Motzkin numbers: Difference between revisions
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So basically we calculate (X<sub>n-1</sub>(1+2n)+3X<sub>n-2</sub>(n-1)) and divide that by 2+n and append this new value to the list. For n we use the list length. For X<sub>n-1</sub> we use the last element of the list. And, for X<sub>n-2</sub> we use the second to last element of the list. For the task we repeat this list operation 40 times, starting with the list 1 1 and check to see which elements of the resulting list are prime. Because these values get large, we need to use arbitrary precision integers for our list values. |
So basically we calculate (X<sub>n-1</sub>(1+2n)+3X<sub>n-2</sub>(n-1)) and divide that by 2+n and append this new value to the list. For n we use the list length. For X<sub>n-1</sub> we use the last element of the list. And, for X<sub>n-2</sub> we use the second to last element of the list. For the task we repeat this list operation 40 times, starting with the list 1 1 and check to see which elements of the resulting list are prime. Because these values get large, we need to use arbitrary precision integers for our list values. |
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