Problems regarding properties of the Floyd's Triangle

Question

Setup

Have a look at the Floyd's Triangle:-

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 
92 93 94 95 96 97 98 99 100 101 102 103 104 105 
106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 
121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 
...

Here, all the Natural Numbers $(\mathbb{N})$ are written consecutively, row-wise. Each row has a length of $n$, where $n$ increments by $1$ in each succession.

I noticed several properties of this triangle, related to a few of which I have questions. So, I have broken this post down into $3$ parts for related questions that I found the most perplexing.

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Part 1

Let's take the sum of each row. Their sequence would be:-

1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, ...

which is given by the formula $\frac{n(n^2 + 1)}{2}$.

If we take the averages of all the $n$th and $(n + 2)$th terms (where $n$ is an odd number), we get the sequence:-

(1 + 15) / 2, (15 + 65) / 2, (65 + 175) / 2, ...

or,

8, 40, 120, 272, 520, 888, ...       --- (i)

Then, taking the sequence of the middle or the even $n$th terms [($n + 1$)th terms, where $n$ is odd], we get this:-

5, 34, 111, 260, 505, 870, ...       --- (ii)

Subtracting the sequence (ii) of middle terms from the sequence (i) of the averages of the neighbouring extremes, we get:-

8 - 5, 40 - 34, 120 - 111, 272 - 12, 520 - 505, 888 - 870, ...

or,

3, 6, 9, 12, 15, 18, ...

Why are multiples of 3 showing up here?

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Part 2

(Here's a smaller version of the same triangle, for your ease of viewing):-

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 *
...

Have a look at the "hypotenuse" or the right-most numbers of the rows of the triangle. Notice that they are triangular numbers, given by the formula $\frac{n(n + 1)}{2}$.

Take a look at the pairs of $2$nd and $3$rd, $5$th and $6$th, $8$th and $9$th, ... $(3n - 1)$th and $(3n)$th of these triangular numbers (shown above by *s). That is,

(3, 6), (15, 21), (36, 45), (66, 78), ...

Observe that, yet again, all of these numbers are divisible by 3. Written as multiples of 3, they are:-

(3 * 1, 3 * 2), (3 * 5, 3 * 7), (3 * 12, 3 * 15), (3 * 22, 3 * 26), ...

Here, the differences between the multiplicands in individual pairs are:-

2 - 1, 7 - 5, 13 - 12, 26 - 22, ...

or,

1, 2, 3, 4, 5, ...

And their sums are:-

2 + 1, 7 + 5, 13 + 12, 26 + 22, ...

or,

3, 12, 27, 48, ...

or,

3 * 1^2, 3 * 2^2, 3 * 3^2, 3 * 4^2,...

Also, observe the difference between the first term of the $(n + 1)$th pair, and the last term of the $n$th pair (where $n \in \mathbb{N}$):-

15 - 6, 36 - 21, 66 - 45, ...

or,

9, 15, 21, ...

or,

3 * 3, 3 * 5, 3 * 7, ...

The multiplicands of the differences are consecutive odd numbers.

Moreover, the terms that we left out (denoted by -s):-

1, 10, 28, 55, 91, 136, ...

also follow a pattern. The differences between this sequence's consecutive terms are:-

10 - 1, 28 - 10, 55 - 28, 91 - 55, 136 - 91, ...

or,

9, 18, 27, 36, 45, ...

What is the reason behind these specific numbers (and mainly $3$) popping up? What is the intuitive reason and / or proof for why things are the way they are? 

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Part 3

(The same triangle reproduced again):-

@ * - 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
      92 93 94 95 96 97 98 99 100 101 102 103 104 105
      106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
    - 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136
      ...

This time, observe the "perpendicular" or the left-most numbers. The following are the perfect squares that show up (shown by -s):-

1, 4, 16, 121, 529, 4096, 17956, 139129, 609961, 4726276, 20720704, ...

Their respective square roots are:-

1, 2, 4, 11, 23, 64, 134, 373, 781, 2174, 4552, ...

The differences between consecutive terms of this sequence are:-

1, 2, 7, 12, 41, 70, 239, 408, 1393, 2378, ...      ---- (iii)

And the differences between this sequence's (iii) terms are:-

1, 5, 5, 29, 29, 169, 169, 985, 985, ...

Why do only these numbers pop up, that too in pairs of two?

Let's look at the vertical gap between two consecutive numbers:-

0, 1, 2, 9, 16, 57, 98, 337, 576, 1969, 3362, ...

For example, the first perfect square -- $1$, shows up in the first row. So, there is $0$ gap between the row containing the previous perfect square and that of $1$. Similarly, $4$ comes $1$ row after $1$, so it has a gap of $1$, $16$ comes $2$ rows after $4$, so it has a gap of $2$, $121$ comes $9$ rows after $16$, so it has a gap of $9$, etc.

This sequence's differences are:-

1, 1, 7, 7, 41, 41, 239, 239, 1393, 1393, ...

