gaps in coprimes of primorials

This also implies that you can find the second half of coprimes of a primorial if you know all the coprimes less than the primorial magnitude/2 (ie for primorial 210 it is 210/2 = 105.

Also if you know just half of the coprimes, you can check the corresponding position and fill it in, so you can know a random any half of coprimes and fill in the rest, ie if you know 13 is coprime for 210, then also you can know that 210-13 is coprime etc.

cheers, Jamie

The mirrored gap sequence seems to work for any number not just primorials.

If you take the coprimes and organize them into two columns, ie for coprimes of 30:

1 29 7 23 11 19 13 17

the sum of the two columns for each row is always 30, ie 1+29, 7+23,

11+19, 13+17

For any natural number n these columns can be generated with the numbers coprimes adding to n.

Coprimes organized into two columns for different numbers n greater than

2, showing unique all unique products:

n=3 column1 column2 product

1 2 2

n=4

1 3 3

n=5

1 4 4 2 3

n=6

1 5 5

n=7

1 6 6 2 5 10 3 4 12

n=8

1 7 7 3 5 15

n=9

1 8 8 2 7 14 4 5 20

n=10

1 9 9 3 7 21

n=11

1 10 10 2 9 18 3 8 24 4 7 28 5 6 30

n=12

1 11 11 5 7 35

n=13

1 12 12 2 11 22 3 10 30 4 9 36 5 8 40 6 7 42

n=14

1 13 13 3 11 33 5 9 45

n=15

1 14 14 2 13 26 4 11 44 7 8 56

n=16

1 15 15 3 13 39 5 11 55 7 9 63

n=17

1 16 16 2 15 30 3 14 42 4 13 52 5 12 60 6 11 66 7 10 70 8 9 72

cheers, Jamie

Here's a picture of a spreadsheet showing an interesting pattern for primorials coprimes when organized into two columns so that all pairs of coprimes add to the primorial size (30 and 210 in the spreadsheet)

The difference (gap) in consecutive products of the coprime pairs (column E) has greatest common divisor 12 for both primorial 30 and primorial 210 which is interesting, so that all gaps in the products of the pairs of coprimes are multiples of 12. Any idea why that is?

cheers, Jamie

of 12

Let v stand for a primorial of 6 or more (eg 6, 30, 210...)

The difference d between any two "products of coprime pairs" can be expressed as d = q*(v-q) - p*(v-p), where none of those four factors are divisible by 2 or 3. We want to show that d mod 12 is 0, which is equivalent to showing that d is a multiple of 3 and a multiple of 4.

Expanding, we have d = q^2-p^2 + v*(q-p). Because q-p is even and v is a multiple of 6, v*(q-p) is a multiple of 12 and need not be considered further. That is, d mod 12 == q^2 - p^2 == (q+p)*(q-p).

Because q+p and q-p are both even, d mod 12 is a multiple of 4.

Now write p = 3u+a, where a is 1 or 2, and q = 3w+b, similarly. Consider the four (b,a) cases: (1,1): q-p = 3w+1-3u-1 == 0 mod 3 so (q+p)*(q-p) == 0 mod 3 (1,2): q+p = 3w+1+3u+2 == 0 mod 3 so (q+p)*(q-p) == 0 mod 3 (2,1): q+p = 3w+2+3u+1 == 0 mod 3 so (q+p)*(q-p) == 0 mod 3 (2,2): q-p = 3w+2-3u-2 == 0 mod 3 so (q+p)*(q-p) == 0 mod 3 Thus, d mod 12 is a multiple of 3. As a multiple of both 3 and 4, d mod 12 is 0, as we wanted to show.

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jiw