ganesh wrote:See your doctor if you have difficulty breathing, chest pain, persistent fever of 102 F (39 C) or higher, or persistent cough, especially if you're coughing up pus.

What is a pus?

Pus is a yellow-white, more or less viscid substance produced by suppuration and found in abscesses, sores, etc., consisting of a liquid plasma in which white blood cells are suspended.

]]>Summary: lozenge tilings are beautiful combinatorial objects with interesting properties. In my opinion, they should be much better known. (It took some asking around even to find out the name for what I was exploring.) The best known early work related to them may be MacMahon's formula for counting plane partitions in his text Combinatory Analysis (1916). Skimming this, I don't see where he notes the equivalence to lozenge tilings, but it was certainly well established by then. The equivalence between lozenge tilings and the flat projection of a plane partition is old enough to be used by Romans in their mosaics and the 1982 arcade game Q*bert (omitting links, which would be nice for those who don't know what I'm referring to). In both of those cases, the rhombuses are tiled regularly (Rhombille tiling, which you can read about at wikipedia) but the shape admits infinitely many irregular tilings as well.

Here is an earlier write-up I placed at conwaylife.com, which is really not the right place at all. It goes into more detail and includes pictures.

What is cool about lozenge tilings? What I find most interesting are the number of seemingly different representations that are all equivalent. Some equivalences may seem trivial (especially as you start to work with them) but they are all useful "coordinate systems" for these things, presenting them in a different light. Some easier to visualize, to work with on paper, or to write computer code to analyze.

(1) Take two triominoes, labeled 0-0-1 and 1-1-0. Use as many duplicates as you like, but match them up according to rules so that any six adjacent corners agree in number.

(2) On a hex grid, fill in the cells with 0s and 1s, but make sure that no three adjacent hexes all have the same number (so again, they must have two 0s and a 1 or two 1s and a 0).

(3) On a hex grid once again, find a bipartite matching of the edges between vertices. I.e. Each vertex is matched to exactly one of the three adjacent vertices.

(4) Tile the plane with lozenges (60°-120° rhombuses)

(5) Cover the plane with stacks of cubes such that the height of stacks is non-decreasing with increasing x or y position. Now project the cube edges onto the plane x+y+z=0.

One thing I liked enough that I made cardboard tiles was a lozenge with curved sides that can be flipped between an "s" and "z" orientation. When you tile with these, each side having a different color, the orientations make nice looking patterns. The orientations actually turn out to be the cube heights mod 2 (as do the 0-1 numbers in the first two representations). I did most of this before realizing the connection to plane partitions (5) and I was puzzling over the triomino representation (1) before even realizing I was reinventing lozenge tilings.

Disclaimer: I have not done anything like a complete literature search. I am not claiming anything above is new, and in fact I believe it is all very well known to those who have studied lozenge tilings.

]]>Proof:

the Euler characteristic is

F (Faces)

E (Edges)

V (Vertices)

lets say that F, E and V are all we know

the angle defects can be represented as

The sum of the angles around each Vertex are

now

and so on....

Add them up

now we know V so we need

This represents the sum of all the angles in all the faces

The angles in an N sided polygon are:

or

If the number of sides in each face are given by

then the the sum of the angles are:

and so on...

Adding

is the sum of the number of edges in all the faces, and since there are two faces to an edgeso substituting that in

and therefore

which simplifies to

rearranging

]]>Where

Consider when p=4, n=8

Smallest solution

]]>So far the scientists have only discovered by preservation of motion,

as the motion wave transfers gradually from the root to the tip, the speed accelerates as the whip gets thinner.

However why the tip of the whip gets a sudden jump of speed as to break the sound, remains unsolved.

I believe it is a singular point problem?

Any mathematicians to theorize it?

]]>My argument is that because divisors of the Mersenne number

can’t be < p if p is a prime number. Therefore if 2p +1 is a divisor of it has no divisors as p is > the square root of 2p + 1. This will therefore make 2p + 1 a prime number.Is this proof correct?

]]>**Every integer >5 can be expressed as the sum of 3 primes**

Because I find that really **beautiful**, but then I realised that this was still just a conjecture.

Whoops!

