Find Formula for nth Term in Tuff Sequence

[Wilmer]

This sequence is similar to Sloane's #A010551; one further multiplication is made...

n= 0  1  2  3  4  5  6   7   8    9   10    11     12
   1  1  1  1  2  4  8  24  72  216  864  3456  13824...

Find a formula for the nth term.

 

[Nono713]

This sequence is similar to Sloane's #A010551; one further multiplication is made...

n= 0  1  2  3  4  5  6   7   8    9   10    11     12
   1  1  1  1  2  4  8  24  72  216  864  3456  13824...

Find a formula for the nth term.

Spoiler

This sequence is generated by:
$$a_n = \tau_n a_{n - 1} ~ \text{with} ~ a_0 = 1$$
Where $\tau_n$ is given by:
$$\tau_n = \begin{cases} 1 ~ ~ &\text{if} ~ 1 \leq n \leq 3 \\ 2 ~ ~ &\text{if} ~ 4 \leq n \leq 6 \\ 3 ~ ~ &\text{if} ~ 7 \leq n \leq 9 \\ &\cdots \end{cases}$$
It's easy to see that $\tau_{3n} = n$ and since every three elements 1, 2, 3, then 4, 5, 6, then 7, 8, 9, etc. are the same, we have $\tau_{3n - 2} = \tau_{3n - 1} = n$ and $\tau_{3n + 1} = \tau_{3n + 2} = n + 1$.
Now observe that:
$$a_{3n} = \tau_{3n} a_{3n - 1} = \tau_{3n} \tau_{3n - 1} a_{3n - 2} = \tau_{3n} \tau_{3n - 1} \tau_{3n - 2} a_{3(n - 1)} = n^3 a_{3(n - 1)}$$
So that by induction we have:
$$a_{3n} = (n!)^3$$
Then:
$$a_{3n + 1} = \tau_{3n + 1} a_{3n} = (n + 1) (n!)^3$$
And:
$$a_{3n + 2} = \tau_{3n + 2} a_{3n + 1} = \tau_{3n + 2} \tau_{3n + 1} a_{3n} = (n + 1)^2 (n!)^3$$
And so we can conclude:
$$a_n = \begin{cases} (m!)^3 ~ ~ &\text{if} ~ n = 3m \\ (m + 1) (m!)^3 ~ ~ &\text{if} ~ n = 3m + 1 \\ (m + 1)^2 (m!)^3 ~ ~ &\text{if} ~ n = 3m + 2 \end{cases}$$
Or, more directly:
$$a_n = \left ( \left \lfloor \frac{n}{3} \right \rfloor + 1 \right )^{n \bmod{3}} \left ( \left \lfloor \frac{n}{3} \right \rfloor ! \right)^3$$
I suppose this can be easily generalized to any similar $\tau_n$ sequence of the form 1, 1, $k$ times, 2, 2, $k$ times, etc.. but I haven't tried.​

 

[Wilmer]

Nice, Bac! Mine's simpler:

Spoiler

nth term = a * b * c, where:
a = [FLOOR(n/3)]!
b = [FLOOR((n+1)/3)]!
c = [FLOOR((n+2)/3)]!

 

[Nono713]

Nice, Bac! Mine's simpler:

nth term = a * b * c, where:
a = [FLOOR(n/3)]!
b = [FLOOR((n+1)/3)]!
c = [FLOOR((n+2)/3)]!

Ah, yep, that works too. Not bad :)

 

[Fallen Angel]

Hi,

You ask for one formula, I bring you a non countable set of formulas

Spoiler

Let $c=(1,1,1,1,2,4,8,24,72,216,864,3456,13824,r)^{T}\in \Bbb{R}^{14}$ with $r\in \Bbb{R}$.

Define $A=\left(\begin{array}{ccccc}
1 & 0 & 0 & \ldots & 0\\
1^{0}& 1 & 1^{2} & \ldots & 1^{13}\\
2^{0}& 2 & 2^{2} & \ldots & 2^{13}\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
13^{0} & 13 & 13^{2}& \ldots & 13^{13}\end{array}\right)$

Call $b=(b_{0}, b_{1},\ldots , b_{13})^{T}\in \Bbb{R}^{14}$, the solution of the linear equations

$Ab=c$

And now consider the polynomial $p_{r}(x)=\displaystyle\sum_{i=0}^{13}b_{i}x^{i}$.

Then $p_{r}(0)=1, p_{r}(1)=1, \ldots , p_{r}(12)=13824, p_{r}(13)=r$.

And works for every $r\in \Bbb{R}$

The n-th term will be now $p_{r}(n)$