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Thread: Tuff sequence...

  1. #1
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    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.
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  2. #2
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    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

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

    Tuff sequence...-clipboard01.jpg

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  3. #3
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    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)
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  4. #4
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    Re: Tuff sequence...

    Quote Originally Posted by Wilmer View Post
    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 $k$
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