1. ## Tuff sequence...

This sequence is similar to Sloane's #A010551; a 3rd multiplication is made...
Code:
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.

2. ## Re: Tuff sequence...

Spoiler:

$1=1^0$

$1=1^1$

$1=1^2$

$1=1^3$

$2=2^1 \times 1^3$

$4=2^2\times 1^3$

$8=2^3 \times 1^3$

$24=3^1\times 2^3 \times 1^3$

$72 = 3^2 \times 2^3\times 1^3$

$216 = 3^3 \times 2^3\times 1^3$

$864 = 4^1 \times 3^3 \times 2^3\times 1^3$

$3456=4^2 \times 3^3 \times 2^3\times 1^3$

$13824=4^3 \times 3^3 \times 2^3\times 1^3$

Let $n=3k + r$

the $nth$ term of your sequence is given by

$(k+1)^r \displaystyle{\prod_{i=2}^k}i^3$

3. ## Re: Tuff sequence...

Thanks Romsek. Mine is:

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

Someone else got this cutie:
nth term = (a+1)^(n MOD 3) * (a!)^3
where a = FLOOR(n/3)

4. ## Re: Tuff sequence...

Originally Posted by Wilmer
Thanks Romsek. Mine is:

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

Someone else got this cutie:
nth term = (a+1)^(n MOD 3) * (a!)^3
where a = FLOOR(n/3)
that 2nd one is equivalent to my form.

my product term is just $(k!)^3$

and their $a$ is my $k$