Compute the kangaroo sequence

Coffeescript, 19 bytes

(n)->(n*n-1<<n-1)+1

Edit: Thanks to Dennis for chopping off 6 bytes!

The formula for generating Kangaroo numbers is this:

Explanation of formula:

The number of 1's in K(n)'s final sum is 2^(n - 1) + 1.

The number of n's in K(n)'s final sum is 2^(n - 1), so the sum of all the n's is n * 2^(n - 1).

The number of any other number (d) in K(n)'s final sum is 2^n, so the sum of all the d's would be d * 2^n.

• Thus, the sum of all the other numbers = (T(n) - (n + 1)) * 2^n, where T(n) is the triangle number function (which has the formula T(n) = (n^2 + 1) / 2).

  Substituting that in, we get the final sum

    (((n^2 + 1) / 2) - (n + 1)) * 2^n
  = (((n + 1) * n / 2) - (n + 1)) * 2^n
  = ((n + 1) * (n - 2) / 2) * 2^n
  = 2^(n - 1) * (n + 1) * (n - 2)
  

When we add together all the sums, we get K(n), which equals

  (2^(n - 1) * (n + 1) * (n - 2)) + (2^(n - 1) + 1) + (n * 2^(n - 1))
= 2^(n - 1) * ((n + 1) * (n - 2) + n + 1) + 1
= 2^(n - 1) * ((n^2 - n - 2) + n + 1) + 1
= 2^(n - 1) * (n^2 - 1) + 1

... which is equal to the formula above.

Jelly, 6 bytes

²’æ«’‘

Uses the formula (n² - 1) 2^{n - 1} + 1 to compute each value. @Qwerp-Derp's was kind enough to provide a proof.

Try it online! or Verify all test cases.

Explanation

²’æ«’‘  Input: n
²       Square n
 ’      Decrement
  æ«    Bit shift left by
    ’     Decrement of n
     ‘  Increment

Java 7, 35 bytes

int c(int n){return(n*n-1<<n-1)+1;}