Comments by "John Berry" (@user-ud6ui7zt3r) on "The Oldest Unsolved Problem in Math" video.
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It would seem, from inspection, that the summation results initially produce two odd numbers, followed by two even numbers, and this basic pattern repeats...
Example:
Using 1, 3, 6, 10, 15, 21, 28, 36... note that...
1, 3, both odd, followed by
6, 10 both even, followed by
15, 21 both odd, followed by
28, 36 both even .
This suggests a second pattern, which can be "layed over" the thus-far presented pattern.
In other words...
(odd) + odd = even
(1+2) + 3 = 6
...and next...
(even) + even = even
(1+2+3) + 4 = 10
...and next...
(even) + odd = odd
(1+2+3+4) + 5 = 15
...and next...
(odd) + even = odd
(1+2+3+4+5) + 6 = 21
Extracting the paradigm of this 2nd Pattern, we get...
(odd) + odd = even
(even) + even = even
(even) + odd = odd
(odd) + even = odd .
👆Let's call this column the parentheses column, Pc.
□□□□□□👆Similarly, let's call this column the NO parentheses column, NPc.
From inspection, it would seem that the identified EVEN perfect numbers only happen for the case...
(odd) + odd = even . Additionally, such perfect numbers only seem to arise when
the number appearing in the NPc column is an integer factor of the associated sum that appears NEXT TO it, in the Pc column. For example...
□□□□ ᴾ ᶜ □□□□□ ᴺ ᴾ ᶜ
□□□(1+2) +□□□□□3 [👈note that ᴺ ᴾ ᶜ 3 is an integer factor of ᴾ ᶜ 3 ]
(1+2+3+4+5+6) + 7 [👈note that ᴺ ᴾ ᶜ 7 is an integer factor of ᴾ ᶜ 21 ]
(1+2 • • • 29+30)+31 [👈note that ᴺ ᴾ ᶜ 31 is an integer factor of ᴾ ᶜ 465].
[note that...
31 × 15 = 465
...and that...
(1+2 • • • 29+30) = 465
...and that...
(1+2 • • • 29+30) + 31 = 496 ]
Using...
(1+2 • • • 29+30) + 31
...as a model, we could rewrite this as...
( ᴾ ᶜ sum ) + ᴺ ᴾ ᶜ .
We can then factor out ᴺ ᴾ ᶜ to yield...
ᴺ ᴾ ᶜ ( odd + 1 )
...or...
ᴺ ᴾ ᶜ × ( even )
...or...
odd × even .
And here's another cute elucidation!...
Going back to...
□□□□ ᴾ ᶜ □□□□□ ᴺ ᴾ ᶜ
(1+2 • • • 29+30) + 31 = 496
...note that the LARGEST INTEGER in the ᴾ ᶜ sum is the number 30.
Taking HALF of 30, we get 15 .
Wouldn't ya know it!... 15 happens to be the crucial number in...
31 × 15 = 465 [👈which is the sum of (1+2 • • • 29+30) ].
Let's try this for the other ones!
Going back to...
□□□□ ᴾ ᶜ □□□□□ ᴺ ᴾ ᶜ
(1+2+3+4+5+6) + 7 = 28
...note that the LARGEST INTEGER in the ᴾ ᶜ sum is the number 6.
Taking HALF of 6, we get 3.
Wouldn't ya know it!... 3 happens to be the crucial number in...
7 × 3 = 21 [👈which is the sum of (1+2+3+4+5+6) ].
Look at that!! It worked AGAIN! Now, going back to...
□□□□ ᴾ ᶜ □□□□□ ᴺ ᴾ ᶜ
□□□(1+2) +□□□□□3 = 6
...note that the LARGEST INTEGER in the ᴾ ᶜ sum is the number 2.
Taking HALF of 2, we get 1.
Wouldn't ya know it!... 1 happens to be the crucial number in...
3 × 1 = 3 [👈which is the sum of (1+2) ].
But all of this is to be expected...
Going back to...
(1+2 • • • 29+30) + 31 = 496
...note that we can form pairs within the ᴾ ᶜ sum.
For example...
(1 +30) = 31
(2 +29) = 31
(3 +28) = 31
□□□ •
□□□ •
□□□ •
(13+18) = 31
(14+17) = 31
(15+16) = 31
So, of course, the ᴾ ᶜ sum always has the ᴺ ᴾ ᶜ integer as a factor.
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