Numbered Triangle: (lack of)Prime Patterns

[stanek]

My obsession with a numbered triangle continues. Assume we arrange numbers like this.

<font class="small">Code:</font><hr /><pre>
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 52 53 54 55
56 57 58 59 60 61 62 63 64 65 66
67 68 69 70 71 72 73 74 75 76 77 78
79 80 81 82 83 84 85 86 87 88 89 90 91
</pre><hr />
There appears to be a pattern between diagonal columns that can contain prime numbers to those that never will. The 2,3,5,7 are anomalies because no primes exist past these numbers. I created a script to quickly check for primes down to the billionth row to satisfy my definition of no primes in these diagonal rows.

1 3-never 0 prime rows in between
2 5-never 1 prime rows in between
4 8-primes
7 12-never 2 prime rows in between
11 17-primes
16 23-primes
22 30-never 3 prime rows in between
29 38-primes
37 47-primes
46 57-primes
56 68-never 4 prime rows in between
67 80-primes
79 93-primes
92 107-primes
106 122-primes
121 138-never 5 prime rows in between

The gap between a diagonal column never containing a prime starts at 0 and increases by one. I've tested this for a few thousand rows and it holds true.

My question is why?
What is happening here?

Obviously I assume this has to do with ordering the numbers in such a fashion. But for the life of me I cannot understand why this is happening.

Any nudge in the right direction would be appreciated.

Thanks

[JennEric Screen Name]

Here is a semi-obvious observation. The right-hand diagonal consists of the sequence 1, 1+2, 1+2+3, ... , 1+2+ ... +N, ....

So the jth entry is the sum of all integers from 1 to j.

This sum is equal to j*(j+1)/2, which proves that no entry in this diagonal can be prime, (since either {j/2 and j+1} are both integer factors of the product or {j and (j+1)/2} are integer factors of the product).

You may be able to arrive at similar results for the other non-prime diagonals. For instance, the 2, 5, ... diagonal has entries that are one less than the 1, 3, ... diagonal. The formula would be j*(j+1)/2 - 1. If you look at this graphically, I think you will see that a j by (j+1)/2 rectangle with a 1x1 area missing from a corner can be rearranged into a (j-1)/2 by (j+1) rectangle if you leave out a 1x1 area. Meaning that j*(j+1)/2-1 is the product of two integers. This might only work half the time, with the other half being when j*(j+1)/2 -1 is even, and therefore not prime.

[Silent A]

General equation for row 'i' and element 'j' is:

= (j+i-1)(j+i)/2 - i + 1

e.g. row 6, element 4

i = 6, j = 4

= (4+6-1)(4+6)/2 - 6 + 1
= (9)(10)/2 - 5
= 45 - 5
= 40

[bigpooch]

The last number of the nth row is just C(n+1,2) = n(n+1)/2
and as mentioned, this won't be prime for n&gt;=3 and is called
a "triangular number", which gives the number of elements of
the triangle constructed so far.

A triangular number is simply a sum 1+2+...+m for some
positive integer m. You can see immediately that the
difference between consecutive triangular numbers m(m+1)/2
and (m+1)(m+2)/2 is (m+1).

The next to last number is n(n+1)/2 - 2/2 = (n^2 + n - 2)/2
or (n+2)(n-1)/2. Thus, for n large enough, you can see it
is composite with factors (not necessarily prime) of either
(n+2)/2 and (n-1) or (n+2) and (n-1)/2 depending on the
parity of n.

This should give you an idea immediately.

In the same way, you can see that for a positive integer k
between 1 and (n-1) inclusive, the product of the two
integers (n-(k-1)) and (n+k) is

(n+k)(n-(k-1)) = n^2 + n - k(k-1).

or (n^2 + n)/2 - k(k-1)/2 will be composite as long as n
is "large enough".

Thus, it should be clear that the diagonals that are a
"triangular" number of diagonals away from the last diagonal
(when the diagonals are from the upper left to lower right)
will have composite entries with the possible exception of
the first few entries near the top.

Thus, not only is the "last diagonal" essentially composite,
but also the last but one, the last but third, etc. The
reason that the "gaps" are increasing by exactly one is that
they simply reflect the differences between consecutive
triangular numbers.

In any case, if t1 and t2&lt;t1 are triangular numbers, t1-t2
is composite except in cases where t1, t2 or t1-t2 are
"small" (including when t2 and t1 are consecutive triangular
numbers).

The more interesting result (harder to prove) is that each
of the "other diagonals" have infinitely many primes.

[jay_shark]

The formula for the ith row jth column should be :

ic2+j = (i-1)(i-2)/2 + j for i&gt;1 and j for i=1 ;i&gt;=j

ie 6th row 3 column is 6c2+3 = 18

3row 2 column is 3c2+2 = 5

[jay_shark]

EDIT

ic2 +j = i(i-1)/2 +j and NOT (i-1)(i-2)/2

Silent A , your formula doesn't work .
The 6th row 4th element should be 19 but according to your formula you get 40 .

It should be 6c2+4 = 19

[Silent A]

[ QUOTE ]
EDIT

ic2 +j = i(i-1)/2 +j and NOT (i-1)(i-2)/2

Silent A , your formula doesn't work .
The 6th row 4th element should be 19 but according to your formula you get 40 .

It should be 6c2+4 = 19

[/ QUOTE ]

You misunderstood how I'm defining a "row" and an "element".

The OP is interested in the diagonal rows. As such, the first row is:

1, 3, 6, 10, 15, 21, ...

and the 6th row is:

16, 23, 31, 40, 50 ...

EDIT: I should have called them "columns" not "rows".

[FortunaMaximus]

x = 25, y = 33, z = 41, t = x.

It's an old Azteca riddle. Temporal morphlingustics.

x is itself a variable function and its valence is affected by a magnetic zeropoint interrupt as it draws closer to zero. The fluid valence would be within t = [y/z], t being self-referential within several colorspeeds.

[FortunaMaximus]

[ QUOTE ]
x = 25, y = 33, z = 41, t = x.

It's an old Azteca riddle. Temporal morphlingustics.

x is itself a variable function and its valence is affected by a magnetic zeropoint interrupt as it draws closer to zero. The fluid valence would be within t = [y/z], t being self-referential within several colorspeeds.

[/ QUOTE ]

Dimensional encoding is 9 + 1 = 91/v, for the valence to pace the growth of elemental enstructuralcoding.

[blah_blah]

[ QUOTE ]
x = 25, y = 33, z = 41, t = x.

It's an old Azteca riddle. Temporal morphlingustics.

x is itself a variable function and its valence is affected by a magnetic zeropoint interrupt as it draws closer to zero. The fluid valence would be within t = [y/z], t being self-referential within several colorspeeds.

[/ QUOTE ]

goddamn it how is it possible to be as horrible of a poster as you are. seriously, just don't post about things you clearly know nothing about (which includes, but is not limited to, math and poker).