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Around the table

2022-07-01T10:21:27.185Z


The problems of diners who sit around a table meeting certain requirements constitute an inexhaustible vein


The eighth term of the Sylvester sequence, as we saw last week, has 27 digits, and since it starts with 1 and each term is approximately the square of the previous one, the ninth term will have 2 x 27 – 1 = 53 digits.

Regarding the series of the inverses of the terms of the Sylvester succession, here is what Salva Fuster comments:

“We start from a square with an area of ​​1 and make rows of squares as large as possible, but on the condition that each new row does not complete the initial square.

In this way we have:

- Row 1: 2 squares of side 1/2.

We cannot take a single square of side 1, since the initial square is completed.

- Row 2: 3 squares of side 1/3 (1/2+1/3 is less than 1).

We cannot take again two squares of side 1/2, because the initial square is completed.

- Row 3: We cannot take squares of 1/4, or 1/5, or 1/6, but we can take 1/7, which is the largest possible size that is still less than 1 (1/2+1 /3+1/7=41/42).

- When following, the sizes are determined by the inverses of the Sylvester sequence.

Therefore, the sum of the inverses of the Sylvester sequence approaches 1″.

And that is precisely what last week's clue-figure graphically tells us.

And regarding the possibility of expressing any rational number in the form of an Egyptian fraction, here is the comment by Manuel Amorós:

“We want to express p/q as an Egyptian fraction.

Being p less than q.

q = pn+r

1/(q/p) = 1/(n+r/p)

1/(n+1) < 1/(q/p) < 1/n

p/q = 1/(n+1)+A

A = p/q-1/(n+1) = (pr)/(q(n+1))

The numerator of fraction A is less than the initial one.

If we continue to apply the procedure, there will come a time when all the addends will have a numerator of 1″.

The usual round table

At the last moment, an interesting problem on the old topic of diners sitting around a table appeared in the comments section:

We have a round table with n equally spaced chairs where n people are going to sit.

People are arranged in a row to be seated sequentially.

The first sits on any chair, the second sits at distance 1 from the first to her left (that is, next to the first), the third sits at distance 2 from the second also to the left, and so on. successively.

Find the values ​​of n for which all the diners can be seated fulfilling the conditions.

But first let's tackle an easier one, to warm up neurons:

In how many different ways can three ill-matched couples sit around a table so that no one is next to their spouse?

What if, in addition to ill-suited, they are heterosexual couples and you want to maintain the traditional boy-girl alternation?

What if instead of three marriages there are four, five, six...?

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Source: elparis

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