]]>“ If a Sophie Germain prime p is congruent to 3 (mod 4), then it’s matching safe prime 2p + 1 will be a divisor of the Mersenne number 2^p - 1.”

And:

“ Fermat's little theorem states that if p is a prime number, then for any integer a, the number a^p − a is an integer multiple of p.”

However if an integer, 2p + 1, where p is a prime number, is a divisor of the Mersenne number 2^p - 1, then 2p + 1 is a safe prime and p it’s matching Sophie Germain prime.

Divisors of the Mersenne number 2^p - 1 can’t be < p if p is a prime number. Therefore if 2p +1 is a divisor of 2^p - 1 it has no divisors as p is > the square root of 2p + 1. This will therefore make 2p + 1 a safe prime and p it’s matching Sophie Germain prime.

For example 11 which is prime, (11*2) + 1 = 23. 2^11 - 1 is divisible by 23 making 11 a Sophie Germain prime and 23 it’s matching safe prime.

]]>Hi Βεν,

Nice contribution! Yes, the repeated iteration of the cosine function converges: actually, it converges to thefixed pointof the cosine function,

That makes so much sense! I also just tried

and it converged to its fixed point, 0. Tan has a fixed point but it is not attractive; sinh becomes a vertical line on 0 which means there is no attractive point, 0; cosh has no fixed point but it converges to ∞; 0 is tanh's attractive fixed point; cot has no attractive point due to it being tan with the x axis shifted; sec is related to tan because it shares half of its discontinuous points with tan suggesting it has no attractive point, and it does have no attractive point; csc is sec with the x axis shifted so it has no fixed point.]]>Welcome to the forum.

Yes, you should be doing your chem engg project. How long did this take?

forum rules wrote:

No Swearing or Offensive Topics. Young people use these forums, and should not be exposed to crudeness.

My first reaction was to delete it completely but then we lose all your posts. I think you should do the right thing and edit it yourself to something more wholesome.

Thanks,

Bob

]]>Mersenne Number= 2^11 -1

Factors = 23 and 89

2^11 -1 takes eleven goes to get to zero if I continually minus one and divide by 2. Watch what happens to 23 and 89 acting as remainders as I do the same to them.

2^11 -1

2^10 -1 First go

2^9 -1 Second go

2^8 -1 Third go

2^7 -1 Fourth go

2^6 -1 Fifth go

2^5 -1 Sixth go

2^4 -1 Seventh go

2^3 -1 Eighth go

2^2 -1 Ninth go

2-1 Tenth go

1-1=0 Eleventh go

23

(23-1)/2 = 11 First go

(11-1)/2 = 5 Second go

(5-1)/2 = 2 Third go

23+2 = 25, (25-1)/2 = 12 Fourth go

23+12 = 35, (35-1)/2 = 17 Fifth go

(17-1)/2 = 8 Sixth go

23+8 = 31, (31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

89

8(89-1)/2 = 44 First go

89+44 = 133, (133-1/2) = 66 Second go

89+66 = 155, (155-1)/2 = 77 Third go

(77-1)/2 = 38 Fourth go

89+38 = 127, (127-1)/2 = 63 Fifth go

(63-1)/2 = 31 Sixth go

(31-1)/2 = 15 Seventh go

(15-1)/2 = 7 Eighth go

(7-1)/2 = 3 Ninth go

(3-1)/2 = 1 Tenth go

(1-1)/2 = 0 Eleventh go

They both take eleven goes to get to zero too.

Working backwards it can be seen that this must be the case. Starting with zero I multiply by 2 and add one continually until I get to 2^11 -1. The same must be done to the** remainders **which will then become 23 and 89 or zero, the factors of 2^11 -1.

Let the number of goes an odd number, y, takes to get to zero by continually minusing one and dividing by 2, (adding y when even) =z.

The z for 2^11 -1, 23 and 89 =11.

**Whatever z equals for y, 2^z -1 must be factorable by y. **When we use the method for 2^z -1, we never add 2^z -1 as it never equals an even number. This could potentially alter remainders for potential factors, but for Mersenne Numbers, this is not the case.

Does anyone know what “z” a composite can’t be?

]]>