Combing the Primes

Let’s try to give to the prime numbers another “address” than their usual one. Today’s address is their rank in the Naturals:

1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 ...

We will use a series of multiple “combs” to locate the primes.

Our 1st comb has 2 spaces between each tooth. We place the first tooth on zero:

0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 ...

|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|_____|___ ...

Our 2nd comb has 4 spaces between each tooth. We place the first tooth on the first previously uncombed integer yet -- one:

0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 ...

|___________|___________|___________|___________|___________|___________|___________|___________|___________|___________|___________|___________|___ ...

Our 3rd comb has 8 spaces between each tooth. We place the first tooth on the first uncombed integer yet, as usual -- three:

0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 ...

|_______________________|_______________________|_______________________|_______________________|_______________________|_______________________|___ ...

Our 4th comb has 16 spaces between each tooth. We place the first tooth on the first uncombed integer yet -- seven:

0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 ...

|_______________________________________________|_______________________________________________|_____________________________________ ...

Our 5th comb has 32 spaces between each tooth. We still place the first tooth on the first uncombed integer yet -- fifteen:

0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 ...

|_______________________________________________________________________________________________|_____________ ...

... etc. As one knows, this way of defining combs is the most economical one in order to punch every integer with the minimum set of different combs. (Most of you have already noticed that the nth comb has its first tooth on [(2^(n-1))-1] and has [(2^n)-1] spaces between each tooth).

Let's now give to each prime a P(n,y) address with n=(the nth comb) and y=(the yth tooth on that comb). We’ll then produce this table:

2-->(1,2)       23-->(4,2)        59-->(3,8)

3-->(3,1)       29-->(2,8)        61-->(2,16)

5-->(2,2)       31-->(6,1)        67-->(3,9)

7-->(4,1)       37-->(2,10)       71-->(4,4)

11-->(3,2)       41-->(2,11)       73-->(2,19)

13-->(2,4)       43-->(3,6)        79-->(5,3)

17-->(2,5)       47-->(5,2)        83-->(3,11)

19-->(3,3)       53-->(2,14)       97-->(2,25)      etc.

Jacques Tramu computed and drew the first hundred primes in (n,y) Cartesian coordinates: does something interesting appear?

I’m not sure... ;-?

Best,

É.