Notice that this looks like the sequence (iii) [i.e., of the differences between consecutive terms of the square roots of the perfect squares, found in the left-most perpendicular column of the triangle.]

1, 2, 7, 12, 41, 70, 239, 408, 1393, 2378, ...
1, 1, 7, 7, 41, 41, 239, 239, 1393, 1393, ...

The only variation being that all of the even $n$th terms are replaced by the term that precedes them.

Why is this?

Let's observe the terms that got "ignored" [even $n$th terms]:-

2, 12, 70, 408, 2378, ...

Dividing this by 2,

1, 6, 35, 204, 1189, ...

Here, after a bit of scrutiny, we observe that the formula for the $n$th term of this sequence would be:-

\begin{align*} T_n = 6 \cdot T_{n-1} - T_{n-2} \end{align*}

where $T_0 = 0$ and $T_1 = 1$.

For example, $T_2$ will be 6 * 1 - 0, $T_3$ will be 6 * 6 - 1, $T_4$ will be 6 * 35 - 6, etc.

What is the reason behind this exact formula here?

Searching the sequence up in OEIS, we get this, having the exact formula we just noticed. But it is also given there that $T_n ^ 2$ is a triangular number.

Why is this true?

Now let's look at the ones that did not get "ignored" [odd $n$th terms]:-

1, 7, 41, 239, 1393, ...

We observe that the formula for the $n$th term of this sequence would be the same as the previous one:-

\begin{align*} T_n = 6 \cdot T_{n-1} - T_{n-2} \end{align*}

where $T_0 = -1$ and $T_1 = 1$.

Note that this time, $T_0 = -1$, and not $0$.

The same questions emerge again, why is this true? What are the relationships among these numbers and formulae, and what more is special about them?

Let's now look for primes in this perpendicular (shown by *s):-

1, 2, 7, 11, 29, 37, 67, 79, 137, 191, 211, 277, 379, 631, 821, 947, 991, 1129, 1327, 1597, 1831, 2017, 2081, 2347, 2557, 2851, 2927, 3571, 3917, 4561, 4657, 4951, 5051, 5779, 6217, 6329, ...

The first few of them come clumped in pairs, but then the pattern changes and becomes confusing at best. I tried seeing if there is a pattern in the fact that some are equal to $3$ (shown by !s), while some are equal to $1$ (shown by @s) (mod $4$), but no luck again. There were many such false alarms, where the patterns ultimately broke apart.

Could you try and explain why these primes show up, and what are some interesting relations / properties that you can come up with?

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I know this is a really long question, so it would be really helpful if you could answer or extend any aspect of any problem. Moreover, more properties and relationships that you can find related to this triangle are really appreciated. Thanks a lot for giving your time.

Answer 1

Let me discuss Part 1, and perhaps it'll give you an idea of how to approach others. First, I want to denote by $i_n$ the initial number of the $n$th row of your table, and $e_n$ then final number in the $n$th row, so that \begin{align} i_1 &= 1& e_1 &= 1\\ i_2 &= 2& e_2 &= 3\\ i_3 &= 4 & e_3 &= 6 \end{align} and so on. In general, $i_{n+1} = e_n + 1$, of course, and $e_n = \frac{n(n+1)}{2}$, as you have observed.

The sum of the $n$th row is \begin{align} \sum_{k = i_n}^{e_n} k &= (i_n + 0) + (i_n + 1) + \ldots + (i_n + (e_n - i_n)) \\ &= i_n + i_n \ldots + i_n + (0 + 1 + \ldots + (e_n - i_n)) \\ \end{align} Now the $n$th row contains $n$ items, so the sum above becomes \begin{align} \sum_{k = i_n}^{e_n} k &= n(i_n) + (0 + 1 + \ldots + n-1) \\ &= n(i_n) + \frac{(n-1)n}{2} \\ &= n\left( \frac{(n-1)n}{2} + 1 \right) + \frac{(n-1)n}{2} \\ &= n \frac{(n-1)n}{2} + n + \frac{(n-1)n}{2} \\ &= (n+1) \frac{(n-1)n}{2} + n \\ \end{align} Let's call that sum $S_n$, and compute $2S_n$: \begin{align} S_n = \sum_{k = i_n}^{e_n} k &= (n+1) \frac{(n-1)n}{2} + n \text{ so} \\ 2S_n &= (n+1)(n-1)n + 2n\\ 2S_n &= (n^2 - 1)n + 2n\\ 2S_n &= n^3 - n + 2n\\ 2S_n &= n^3 + n\\ 2S_n &= n(n^2 + 1)\\ S_n &= \frac{n (n^2 + 1)}{2} \end{align} as you found. Let's call $$ Q_n = \frac{S_n + S_{n+2}}{2} $$ which is your sequence called (i) .. almost. (We'll only want the odd terms) The formula can be written out with some algebra \begin{align} 4Q_n &= 2S_n + 2S_{n+2}\\ &= n(n^2 + 1) + (n+2)((n+2)^2 + 1) \\ &= n^3 + n + (n+2)^3 + (n + 2) \\ &= n^3 + n + n^3 + 6n^2 + 12n + 8 + n + 2 \\ &= 2n^3 + 6n^2 + 14n + 10 \text { so }\\ 2Q_n &= n^3 + 3n^2 + 7n+ 5 \end{align} The averages you want are the 1st, 3rd, etc items in this sequence, namely $$ A_n = Q_{2n-1} $$ (which tells me I should have started my indexes at zero, but it's too late now), so that \begin{align} 2A_n &= (2n-1)^3 + 3(2n-1)^2 + 7(2n-1)+ 5 \\ &= (2n)^3 - 3(2n)^2 + 3(2n) - 1 + 3\left(4n^2 - 4n + 1) \right) + 14n - 7 + 5 \\ &= 8n^3 - 12n^2 + 6n - 1 + 12n^2 - 12n + 3 + 14n - 2 \\ &= 8n^3 +8n \\ \end{align}

