Question: I write the following infinite table: $$T= \text{ }\begin{matrix} 1\\ 1&1\\ 1&2&1\\ 1&1&3&1\\ 1&1&1&4&1\\ 1&2&1&1&5&1\\ 1&1&3&1&1&6&1\\ 1&1&1&4&1&1&7&1\\ 1&1&1&1&5&1&1&8&1\\ 1&\color{red}{2}&1&1&1&6&1&1&9&1\\ 1&1&3&1&1&1&7&1&1&10&1\\ 1&1&1&4&1&1&1&8&1&1&11&1\\ \vdots\\ \end{matrix}$$
I write $t_{ij}$ for the entry in the $i-$th row and $j-$th column of $T.$ What is a closed formula for the value of $t_{ij}$ in only the indeterminants $i$ and $j $ ?
The $j-$th column of $T$ might be interpreted as $k$ $1'$s followed by $j.$ For example column 2 might be written as $1,2,1,1,2,1,1,1,2,1,1,1,1,\ldots,$ which can be found in the OEIS as A042974(N). In particular I have without proof: $$ t_{i2}=\bigg\lfloor\frac{1}{2}\bigg(3-\cos\bigg[\pi\sqrt{8i+9}\bigg]\bigg)\bigg\rfloor $$ For example: $$ t_{10,2}=\bigg\lfloor\frac{1}{2}\bigg(3-\cos\bigg[\pi\sqrt{8*10+9}\bigg]\bigg)\bigg\rfloor=\bigg\lfloor\frac{1}{2}\bigg(3-\cos\bigg[\pi\sqrt{89}\bigg]\bigg)\bigg\rfloor=\bigg\lfloor\frac{1}{2}(4)\bigg\rfloor=\color{red}{2} $$ This formula might be leveraged to calculate the more general term $t_{ij}.$
The following are conjectural observations:
- It appears the sum of the entries on the $i-$th row of $T$ equals $A006463(N+1).$ A motivation for this sequence is shown in the reference: R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; Problem 7.2(f). In particular it appears the row sums of $T$ are the length of the longest maximal chain in $Par(N);$ where $Par(N)$ denotes the set of all partitions of $N$ with dominance ordering.
- It appears the sum of the non-unit entries, those entries not equal to 1, on the $N-$the row of $T$ equals $A023546(N+1).$
- It appears the sum of unit entries, those entries equal to 1, on the $N-$th row of $T$ equals $A122797(N+1).$
- It appears following 1,2 and 3 this equation holds: $$A006463(N+1)=A023546(N+1)+A122797(N-1)$$
Note I am not asking for formulas for 1,2,3 and 4.