14, 2016 7:38 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Primes, sequences and infinite

Thank you anyway. Results do exist : asymptotic limit of S(p) and A(p). Each one of them was obtained using known results. That is not my purpose. I will do it by myself (I mean the expansion of A(p) in function of S(p)) even if it is going to take long time. I know what I`m looking for. The result ...

- Sun Aug 14, 2016 5:13 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Primes, sequences and infinite

I did the computation of A(p) and S(P) up to the 10000th prime : 10000th prime = 104729 When A(p)=10000 S(P)=2.70925824879732 If I could find closed formula for A(p)=g(S(P)) for some given p (g is some function to find) then I could after simplification find more precisely the asymptotic limit of S(...

- Sun Aug 14, 2016 4:52 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Primes, sequences and infinite

It's ln(ln(n)). You can find that information on the linked wiki page. Thank you.. Everything I can read on wiki or on internet I did it before posting. Here is what I want to expand. I need to expand A(p) in function of S(p) here are the first elements : A(p)=(2*S(P))+(S(p)-(1/2))+(S(p)-(1/2+1/3))...

- Sun Aug 14, 2016 3:29 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Primes, sequences and infinite

In other words how can we express A(P) in S(p)?

- Sun Aug 14, 2016 3:28 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Primes, sequences and infinite

I did it long time ago.

What I mean by limit is asymptotic limit.

I thought it was obvious.

Sorry for not mentioning it.

- Sun Aug 14, 2016 2:51 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Primes, sequences and infinite

Hi,

Let

A(p)=(1/2+1/2)+(1/3+1/3+1/3)+(1/5+1/5+1/5+1/5+1/5)+....+(1/p+1/p+...1/p) p times

S(P)=1/2+1/3+1/5+1/7+....1/p

A(p)=pi(p) where pi(p) is the counting function of primes

What is the exact limit of S(p) when p goes to infinite?

Thank you for any comment.

Let

A(p)=(1/2+1/2)+(1/3+1/3+1/3)+(1/5+1/5+1/5+1/5+1/5)+....+(1/p+1/p+...1/p) p times

S(P)=1/2+1/3+1/5+1/7+....1/p

A(p)=pi(p) where pi(p) is the counting function of primes

What is the exact limit of S(p) when p goes to infinite?

Thank you for any comment.

- Fri Aug 12, 2016 12:34 pm UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

If this was their solution I really think that is dumb one. A has 3 cards 1,6,7 He does not know what B has. He could write down 6 numbers : 1,2,3,5,6,7 by claiming my three cards are among those 6 cards which is true B has 3 cards : 2,3,5 He will then deduce that C holds the card 4 hence he will c...

- Fri Aug 12, 2016 12:29 pm UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

- Fri Aug 12, 2016 12:22 pm UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

Here you can read one the best pdf documents about the russian card problem :

http://cacr.uwaterloo.ca/~dstinson/Wils ... tinson.pdf

http://cacr.uwaterloo.ca/~dstinson/Wils ... tinson.pdf

- Fri Aug 12, 2016 11:49 am UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

Anyway there are soiutions other than the Fano plane (Fano plane was given as solution).

Please google "russian cards pdf" before trying to answer my unfinished solution.

Please google "russian cards pdf" before trying to answer my unfinished solution.

- Fri Aug 12, 2016 11:45 am UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

If this was their solution I really think that is dumb one. A has 3 cards 1,6,7 He does not know what B has. He could write down 6 numbers : 1,2,3,5,6,7 by claiming my three cards are among those 6 cards which is true B has 3 cards : 2,3,5 He will then deduce that C holds the card 4 hence he will c...

- Fri Aug 12, 2016 10:26 am UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

I was talking about the solution given on some websites. The solution is using Fano plane.

Just google "russian cards problem". There are many pdf documents about this problem.

The problem has in fact many solutions. In my point of view it is still open for finding the best solution.

Just google "russian cards problem". There are many pdf documents about this problem.

The problem has in fact many solutions. In my point of view it is still open for finding the best solution.

