Partial sums of Nicomachus' Triangle rows produce Stirling numbers of the 2nd kind?

Question

I took partial sums of this triangle OEIS A036561 and found Stirling numbers of the 2nd kind.

At OEIS A000392, at the mid-point of the comments section, is a conjecture. I think it's what I found. I need a second pair of eyes to confirm it. Thanks to Peter Taylor for his nice proof.

View the triangle here.

How we build the triangle: We seed the first element with $\{\{1\}\}.$ Then we create the triangle by multiplying the first element of the previous row by 2 and all elements in the previous row by 3 to create the new row. Then we sum the first row, then first and second, etc.

Mathematica code:

a = {{1}};
Table[AppendTo[a, Flatten[{(a[[n, 1]]*2), (a[[n]]*3)}]], {n, 1, 10}]; a
Table[Sum[Total[a[[n]]], {n, 1, m}], {m, 1, Length[a]}]
Table[StirlingS2[n, 3], {n, 0, 13}]

{{1}, {2, 3}, {4, 6, 9}, {8, 12, 18, 27}, {16, 24, 36, 54, 81},
{32, 48, 72, 108, 162, 243}, {64, 96, 144, 216, 324, 486, 729}, {128,
192, 288, 432, 648, 972, 1458, 2187}, {256, 384, 576, 864, 1296,
1944, 2916, 4374, 6561}, {512, 768, 1152, 1728, 2592, 3888, 5832,
8748, 13122, 19683}, {1024, 1536, 2304, 3456, 5184, 7776, 11664,
17496, 26244, 39366, 59049}}

{1, 6, 25, 90, 301, 966, 3025, 9330, 28501, 86526, 261625} (* offset 3 *)

{0, 0, 0, 1, 6, 25, 90, 301, 966, 3025, 9330, 28501, 86526, 261625}

When we seed with $\{\{3\}\}$ we get S$(n,3)*3$. OEIS A094033
In fact, we can seed with $x= 5,7,9...$ or any odd number and get S$(n,3)*x.$
Even numbers for $x$ also work, but they change the parity of the right diagonal so we won't use them.

My question is: Do the multiples change the partitioning? Per Peter Taylor, the answer in no.

Answer 1

Given that the triangle is built up by multiplication of the previous row, using a seed greater than 1 just multiplies every element in the triangle by the seed. Therefore the partial sums of the triangle are also multiplied by the seed. I don't think there's any interpretation in terms of set partitions.

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PS The conjecture is

Let $S(1)=\{1\}$ and, for $n>1$, let $S(n)$ be the smallest set containing $x$, $2x$ and $3x$ for each element $x$ in $S(n-1)$. Then $a(n)$ is the sum of the elements in $S(n)$. - John W. Layman, Nov 21 2007

where $a(n) = {n \brace 3}$. It's not hard to prove.

Lemma: $S(n)$ corresponds to the first $n$ rows of Nicomachus' triangle

By induction. From the definition of Nicomachus' triangle, this corresponds to the statement that $$S(n) = \bigcup_{i=0}^{n-1} \{2^{i-j}3^j \mid j \in \mathbb{N} \wedge 0\le j\le i \}$$

$S(1) = \{2^0 3^0\}$ checks out.

$S(n+1) = S(n) \cup 2 S(n) \cup 3 S(n)$ can be checked with a bit of fiddling.

Then the sum of the elements in $S(n)$ is $$\sum_{i=0}^{n-1} \sum_{j=0}^i 2^{i-j}3^j = \sum_{i=0}^{n-1} 2^i \sum_{j=0}^i \left(\frac32\right)^j = \sum_{i=0}^{n-1} 2^i \left(2^{-i}3^{i+1}-2\right) = \sum_{i=0}^{n-1} \left(3^{i+1}-2^{i+1}\right)$$

which corresponds to the comment on A000392:

With offset $0$, this is $9*3^n/2-4*2^n+1/2$, the partial sums of $3*3^n-2*2^n=A001047(n+1)$ - Paul Barry, Jun 26 2003