# Longer and Longer: Billy's Proof

Thank you to Billy for sending us his thoughts on Longer and Longer

Let a<b<c. In order to be able to combine the rods to the largest number of consecutive lengths, no combination should be "wasted", i.e. a certain combination should not be equal to another possible combination.

For a,b,c there are the following combinations:
a,
b,
c,
a+b,
a+c,
b+c,
c-b,
c-a,
b-a,
c-a+b,
c-b+a
either c-b-a for a+b<c or a+b-c for a+b>c
and a+b+c.

If there is a set of lengths for which there are no "wasted" combinations, the largest number of consecutive lengths is a+b+c=13. The four largest numbers have to be c+a+b, c+b-a, c+b, c+a. Thus you have to be able to combine a,b to the four consecutive lengths 1,2,3,4.

Set a=1 => b=3 (because for b=2 there would be the "wasted" combination b-a=a) You can combine a,b to 1,2,3,4 (so the determined requirement is met).

=> c=9 because the smallest combination in which c is involved c-a-b has to be equal 5. The set (1,3,9) makes up 1 = a 2 = b-a 3 = b 4 = b+a 5 = c-a-b 6 = c-b 7 = c+a-b 8 = c-a+b 9 = c 10 = c+a 11 = c-a+b 12 = c+b

Hence, (1,3,9) is the set for the largest number of consecutive lengths.

You can use the set (a,b,c) to make all lengths up to 13. Thus you can subtract 13 from d to get the subsequent length 14.

$d=2\times13+1$ or  $d=2\times(a+b+c)+1$

The length $X_p$ of the set $(X_1,X_2,X_3,...,X_n)$ is one plus two times the sum of all precedent lengths $(X_1+X_2+...+X_{p-1})$.

This can be simplified to $X_p=3^{(p-1)}$

Respectively, the largest number of consecutive lengths which can be made with a set of n lengths $(X_1,X_2,...,X_n)$ is $X_n+(X_n -1)/2=1.5 \times X_n -0.5$ . So (1,3,9,27) makes up all lengths up to 40, (1,3,9,27,81) to 121 and so on.

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