Let $\lambda=(\lambda_1,\cdots,\lambda_n)$ be a partition, with $|\lambda|:=N$. Attach an extra box to $\lambda$ to the right end of the $r$'th row. In coordinate form, the last box on row $r$ has label $(r,\lambda_r)$, so the added box has coordinates $(r,\lambda_r+1)$. Now look at "almost-standard young tableaux at row $r$" of this shape, to be defined as follows: our partition is now $|\lambda|+1$, where all the boxes in the original diagram of $\lambda$ are increasing right and down BUT, enforce that the value in the added box is bigger than $(r,\lambda_r)$, but *not necessarily* bigger than any box above it. This is all best shown by an example:

Take $\lambda=(4,3,2,1)$, giving $|\lambda|=10$, and fix $r=2$. We now have $|\lambda|+1=11$ and so the following are examples of almost standard tableaux:

$$\begin{array}{cccccc} 1 & 2 & 4 & 6 & \ \\ 3 & 5 & 7 &\leftrightarrow & 8 \\ 9 & 11 & \ & \ & \ \\ 10 & \ &\ & \ & \ \end{array}$$

$$$$

$$\begin{array}{cccccc} 1 & 3 & 6 & 11 & \ \\ 2 & 4 & 9 &\leftrightarrow & 10 \\ 5 & 7 & \ & \ & \ \\ 8 & \ &\ & \ & \ \end{array}$$

where the double arrows show that the added box attaches to the end of the second row. Notice that in the second example, $9<10$ as required but $11>10$ is OK since the added box is "coupled" only to the second row. There are a total of 1850 such tableaux. Compare this to the number of SYT of shape $(4,4,3,2,1)$ which is 1320 and is naturally less than the number of the almost standard tableaux.

I'm interested in counting the number of such tableaux. When $r=1$, we're back to Standard Young Tableaux so everything is easy via the hook formula. Numerically, for $r>1$ things get trickier. Experimentally, there's no obvious hook formula for these tableaux. I'd be very satisfied if this is tractable for staircase shapes $(n,n-1,\cdots,1)$.

These tableaux also satisfy the same kind of recurrences as SYT: you can take away corner boxes (where the maximum value is) and sum over all possible tableaux to get the total. So it *looks* like a good idea might be to define schur functions in terms of these tableaux. Unfortunately, to calculate such Schur functions, I believe one needs at least some kind of involution on these tableaux.

One idea I have is that when $\lambda_{r-1}>\lambda_r$, you can split into cases where the added box $(r,\lambda_r)$ has value greater/less than the box above it $(r-1,\lambda_r+1)$. In the case of greater, we're back to SYT. It then amounts to calculating the less case, which again seems difficult. Perhaps one can move the $(r-1,\lambda_r+1)$ box to the right of the $(r,\lambda_r+1)$ box. From the second example above we get:

$$\begin{array}{cccccc} 1 & 3 & 6 & \ \\ 2 & 4 & 9 & 10 & 11 \\ 5 & 7 & \ & \ &\ \\ 8 & \ &\ & \ & \ \end{array}$$

So there's some inclusion-exclusion going on here that might reduce these tableaux to truncated tableaux . Unfortunately this idea seems to only work for $r=2$. Another idea is to take the value in the added box $(r,\lambda_r+1)$ and do RSK with it by pushing into the tableaux $\lambda$. This seems to have the effect of taking almost standard young tableaux at row $r$ to one's at rows $1,\cdots,r$ and gives a kind of Pieri rule for the aformentioned tableaux. In the above examples, RSK on the added box gives:

$$\begin{array}{cccccc} 1 & 2 & 4 & 6 & 8 \\ 3 & 5 & 7 & \ & \ \\ 9 & 11 & \ & \ &\ \\ 10 & \ &\ & \ & \ \end{array}$$

$$\begin{array}{cccccc} 1 & 3 & 6 & 10 \ \\ 2 & 4 & 9 & 11 \\ 5 & 7 & \ & \ &\ \\ 8 & \ &\ & \ & \ \end{array}$$

Unfortunately, I don't see how one might actually get a recurrence in terms of $1,2,\cdots,r$ almost-standard tableaux. This is because counting all possible RSK shapes won't ensure the added box has a value bigger than $(r,\lambda_r)$.

The only similar types of tableaux that I've seen are called composition tableaux (mainly because they allow for non-monotonically decreasing partitions), which come up in definitions of quasisymmetric functions. However, the rules for those tableaux are more general than just SYT.