Frustration Rules Board Games

Enumeration of the first

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Meanwhile, here is a simple algorithm gives the following first :

- Without making divisions.
- Not block multiples.

aim is to build all the primal divisors of N (a divisor is an integer primal of dividing one and only one n such that n / d always gives a first if n is greater than d ^ 3).
This amounts to construct all cycles n according to two principles:

1) The hierarchy
2) Moving

The result is the "primary reason" for any integer. Either result

N integers.
was placed first the "1". We took the square of 2 (4).

1 2 3 4 1 1 1 2


then placed the "2" to the next square (9). Complete with the "1".

4 5 6 7 8 9
2 1 2 1 2 3

then placed the "3" to the next square (16). Complete with the "2" and "1".

9 / 3
10 / 2
11 / 1
12 / 3
13 / 1
14 / 2
15 / 3
16 / 4

then placed the "4" to the next square (25) . Complete with the "3", "2" and "1".

16 / 4
17 / 1
18 / 3
19 / 1
20 / 4
21 / 3
22 / 2
23 / 1
24 / 4
25 / 5

Etc.

We see, over time, that "holes" are formed within a given sequence, reproducing the first differences multiplied by d. holes then "1" provides the usual succession of the former (1,2,3,5,7,11,13,17 holes ...), then "2" gives the succession of first increased by 2 (10,14,22,26,34,38,46 holes ...), then "3" give the succession of the first multiplied by 3 (33,39,51,57,69,87. ..), and so on. It appears that the "primary reason" is not just the "first" but all integers. The set N is the set of cycles n sorted by the first reason. He did not, so, the dichotomous attributed to him in ages as soon as one approaches the issue of prime numbers, look popularized by the "sieve of Eratosthenes", which means that N is formed of "first" and "multiple". However, all integers are sorted according to multiple primary reason, not just the first, the usual result is that the sequence of multiples of 1 sorted according to this same reason, isolated from the other suites.
For a given cycle, the primary reason for forming successors for cycles, which exert a stress position on lower cycles. It is therefore impossible to construct isolation the reason for a given cycle (otherwise we would quickly reach a very large prime numbers).
Ex: the cycle "2" changes from rank 10 to rank 14, because the cycle "3" has priority over the cycle "2" in row 12. In other words, "2" is not a divisor of 12 primal. Similarly, the cycle "4" has priority over the cycle "3" to rank 24, which passes Range 27: 3 is not a divisor of 24 primal.
The first order of apparently non-periodic, is actually the result of a stress position exercised by any n its predecessors - not the order that appears when all the "many" were eliminated.
In a sense, there is no difference between the order of integers and order of the former, the latter appears only if you isolate a given cycle of all integers.
other words: it is because a whole is unique there is a prime order.

I make here the theorem resulting from this algorithm:

Either an integer n and a prime p.
Let d be the largest divisor of n whose square is less than or equal to n. If n is greater than d ^ 3, then n / d = p.

Ex:
n = 68. Most
d if d ^ 2 d ^ 3 <> (= 64) then 68 / 4 = 17 (first).
I recall that after the of (primal divisors of N) obtained by the algorithm. The next problem was to calculate d for any integer. It's obviously very difficult, even if we know that n is composite.