Monday, May 31, 2004
prime
why is the distribution of primes so hard to find?
1 2 3 4 5 6 7 8 9 10
1112 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
3132 33 34 35 36 37 38 39 40
4142 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
6162 63 64 65 66 67 68 69 70
7172 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100
Except 2, every prime is odd.
A prime larger than 10 must end with {1,3,7,9}
Prime occurs irregularily.
There are infinite number of primes. (If there's a finite set of primes, we can have a product of the elements in the set + 1 which is not divisible by any of the elements).
So, how would I generalize occurance of primes if it's irregular?
Neglecting the first row, each row can have max of 4 primes.
02nd row has 4 primes (-1 -3 -7 -9).
03rd row has 2 primes ( -3 -9).
04th row has 2 primes (-1 -7 ).
05th row has 3 primes (-1 -3 -7 ).
06th row has 2 primes ( -3 -9).
07th row has 2 primes (-1 -7 ).
08th row has 3 primes (-1 -3 -9).
09th row has 2 primes ( -3 -9).
10th row has 1 prime! ( -7 ).
11
31
41
61
71
Except 2, every prime is odd.
A prime larger than 10 must end with {1,3,7,9}
Prime occurs irregularily.
There are infinite number of primes. (If there's a finite set of primes, we can have a product of the elements in the set + 1 which is not divisible by any of the elements).
So, how would I generalize occurance of primes if it's irregular?
Neglecting the first row, each row can have max of 4 primes.
02nd row has 4 primes (-1 -3 -7 -9).
03rd row has 2 primes ( -3 -9).
04th row has 2 primes (-1 -7 ).
05th row has 3 primes (-1 -3 -7 ).
06th row has 2 primes ( -3 -9).
07th row has 2 primes (-1 -7 ).
08th row has 3 primes (-1 -3 -9).
09th row has 2 primes ( -3 -9).
10th row has 1 prime! ( -7 ).
0 Comments:
Post a Comment
<< Home