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 The next cube (Posted on 2022-10-19)
True story: I was teaching my 11-year old about recursion and this came up... The integer squares, f(n) = n^2 can be generated recursively:
f(1) = 1, f(n) = f(n-1) + 2 n - 1.
So, how does one similarly generate the integer cubes through addition of terms in n?

Bonus 1: How about f(n) via recursion only using previous terms, without using n alone?

Bonus 2: What about for higher powers? Does n^4 have a solution made by adding an integer coefficient cubic in n? How else?

Bonus 3: Is there a general solution to make n^j, for all j using recursion?

 See The Solution Submitted by Steven Lord Rating: 5.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Simple (spoiler) | Comment 1 of 3
Let g(n) = n^3

Then g(n) = n^3 + (g(n-1) - (n-1)^3)
= g(n-1) +3n^2 - 3n + 1
= g(n-1) +3n*(n-1) +1

Bonus 1
f(n) = f(n-1) + (f(n-1) - f(n-2) + 2) =
= 2f(n-1) - f(n-2) +2

Bonus 2 + Bonus 3
Let j(n) = n^j =
Then j(n) = n^j - (n-1)^j + j(n-1)
The expression (n^j - (n-1)^j) can be expanded.
Its coefficients have alternating sign and can be read off Pascal's table

 Posted by Steve Herman on 2022-10-19 10:30:38

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