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