- Thu Aug 11, 2016 4:54 pm UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

If this was their solution I really think that is dumb one. A has 3 cards 1,6,7 He does not know what B has. He could write down 6 numbers : 1,2,3,5,6,7 by claiming my three cards are among those 6 cards which is true B has 3 cards : 2,3,5 He will then deduce that C holds the card 4 hence he will cl...

- Thu Aug 11, 2016 4:14 pm UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

While searching on internet I read the solution. I`m not convinced by the solution as the word statement is not clear at all. In french we say "c`est du n`importe quoi?". Anyway it happens to me many times the solution is not convincing at all. And if is a solution then there are many othe...

- Thu Aug 11, 2016 3:10 pm UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

It does not seem to work because I did not finish my demonstration.

I did not say that once A say sum odd or even what B would say.

I`m still sick but in 2 or 3 days I will post all my demonstration.

Thank you for commenting.

I did not say that once A say sum odd or even what B would say.

I`m still sick but in 2 or 3 days I will post all my demonstration.

Thank you for commenting.

- Thu Aug 11, 2016 1:12 am UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

No need to use modulo. Odd and even sums could solve the problem. A (or B) when they claim the sum is Even (or Odd) there are 4 possibilities : If the sum of the 3 cards is even then A (or B) would have : 1. Odd-Odd-Even 2. or Even Even Even As there are 3 Even numbers (2,4,6) and 4 Odd numbers (1,3...

- Thu Aug 11, 2016 12:34 am UTC
- Forum: Logic Puzzles
- Topic: Logic Puzzles - Admittedly, not naturally equipped to solve them...
- Replies:
**17** - Views:
**11537**

### Re: Logic Puzzles - Admittedly, not naturally equipped to solve them...

Do not worry about IQ (low or high is not important) because solving puzzles has nothing to do with IQ. I have seen people solving hard puzzles only because they worked hard to know some tricks to recognize some patterns and so on. The same prevail with chess game. I have seen "dumb" peopl...

- Tue Aug 09, 2016 4:11 pm UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

Here is the starting statement assuming that A is the first to make a statement : - A sum his 3 cards and say the sum of my 3 cards is ODD or EVEN - Hence B by watching his 3 cards could know if the card of C is EVEN or ODD as the sum of the 7 cards is EVEN. B know for sure if C has (2,4,6) or (1,3,...

- Tue Aug 09, 2016 11:52 am UTC
- Forum: Logic Puzzles
- Topic: Sharing secret information publicly
- Replies:
**27** - Views:
**7069**

### Re: Sharing secret information publicly

If you mean that C could not know at least one card hold by A or B then I think that is impossible. C know for sure the 6 remaining cards once he knows the card he owns. A and B have to make each one a true statement such as A and B knows what C card have. Once A and B know what card C have then A a...

- Sun Aug 07, 2016 9:06 pm UTC
- Forum: Logic Puzzles
- Topic: Puzzle about odd semi-primes and Euler totient
- Replies:
**8** - Views:
**3719**

### Re: Puzzle about odd semi-primes and Euler totient

Here is another strange : Let us call S(n) the sum of modulo power 2 up to the biggest power < n Let us call phi(n) the Euler totient of n If we sum all the S(n) and all the phi(n) up to 100000 we obtain almost a similar sum : Sum of the S(n)`s = 3107218357 Sum of the phi(n)`s = 3039610754 Ratio sig...

- Sun Aug 07, 2016 4:47 pm UTC
- Forum: Logic Puzzles
- Topic: Puzzle about odd semi-primes and Euler totient
- Replies:
**8** - Views:
**3719**

### Re: Puzzle about odd semi-primes and Euler totient

For n to up 100000 I obtained 100 ECN : 6,10,12,18,20,21,24,27,36,40,48,72,75,80,91,96,98,119,138,144,160,192,196,276,288,320,377,384,392,485,546,552,576,640,768,784,802,1092,1104,1152,1251,1280,1323,1536,1568,1604,2184,2208,2304,2560,2602,3072,3136,3154,3208,4368,4416,4608,5120,5204,6144,6272,6308,...