The corresponding "middle" term $M_n$ is just $S_{2n}$. o \begin{align} 2M_n &= (2n)( (2n)^2 + 1) \\ &= (2n)( 4n^2 + 1) \\ &= 8n^3 + 2n \\ \end{align} Taking the difference, we get \begin{align} 2(A_n - M_n) &= (8n^3 +8n) - (8n^3 + 2n) \\ &= 6n \\ \end{align} so $$A_n - M_n = 3n$$ and that's where your multiple of 3 are coming from.

There's nothing here except naming of things, keeping indexes right, and algebraic simplification. I'm pretty sure you'll find the same thing for the remaining parts.

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I just thought about this a little more, and there's a better (or at least less-algebra) way to think about it. Let's look at rows $3,4,5$; this is the case where $n = 3$, and I'll keep referring back to $n$ as we go.

4  5  6
7  8  9  10
11 12 13 14 15

I'm going to transform this by moving the 15 to the first row, but I'm going to split it into a 7 (that goes in the first row) plus 8 more at the end.

4  5  6  7
7  8  9  10
11 12 13 14    + 8

Note: in general, what I did was take $e_{n+2}$, and split it into $e_{n} + 1$ (that's the 7 I moved) and an extra bit, $e_{n+2} - e_n + 1$ (that's the 8 I tacked on at the end).

Now I add the first and third rows (aiming for an average eventually) and I get

7  8  9  10
15 17 19 21    + 8

And then I divide by 2 to get the average of the first and 3rd row:

7  8  9  10
7.5 8.5 9.5 10.5    + 4

The difference between these is then

0.5 0.5 0.5 0.5 + 4

or $4 (0.5) + 4$.

The key thing to see here is that exactly the same thing works for any three rows; the only things that change are

(a) the multiplier 4: if we start at row $n$, the middle row of our triple has $n+1$ items, and that's what we use in place of the $4$, and

(b) the 'extra' 4 at the end.

Before looking into those, though, How do I know that the differences will always be $0.5$s?

Well, where did that $0.5$ come from? It was the difference between (a) the average of the first elements from row $n$ and $n+2$ and (b) the fist element of row $n+1$. In other words, I took $(i_3 + i_5)/2 - i_4$. Let's write that out \begin{align} H &= (i_3 + i_5)/2 - i_4 \\ 2H &= i_3 + i_5 - 2i_4 \\ &= (e_2 + 1) + (e_4 + 1) - 2(e_3 + 1) \\ &= e_2 + e_4 - 2e_3 \\ &= (1 + 2) + (1 + 2 + 3 + 4) - (1 + 2 + 3) - (1 + 2 +3) \\ &= (1 + 2) - (1 + 2 + 3) + (1 + 2 + 3 + 4) - (1 + 2 + 3) \\ &= -3 + 4 = 1 \end{align}

So $H$ turns out to always be $1/2$.

Thus, the number we're computing looks like $$ (n + 1) \cdot \frac12 + e $$ where the extra term $e$ is half of $e_{n+2} - e_n + 1$. Let's work that out too: \begin{align} e &= \frac12 \left(e_{n+2} - e_n + 1 \right)\\ &= \frac12 \left(\frac{(n+2)(n+3)}2 - \frac{n(n+1)}2 + 1 \right)\\ &= \frac14 \left((n+2)(n+3) - n(n+1) + 2 \right)\\ &= \frac14 \left(n^2 + 5n + 6 - (n^2 + n) + 2 \right)\\ &= \frac14 \left(4n + 8 \right)\\ &= n + 1\\ \end{align} (I'm pretty sure there's a shorter way to see this, but ...)

So now we have a formula. The difference between the average of the sums of rows $n$ and $n+2$ and the sum of row $n+1$ is $$ (n+1)\cdot \frac12 + n+1 $$

But we only consider odd numbers $n$; let's write $n = 2k - 1$, and we get $$ (2k-1 + 1)/2 + (2k-1 + 1) = k + 2k = 3k $$ and there's your "counting up by 3s" thing right there.