Is the following producing a list of all prime numbers without skipping in a consecutive order?

Question

The following consecutive list is made as a table with the following columns :

column 1	column 2	column 3
$A$	$B=(A^2 -1) / 6$	prime factors of $B$

Where $A$ is odd and $A \mod 3 ≠ 0$

Example: $A = 7$ : $B=(7^2-1) / 6 = 8$ : prime factors of $8$ are $1,2$

You will notice that the results start building a sequence of all possible prime numbers, and never skipping a prime number in a consecutive way.

For example: If $19$ is a prime factor result of $A=37$, it means that $17$ had somewhere appeared before, and as if it would been impossible for $19$ to appear after a resulted $13$ while skipping a previous $17$. I have run the experiment until $A=1000$ and it seems to be true. Is there any proof of it being true? or is there a counter example?

Update: per @Ross Millikan comments 1 is not a prime and 5 and 7 appear in the list before 3 (making it the only exception so far). I didn't take into account 3 as a prime because the whole list is generated by the condition that $$ is odd and $ \mod 3 ≠0$, and the reason $1$ was part of the results is just produced by my programming.

To better illustrate, here is a sample list where $A$ is growing consecutively (Where $A$ is odd and $A \mod 3 ≠ 0$):

$A$ : $B=(A^2 -1) / 6$ : prime factors of $B$

5 : 4 : prime factors of 4 are 2

7 : 8 : prime factors of 8 are 2

11 : 20 : prime factors of 20 are 2 , 5

13 : 28 : prime factors of 28 are 2 , 7

17 : 48 : prime factors of 48 are 2 , 3

19 : 60 : prime factors of 60 are 2 , 3 , 5

23 : 88 : prime factors of 88 are 2 , 11

25 : 104 : prime factors of 104 are 2 , 13

29 : 140 : prime factors of 140 are 2 , 5 , 7

31 : 160 : prime factors of 160 are 2 , 5

35 : 204 : prime factors of 204 are 2 , 3 , 17

37 : 228 : prime factors of 228 are 2 , 3 , 19

41 : 280 : prime factors of 280 are 2 , 5 , 7

43 : 308 : prime factors of 308 are 2 , 7 , 11

47 : 368 : prime factors of 368 are 2 , 23

49 : 400 : prime factors of 400 are 2 , 5

53 : 468 : prime factors of 468 are 2 , 3 , 13

55 : 504 : prime factors of 504 are 2 , 3 , 7

59 : 580 : prime factors of 580 are 2 , 5 , 29

61 : 620 : prime factors of 620 are 2 , 5 , 31

65 : 704 : prime factors of 704 are 2 , 11

67 : 748 : prime factors of 748 are 2 , 11 , 17

71 : 840 : prime factors of 840 are 2 , 3 , 5 , 7

73 : 888 : prime factors of 888 are 2 , 3 , 37

77 : 988 : prime factors of 988 are 2 , 13 , 19

79 : 1040 : prime factors of 1040 are 2 , 5 , 13

83 : 1148 : prime factors of 1148 are 2 , 7 , 41

85 : 1204 : prime factors of 1204 are 2 , 7 , 43

89 : 1320 : prime factors of 1320 are 2 , 3 , 5 , 11

91 : 1380 : prime factors of 1380 are 2 , 3 , 5 , 23

95 : 1504 : prime factors of 1504 are 2 , 47

97 : 1568 : prime factors of 1568 are 2 , 7

101 : 1700 : prime factors of 1700 are 2 , 5 , 17

103 : 1768 : prime factors of 1768 are 2 , 13 , 17

107 : 1908 : prime factors of 1908 are 2 , 3 , 53

109 : 1980 : prime factors of 1980 are 2 , 3 , 5 , 11

113 : 2128 : prime factors of 2128 are 2 , 7 , 19

115 : 2204 : prime factors of 2204 are 2 , 19 , 29

119 : 2360 : prime factors of 2360 are 2 , 5 , 59

121 : 2440 : prime factors of 2440 are 2 , 5 , 61

Answer 1

In effect you are asking whether there exists a prime $p$ which is not a factor of $\frac{A^2-1}{6}$ for any $A$ not divisible by $2$ or $3$

Any $A$ not divisible by $2$ or $3$ will be of the form $A=6n+1$ or $6n-1$ for some $n$.

• If $A=6n+1$ then $\frac{A^2-1}{6} = 2n(3n+1)$ which is divisible by $n$
• If $A=6n-1$ then $\frac{A^2-1}{6} = 2n(3n-1)$ which is also divisible by $n$

So just let $n=p$ and $A=6p\pm1$ (your choice): you will then have $\frac{A^2-1}{6}$ divisible by $p$.

For example for $p=2$, $A=11$ or $13$ works; for $p=3$, $A=17$ or $19$ works; for $p=5$, $A=29$ or $31$ works; for $p=7$, $A=41$ or $43$ works; for $p=11$, $A=65$ or $67$ works; and so on. There will always be such examples fitting this pattern on top of many other cases. And you do not need to restrict $n$ to be prime: for example for $n=4$, $A=23$ or $25$ leads to $\frac{A^2-1}{6}$ being divisible by $4$.

So no, there are no such primes.

Answer 2

To add to Henry's answer: your claim that the first appearances of primes (other than 3) come in increasing order is also correct. More specifically, for all primes $p > 3$, the first occurrence of $p$ in the last column of the table is next to either $A = 2p-1$ or $A = 2p+1$. Proof: since $B = (A-1)(A+1)/6$ (and $p$ doesn't divide 6), $p$ divides $B$ if and only if it divides either $A-1$ or $A+1$. The first such odd numbers $A > 1$ with this property are $A = 2p \pm 1$. So $p$ will first appear next to either $A = 2p-1$ (if this isn't a multiple of 3) or $A = 2p+1$ (otherwise).