- Sun Aug 07, 2016 12:27 pm UTC
- Forum: Logic Puzzles
- Topic: Puzzle about odd semi-primes and Euler totient
- Replies:
**8** - Views:
**3719**

### Re: Puzzle about odd semi-primes and Euler totient

I do not have enough odd numbers to see if there is a pattern : Only 9 Eulerian composite numbers (ECN) : 21,27,75,91,119,377,485,1251,1323 I need to compute n up to 1000000 to have more. On Excel it is going to take lot of time. If I discover some pattern for odd numbers it will be then easy to com...

- Sun Aug 07, 2016 11:21 am UTC
- Forum: Logic Puzzles
- Topic: Puzzle about odd semi-primes and Euler totient
- Replies:
**8** - Views:
**3719**

### Re: Puzzle about odd semi-primes and Euler totient

I hope there was no mistake. Here are the 67 Eulerian composite numbers for n up to 10000: 6,10,12,18,20,21,24,27,36,40,48,72,75,80,91,96,98,119,138,144,160,192,196,276,288,320,377,384,392,485,546,552,576,640,768,784,802,1092,1104,1152,1251,1280,1323,1536,1568,1604,2184,2208,2304,2560,2602,3072,313...

- Fri Aug 05, 2016 6:39 pm UTC
- Forum: Logic Puzzles
- Topic: Puzzle about odd semi-primes and Euler totient
- Replies:
**8** - Views:
**3719**

### Re: Puzzle about odd semi-primes and Euler totient

I hope there was no mistake. Here are the 67 Eulerian composite numbers for n up to 10000: 6,10,12,18,20,21,24,27,36,40,48,72,75,80,91,96,98,119,138,144,160,192,196,276,288,320,377,384,392,485,546,552,576,640,768,784,802,1092,1104,1152,1251,1280,1323,1536,1568,1604,2184,2208,2304,2560,2602,3072,3136...

- Fri Aug 05, 2016 6:00 pm UTC
- Forum: Logic Puzzles
- Topic: Puzzle about odd semi-primes and Euler totient
- Replies:
**8** - Views:
**3719**

### Re: Puzzle about odd semi-primes and Euler totient

I computed n up to 10000. If I did not make a programming mistake I obtained only 67 Eulerian composite numbers.

What is strange is that after the Eulerian odd number 1323 which is not semi prime there was only even numbers.

Why? I do not know for now.

What is strange is that after the Eulerian odd number 1323 which is not semi prime there was only even numbers.

Why? I do not know for now.

- Fri Aug 05, 2016 4:02 pm UTC
- Forum: Logic Puzzles
- Topic: Puzzle about odd semi-primes and Euler totient
- Replies:
**8** - Views:
**3719**

### Re: Puzzle about odd semi-primes and Euler totient

If we write in base 2 the number n=pq (where p and q are odd distinct prime) it will be easier to find a way to know if any number n is Eulerian odd semi prime number. Assuming that we could recognize quickly an Eulerian odd semi prime number it will lead to an easy factorization of n. I know that o...

- Fri Aug 05, 2016 1:53 pm UTC
- Forum: Logic Puzzles
- Topic: Puzzle about odd semi-primes and Euler totient
- Replies:
**8** - Views:
**3719**

### Re: Puzzle about odd semi-primes and Euler totient

Without filtering the odd semi-primes I have obtained this list up to 500 : 6,10,12,18,20,21,24,27,36,40,48,72,75,80,91,96,98,119,138,144,160,192,196,276,288,320,377,384,392,485 So 30 numbers (let us call them Eulerian composite numbers) out of 500 (6%). Only 5 are odd semiprimes : 21,91,119,377,485

- Thu Aug 04, 2016 6:19 pm UTC
- Forum: Logic Puzzles
- Topic: Puzzle about odd semi-primes and Euler totient
- Replies:
**8** - Views:
**3719**

### Puzzle about odd semi-primes and Euler totient

Hi, I will start by an example : n=21 n is an odd semi-prime = 3*7 (3 and 7 are distinct prime) The biggest power of 2 immedialy < 21 is = 4 So 2^4=16<21 Let us compute 21 mod (2^k) with k varying from 2 to 4 21=1 mod 2 21=1 mod 4 21=5 mod 8 21=2 mod 16 If we sum the residue modulo : 1+1+5+5=12 (1) ...

