Don't try this if you don't know your maths!

Question

Find the next number in this sequence:

$120, 106, 112, 108, 80, 92, 84,...$

This sequence might be easy for some, but you won't be able to solve it if you don't know a certain subject in maths (That I will reveal if nobody finds the answer in 30 minutes).

First Hint:

Look at the title!

Second Hint:

It doesn't involve any special series or integrals.

Third hint:

It is in the number theory puzzle

Fourth hint:

It is an easy recurrence relation.

Fifth hint:

It uses a well known function

Sixth hint:

120 and 106 are just starting values

Answer 1

The next number in this sequence is,

$80$

Because the sequence is defined as such:

$a_0 = 120,\:a_1 = 106,\:a_n = \phi(a_{n-1} + a_{n-2})$

Where $\phi$ denotes Euler's totient function, defined as the number of integers between 1 and $x-1$ inclusive that are coprime to $x$.

The sequence continues,

$120,\,106,\,112,\,108,\,80,\,92,\,84,\,80,\,80,\,64,\,48,\,48,\,\overline{32,\,32,\,32},\,\dots$

Answer 2

The sequence and its difference sequence are-

$S = 120,\ 106,\ 112,\ 108,\ 80,\ 92,\ 84,...$
$\Delta S= -14,\ 6,\ -4,\ -28,\ 12,\ -8...$

It seems that the difference sequence doubles every 3 terms. So, I would guess that the next three terms in the difference sequence would be

-14, 6, -4, -28, 12, -8, (-56, 24, -16) or (-42,18,-12)

So, the next terms might be:

120, 106, 112, 108, 80, 92, 84, (28, 52, 36) or (42, 60, 48)

Answer 3

These are the known numbers so far:

$$ S = \left(\begin{matrix} 0 & 120 \\ 1 & 106 \\ 2 & 112 \\ 3 & 108 \\ 4 & 80 \\ 5 & 92 \\ 6 & 84 \end{matrix}\right) $$

When investigating the numbers, I noticed a pattern in them:

$$ \Delta S = 112 - 106 = 6 \\ \Delta S = 112 - 108 = 4 $$

$$ \Delta S = 92 - 80 = 12 \\ \Delta S = 92 - 84 = 8 $$

These number seem to be related to each other in groups of 3 and their $\Delta$'s between each other seem to be doubling. Backwards engineering using this, leaves us with:

$$ \Delta S = a - 120 = 2 \\ a = 2 \\ S(-1) = 122 $$

Now let's compare the highest numbers of all the known triplets:

$$ \Delta S = 122 - 112 = 10 \\ \Delta S = 112 - 92 = 20 $$

Which confirms my theory that the pattern must be:

$$ \text{in }\mathbb{Z} \\[12pt] S(x) = \begin{cases} S(x+1) - \left(6 * 2^{\left\lfloor\frac{x-1}{3}\right\rfloor}\right) & \text{if } ((x-1) \text{ mod } 3) = 0 \\ 132 - \left(20 * 2^{\left\lfloor\frac{x-1}{3}\right\rfloor}\right) & \text{if } ((x-2) \text{ mod } 3) = 0 \\ S(x-1) - \left(4 * 2^{\left\lfloor\frac{x-1}{3}\right\rfloor}\right) & \text{if } (x \text{ mod } 3) = 0 \end{cases} $$

The next number in the sequence would be:

$$ \begin{align} S(7) &= S(7+1) - \left(6 * 2^{\left\lfloor\frac{7-1}{3}\right\rfloor}\right) \\ &= \left(132 - \left(20 * 2^{\left\lfloor\frac{8-1}{3}\right\rfloor}\right)\right) - \left(6 * 2^{\left\lfloor\frac{7-1}{3}\right\rfloor}\right) \\ &= \left(132 - \left(20 * 4\right)\right) - \left(6 * 4\right) \\ &= \left(132 - 80\right) - 24 \\ &= 52 - 24 \\ &= 28 \end{align} $$