Would those who will.........consider the following series of number

pairs.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 .........210

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 .........211

take out any pair which includes any number divisible by 3...leaving

1 4 7 10 13 16 19...............208

2 5 8 11 14 17 20...............209

Take out all of the pairs divisible by 5.....leaving

1 7 13 16 ......................208

2 8 11 17 ......................209

Do the same for 7.

The number of gaps formed will be 30, and this can be found by the

Chinese Remainder Theroem (I believe).

I am looking for a formula which will tell me about the largest gap

that will be left between two consecutive pairs.

Obviously for this one example I can use observation and say that

there is at least one gap of size 15, which is the largest.

But I am also interested in the largest gap for the series of pairs

ending in 2.3.5.7.11 (2310) pairs; and of course the series of pairs

ending in 2.3.5.7.11.13 and so on.

I need to place some kind of bound on the largest gap that could

exist between two consective pairs in such series.

Let me return to my first example to show some of my thinking on the

subject.

By the CRT I can see that out of 210 pairs, there remain 30 gaps.

I may approach the subject of the largest one by segregating out one

gap as "that largest" and the other 29 as equivalent or smaller.

I can say that due to the extraction of all pairs containing a number

divisible by 3 the gap between any consecutive pair remaining must be

3 and thus I have 29 gaps which are 3 at least.

Up till then I can say that 3.29 gaps cover an area of 87 and thus

the other gap must be at most 210 - 87 = 123

If I consider the fact that more of the pairs are extracted due to

containing numbers divisible by 5 then, I can say after that

extraction, some gaps will be 3 and others form gaps of 6.

I can say that 210 /(3.5) gaps will be 6 i.e. 14 gaps will be of size

6.

Thus I can say that out of the 29 gaps 14 of them must cover an area

of at least 6 and the remaining 15 can be 3.

Thus I can say that so far the 29 cover an area of

(6.14) + (3.15) = 129

Thus the remaining single gap must at this point be at most

210 - 129 = 81

The difficulty begins when I have to consider other factors.

Can anyone see this line of reasoning being sensible?

Are there alternatives.

For example can one use something of the results for the largest gap

between primes such as:

ยท g(pn) < (1/16597)pn for n > 2010760 (Schoenfeld 1976)

to form an estimate for the largest gap in my series?

Chris