I hope my pinning will allow me to get more results related to Ramsey Theory. Consider coloring (just a function)

and  .  A c-pinning is a set such that
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Let me call    the pinfree set. Set    is free of color  .

I got an elegant way (efficient, algebraic!) to introduce the multiplication of integers. Actually more than that. The common approach works not only for Inteland (integers) but also for all cyclands. Thus the conceptual sequence is:

minusop   ⇒   cycland   Inteland

It’s 2012, and to spend time on foundations of integers may seem silly. Nevertheless it was worth it! Technically, everything is easy, virtually trivial (especially after it’s done). But my Inteland (no more “InteGland”, no more “G”) has conceptual value, it allows regular mathematicians to have a good feeling about it, it has value for education. Inteland means a mild but nice breakthrough since the times of Peano and Hilbert. The topic in the past was so tedious that only logicians cared about it. No textbook on number theory would bother to introduce integers (or natural numbers) thoroughly. They would say this and that about induction and perhaps a few words about commutativity and associativity (?), and otherwise they would rely on the reader’s good intuition of natural numbers.
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Enough of this shameless self-praising. It’s time to write all of it again from scratch. Should I go to Starbucks at Arborland. I am at the Glenco Crossing shopping center, as usually. The one at Arborland is open 24h/day all year except for the X’mass.

The coming Tuesday (in 2 days) will be messed up for me.

OK, plans, or rather vague possibilities:

• Mathematics (not related directly to computers):
• Topology–go back, and seriously, to my universal functions. I had better proofs than those which I had published soon afterwards. It bothers me. But I’d like also to obtain new results. Of course!
• Finish Integland; and start an exposition of number theory (what for? also, which direction to choose–I have a few)?
• Learn or obtain number-theoretic results which lead to a more efficient (computer) search of baroque numbers.
• publish one way or another my old result about combinatorial circles (ECC of equal weight)
• Edit anew, improve and republish my several old results.
• Just write-write-write-… on many of the attractive themes which I have touched upon in the past (OK, that simplified my task of listing my mathematical plans :-)).

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Hm, it was a disappointing, unsatisfactory exercise, so incomplete.

There is a regular Starbucks patron who talks loud and in a voice which cannot be ignored. When he does so I can’t concentrate on anything. Hm, just as I wrote these words he started to talk in a softer voice, less loud, for the first time. We at our online pharmacy store here cialis 5 mg pharmacy provide the FDA approved this drug after analyzing its clinical efficacy and safety. Kamagra comes in many versions like Generic kamagra, canada pharmacy tadalafil , Edegra, Silagra, Penegra etc. A detailed surgical valve analysis is performed to determine the functional type and the super cialis cheap segmental localization. What Changes while Lovemaking Session? Such complications india viagra for sale of aging can ultimately effect on how one performs in bed. But it’ll take me a while before I can get going, especially that from time to time he raises his voice again (not too bad–what happened/ :-)). (So far there is only one such regular around here).

Let  V  be a  v-element set. Let’s call every set of two different triangles which share an edge–a  folder;  and every set of three different triangles which together have only four vertices–a  tent.

Let

c : (V ## 3) → {0 1}

be an arbitrary 2-coloring of triangles. Let

• Δ_e  :=  |c^-1(e)|  be the number of triangles colored  e = 0 or 1;
• integer  B  be the number of tetrahedra which have two faces painted in one, and the other two faces in the other color;
• C  :=  (v ## 4) – B;
• S  be the number of unicolor tents;
• Q  be the number of the remaining tents, i.e.;

  Q  :=  4*(v ## 4) – S

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Let’s assume that (for the given coloring) there are no unicolored tetrahedra, i.e. that

B + C   =   (v ## 4)

Then

• S  =  C
• Q  =  4*B + 3*C

---

Each folder is contained in exactly 2 different tents, and each tent contains exactly 3 different folders. Thus, in particular, the number of folders is

(v ## 2) * ((v-2) ## 2)   =   6 * (v ## 4)

---

(I’m not doing anything, I’m just typing; I am trying to extend my edge coloring approach–in the past I have not obtained any sharp result for the triangles-tetrahedra case, and it’s tantalizing). Each unicolor tent contains 3 unicolor folders (and no other), and each bicolor tent contains 1 unicolor folder, and 2 bicolor folders. Let  F_e  be the number of e-colored folders,  e=1 or 2. Then

• F₁  =  (1/2) * (3*S + Q)
• F₂  =  Q

(My past analysis indicates that if I get enough of the 2-2 balanced colored tetrahedra (on the assumption that there are no unicolored tetrahedra) then I will get the desired contradiction, i.e. the existence of at least one unicolor tetrahedron, hopefully for a low value of  v.

One kind of inequalities comes from the distribution of triangle colors

e₀ + e₁  =  ((v-2) ## 2)

which share edge  e. The more even split the fewer unicolored folders, and more bicolored folders (thus we get a lower bound on the number of the unicolored folders, and an upper bound on the bicolored folders). More is needed.

I just can’t work away from my place. I am wasting a lot of time at Starbucks. I’ll move to a new place in one month, and I hope to do better then.

Let me describe a sequence of fields isomorphic to R via iterations of exp. The multiplication of each of them is the sum of the next one. I could vary the base, make it dependent of the index of the consecutive iteration but first let me keep it simple.

Let:

• e_-2  :=  -∞
• e_-1  :=  0   (thus  e_-1 = e^e_-2)
• e_n  :=  e^e_n-1   for every  n=0 1 …

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so that  e_n = eⁿ  for every  n=0 1 …. We have the standard addition and multiplication operations in

R = (e_-2; ∞)

namely

• #₀  :=  + : (e_-2; ∞)² → (e_-2; ∞)
• #₁  :=  * : (e_-2; ∞)² → (e_-2; ∞)

followed by more advanced admun (add-multiplication) operations:

#_n : (e_n-3; ∞)² → (e_n-3; ∞)

defined by:

exp(x) #_n exp(y)  :=  exp(x #_n-1 y)

for every  n = 2 3 …

Let  R_n  :=  (e_n-2); ∞)  for each  n=0 1…  Then

(R_n  #_n  #_n+1  e_n-1  e_n)

is a field, where:

• R_n  is the set of elements of the field;
• #_n  is the field’s addition operation (it’s commutative and associative);
• #_n+1  is the field’s multiplication operation (it’s commutative and associative);
• e_n-1  is the field addition’s neutral element (“zero”, even if it looks like something else )
• e_n  is the field multiplication’s neutral element (“one”, even if it looks like something else)

The Starbucks is about to close, I need to go (while Starbucks at Arborland is open 24h–should I write there before I move?).

Yes, I went home then, but now I am editing my entry (cheating? :-)).

They’ll close Starbucks in 13 minutes. I’ve spent here half a day and accomplished a big… NOTHING. I hope to be more efficient once I move to a new apartment.

I pushed Integland further at home, on my Eec. It’s good that it still works despite the battery going down the drain. I met the mild challenge presented by Integland (IntegLand?–either way it is a bit ugly). Now it’s just a chore, and I will have to force myself to continue, to finish.

I feel like writing about my old ideas related with the iteration of the isomorphism

exp : R → (0;∞)

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a#b := a^log(b)

Operation # is of course commutative and associative, it’s the multiplication in a disguise.

I should really go now, it’s 22:57.