Я медленно начал расстегивать ширинку на своих джинсах, наслаждаясь просмотром как эта сучка долбит себя вибратором!

Ольку и занялся с ней сексом. Он смотрел на меня с недоумением и даже спросил, не перепила ли я. Но я повторила своё предложение ещё раз и сказала, что я этого хочу сама. Он

а поцелуями покрывать её шейку. Оля, закрыв глаза, прижимала голову Сергея к себе.
что самой Алисе, мы с ней порой часами жеманились, терлись носиками и чмокали друг друга в губки. У
у нас похоже был небольшой перерыв, а потом все как нахлынуло, надо перестраиваться под новые
. Я пальчиками удовлетворяла свою дырочку, глядя на довольное лицо моей подруги, которую шпарит сзади мой муж. Оля стонала и,

Рассказывая очередную довольно неприятную историю произошедшую со мной за время его отсутствия, я вдруг не ожиданно даже для самой себя расплакалась и Виталя растерявшись и незная что же делать прижал меня к себе и
дала ему понять, что безумно его хочу! Он вдруг стал снова нежен, ласково раздвинул мне
опять сидела за ноутбуком отца.
- Да, уже собираюсь. – отозвался я. – Господи, что у тебя с руками??? – |

Я снова намылил руки и стал аккуратно смывать с живота токсин. Груди старался не касаться, член итак был на пределе. Опустив взгляд вниз, я чуть не присвистнул усмехнувшись. Волосы на лобке были аккуратно выбриты буквой V. Очень сексуально. Видимо так бережно охраняемая отцом дочка уже не девочка. Ниже по ногам также были легкие красные разводы.

Anyway, It occurs to me that there are two ways to alter the original problem that may be interesting. In the descriptions that follow, “left” and “right” are relative to looking at the line perpendicularly.
First way: points are allowed to be collinear. When the line turns to meet additional points, the new pivot is now the nth from the left if the original was the nth from the right. Any points outside the (new,old) pivot pair are counted as used, while points between the (new,old) pair are not. The new pivot and old pivot need not be distinct.
Second way: a ray is used. When a point intersects with the ray, that point becomes the new pivot. At a pivot change, the ray reverses direction (so if it was infinite to the left, it’s now infinite to the right – and vice-versa).

I believe the original proof still works for the first modification, but it easily fails for the second. However, during my brief thoughts, I have so far failed to think of an example where the second can’t also be solved.

Thus in this case we return to point P each rotation by pi, so the windmill goes through P infinitely often. But, as noted earlier, at any other stage of the windmill with the line going through a point Q we have the very same properties, so the windmill goes through every point visited infinitely often. Appealing to geometry, a point of S can only switch sides of the line if the line has at some point gone through it (okay, this might be another gap but it seems reasonable). So in this case all the points of S have been visited after a rotation through pi. Hence the result follows in the case a=b.

In the other case, we can suppose WLOG that b=a+1 (otherwise again switch orientation by pi). So after the rotation through pi we have a vertical line through some P’ in S with b points on the left and a on the right. This is equivalent to sliding our original line to the right until it hits the first point of S. Now do the rotation through pi again and we have a vertical line through some P’’ with a points of S on the left and b on the right — but then P’’ must be P by the sideways sliding argument again. So in this case we return to P after a rotation through 2pi. Applying similar arguments as in the previous case, the windmill visits every point of S infinitely often as required.

Just thought I’d let you know that I’ve proven Andrica’s Conjecture by re-examining the Sieve of Eratosthenes:
http://www.ugcs.caltech.edu/~kel/MPP/AndricaConjectureTrue.pdf
The proof shows that, at the n-th step of the sieve, the largest gap generated is at most $2 p_{n-1}$, from which Andrica follows. I call the state of elimination at the n-th step $\kappa_n$.

So, right off the bat, you can find a prime in $N^{1/2}$, even without the Riemann Hypothesis. (Note that this isn’t an asymptotic result, like $N^{0.525}$; it holds for all primes.)

However, I believe we can do a better search in practice. A naive generation of $\kappa_n$ up to $N$ requires $O(log N)$ computation and $O(N)$ storage. At that point, the density of primes is around
$M_n = \Pi_{i=1}^{n} \frac{p_i -1}{p_i} ~ \frac{1}{e^\gamma \log{p_n}}$
But a more sophisticated search would pre-calculate where in $\kappa_n$ $N$ lay and only keep around the requisite intervals, namely those that lie around a Biggest Resolution of $p_B = p_n$. Check out Theorems 2.8 and 2.9 for more details.

Btw, I’m firmly convinced Cramer’s Conjecture is true. I have developed a conditional proof that shows if Cramer’s Conjecture were true, then all the constellation infinity conjectures must be true simultaneously. Were Cramer true, then primes are packed incredibly tight and with much more regularity than we can currently show. The side effect of the conditional proof is that we could not only find _a_ prime after $N$ in log time, we could find the _exact next prime_ in log time.

I hope this helps you in your quest for deterministic prime finding!