What numbers of items cannot be listed sequentially in 4 columns (see details)?

Wow…watching!

What do you mean by “c colums”?
Do you mean I can have a number of columns other than 4?

Lists of either 2 or 3 items cannot be listed in the manner described because the list would fail to use either one or two of the columns.

For six items, the distribution would be 2 – 2 – 1 – 1, which fails the three equal columns requirement.

In general, where c is the number of columns, any number less than c + n (where n = # of equal length columns & c = total # of columns) cannot be listed using the arrangement structure defined in the OQ.

@sahID
Could you not just go 5–1?

I was mistaken about there being just two numbers. There are at least two numbers greater than 4. There is a way of coming up with a lower bound for the general case of the number of columns being c, but I have not come up with a way of listing them all.

Sorry for not having completely thought this through. There will be infinitely many cases. I will show what I came up with tomorrow.

If there are 4 columns, there are 6 numbers that violate the rule: 1, 2, 3, 5, 6, and 9

If there are N columns, there are ( N * ( N – 1 ) ) / 2 numbers that violate the rule.

I suppose 0 items violates the rule also, since the last column must have at least 1 item. So….

If there are 4 columns, there are 7 numbers that violate the rule: 0, 1, 2, 3, 5, 6, and 9.

If there are N columns, there are ( ( N * ( N – 1 ) ) / 2 ) + 1 numbers that violate the rule.

In the usual case when people divide objects into columns, the number of columns is small relative to the number of items and there is no problem. Strange things can happen when the number of columns is large compared to the number of items. For the case of 4 columns, the small numbers of items shown by @Lawn illustrate the kinds of things that can go wrong.

I still have not analyzed this algebraically, but I am pretty sure that as long as the number of columns is less than or equal to the number of rows, things will work out. That is the same as saying that the number of columns is no greater than the square root of the number of items.

For what it is worth, here is the algebra to show that whenever the number of columns is less than or equal to the number of rows, we can always present the items in the schema.
Let c = number of columns and r = number of rows and n = number of items.

For c columns we get the number of rows r by dividing n by c.
This gives n = rc + k, where k is the remainder of the division by c. 0 <= k < c.

If k=0, we are done. We have n items as c columns of r.

Assume that k > 0. Now we have a c+1st column containing k items. I claim that by changing the number of rows to (r+1), we always end up with c columns. We want (c-1) columns of (r+1) plus some non-zero remainder to be in column c.

n = rc + k = (r+1)(c – 1) + k’.
We can always find a k’. What we need to show is that 0 < k’ < r+1. We are saying k’ < (r+1) since we have added an extra row.

Solving for k’:
k’ = r – (c – 1) + k

Since k < c
k’ < r + 1

Since r >= c, r – (c-1) >= 1.
This gives us:
0 < 1 <= k’ < r+1, which is what we wanted to prove.

Here is a slicker way of getting the equation for k’. We start with c columns of length r plus remainder k. Remove one column of length r and add a member from the column to each of the remaining c – 1 columns. That leaves us with r – (c – 1) remaining in the column plus the original remainder of k, so k’ = r – (c – 1) + k.