Posted by: "chrisdarroch" chrisdarr2@... chrisdarroch

Date: Fri Jun 6, 2008 10:23 pm ((PDT))

Just an additional explanation of what I am asking.

Consider a value n = 38.

Consider values x =[13,17,19,23,29,31,37,41,43,47] which are all

indivible(as it were) by any member of A.

***************

Kermit says:

Where A = [2,3,5,7,11]

************

chrisdarroch says:

Consider the results of n+x: [51,55,57,61,75,79,81,85]

*******

Kermit says:

38 + [13,17,19,23,29,31,37,41,43,47]

=[51,55,57,61, 75, 79, 81, 85]

Since A = [2,3,5,7,11], consider representing each of the numbers in x ,

[13,17,19,23,29,31,37,41,43,47]

by the set of remainders when divided by an element in A.

That is,

13 = [1,1,3,6,2] since 13 has remainder 1 when divided by 2, remainder 1 when divided by 3,

remainder 3 when divided by 5, remainder 6 when divided by 7, remainder 2 when divided by 11.

Make a matrix from x using the remainders when divided by elements of A.

A= 2 3 5 7 11

13 1 1 1 6 2

17 1 2 2 3 6

19 1 1 4 5 8

23 1 2 3 2 1

29 1 2 4 1 7

31 1 1 1 3 9

37 1 1 2 2 4

41 1 2 1 6 8

43 1 1 3 1 10

47 1 2 2 5 3

By looking at the column under the 2, we see that if n is odd, every sum will be divisible by 2,

and if n is even, no sum will be divisible by 2.

By looking at the column under the 3, we see that if n is a multiple of 3, that no sum will be divisible by 3,

and that if n has remainder 1 when divided by 3, that the number of 2's under the 3 is the number of sums that

will be divisible by 3.

Etc

Since under 3 there are 1 and 2,

and under 5 there are 1,2,3,4

and under 7 there are 1,2,3,5,6, [ but not 4]

and under 11 there are 1,2,3,4,6,7,8,9,10, [ but not 5]

then we can easily calculate what n to use to make every sum not divisible by any element of A.

n must be divisible by 2.

n must be divisible by 3.

n must be divisible by 5.

n must either be divisible by 7 or have remainder 3 when divided by 7.

n must either be divisible by 11, or have remainder 6 when divided by 11.

Thus n must be one of the four forms.

n1 = 2 * 3 * 5 * 7 * 11 k = 2310 k

n2 = 2310 k + 1260

n3 = 2310 k + 990

n4 = 2310 k + 2250

If n is any of these four forms, then every sum will be not divisible by

any element of A.

********

chrisdarroch says:

All of these results are divisible by at least one member of

A:[2,3,5,7,11] except for 61 and 79 which are primes. The largest gap

within which I have found a prime thus far is 10 lets say, i.e.

between 51 and the first prime of 61.

I might find a relative prime instead of a prime and that gap would

count for example if I added x = 131 to n = 38 I would get 169 which

is divisible by 13 but not any member of A and this is how I define

relatively prime in this case.

Chris

I cannot see that tautology here unless I am using the phrase

Relatively Prime improperly.

*******

Kermit says:

Predictions of gaps can also be determined by looking at the remainder matrix,

but it is much more complicated.