- Mon Aug 01, 2016 11:00 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Theorem and consequences (powers between x^k and (x+1)^k)

Sorry! I have found a reasoning mistake and I need to restate my second theorem by correcting it.

- Sun Jul 31, 2016 3:52 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Theorem and consequences (powers between x^k and (x+1)^k)

As a<x^3<(a+1)^2 and b<y^3<(b+1)^2 then if (a+b)<z^3<(a+1)^2+(b+1)^3 we could conclude that z^3=x^3+y^3

Am I right?

The same prevail for powers n>3 as long as we use the squares as bounds.

False or true?

Thank you for any comment.

Am I right?

The same prevail for powers n>3 as long as we use the squares as bounds.

False or true?

Thank you for any comment.

- Fri Jul 29, 2016 11:06 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Theorem and consequences

The n=3 special case of Fermat was already proven by Euler, long before Wiles came along. I gave the example of n=3 but we could extend to n>3. First : I need to prove that my second theorem is true for n=3. Second : Any power n>3 could be bounded by 2 squares and so on. Thank you for your comment....

- Fri Jul 29, 2016 8:53 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Theorem and consequences

Here is my idea. I expose it as it came to my mind when I posted. Let us take the sum of cubes : x^3+y^3=z^3 (1) We know that the equation (1) does not hold (A.Wiles proved it) The first x^3 will be between 2 squares a^2 and (a+1)^2 The second y^3 will be between 2 squares b^2 and (b+1)^2 If we sum...

- Fri Jul 29, 2016 7:53 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Theorem and consequences

Here is my idea. I expose it as it came to my mind when I posted. Let us take the sum of cubes : x^3+y^3=z^3 (1) We know that the equation (1) does not hold (A.Wiles proved it) The first x^3 will be between 2 squares a^2 and (a+1)^2 The second y^3 will be between 2 squares b^2 and (b+1)^2 If we sum ...

- Fri Jul 29, 2016 4:01 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Theorem and consequences

The idea behind this theorem which is obvious is some sort of personal puzzle. Now that I know that this theorem is true what can I do with this truth? I could add another theorem existing or future to make things more complicated. I have no answer for now. I`m still thinking on how to use it to rea...

- Thu Jul 28, 2016 12:20 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Theorem and consequences (powers between x^k and (x+1)^k)

Hi, Theorem : k and x are positive integers k>1 x>1 Between x^k and (x+1)^k non-inclusive there exist either 0 or 1 numbers expressed as y^(k+1). Example : Between the 2 squares x^2 and (x+1)^2 non-inclusive there is either 0 or at most 1 cube What are the consequences of this theorem if proved true...

- Tue Jul 26, 2016 11:47 am UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Factorials

Here is an algorithm to test quickly if such k exists. 1. Compute (k!)^2 < n! to find the maximal k 2. Compute from 2 to k the number r=n!/(k!)^2 3. If r is the product of 2 consecutive numbers a*(a+1) then the number k corresponding is solution otherwise compute the next value of k. Example : n=6 6...

- Mon Jul 25, 2016 9:45 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Finding k such that (n^k) is the biggest number < (n-1)!

I`m very sorry. It was my fault : instead of reading the line 200 (k=161) on Excel I read 201 (k=162). Very sorry. Anyway when n reach 300 digits and more I do not know if your formula gives the right value of k. Thank you very very much for your help. I need the exact value of k (that is the first ...

- Mon Jul 25, 2016 8:54 pm UTC
- Forum: Mathematics
- Topic: Goahead52's Math Posts
- Replies:
**148** - Views:
**19027**

### Re: Finding k such that (n^k) is the biggest number < (n-1)!

Thank you very much. Now I can use the formula written in Latex. There is a difference of 1 as n grows to 200. Maybe more when n reach 300 digits. Anyway I have to use maybe some properties of sequences instead of Stirling approximation. k=floor (sigma log(i) with i varying from 2 to n-1 divided by ...