Find the area of the smallest rectangle to contain squares of sizes up to n This is a sequence question of the usual type, as applied to OEIS sequence A038666. That is, do either of the following:


Accept no or any input, and output A038666 until the heat death of the universe.
Accept a positive integer as input, and output the \$n\$th term of A038666 or its first \$n\$ terms. (If using \$0\$- instead of \$1\$-indexing, then of course you also have to output 1 on 0 input.)


The \$n\$th term of A038666 is the least area among rectangles that contain non-overlapping squares of sizes \$1\times1,2\times2,\dots n\times n\$ if you're using \$1\$-indexing.

Example:

The smallest-area rectangle which can contain non-overlapping squares of sizes \$1\times1\$ through \$4\times4\$ has dimensions \$7\times5\$:

4 4 4 4 3 3 3
4 4 4 4 3 3 3
4 4 4 4 3 3 3
4 4 4 4 2 2 1
x x x x 2 2 x


Therefore, \$a(4)=7\times5=35\$ (\$1\$-indexed).

Similarly, the least-area rectangle containing non-overlapping squares of sizes \$1\times1\$ through \$17\times17\$ has dimensions \$39\times46\$, so \$a(17)=39\times46=1794\$ (\$1\$-indexed).
JavaScript (ES6), 172 bytes

Slower but shorter version suggestion suggested by @JonathanAllan (also saving 4 bytes in the original answer):



f=(n,A,S=(n,c)=&gt;n&gt;=0?c(n)||S(n-1,c):0)=&gt;S(A,w=&gt;(F=(l,n)=&gt;n?S(w-n,x=&gt;S(A/w-n,y=&gt;l.some(([X,Y,W])=&gt;X&lt;x+n&amp;X+W&gt;x&amp;Y&lt;y+n&amp;Y+W&gt;y)?0:F([...l,[x,y,n]],n-1))):A%w&lt;1)([],n))?A:f(n,-~A)


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Original answer, &nbsp;209 183 178&nbsp; 174 bytes

Returns the \$N\$th term of the sequence, 1-indexed.



f=(n,A,S=(n,c)=&gt;n&gt;=0?c(n)||S(n-1,c):0)=&gt;S(A,w=&gt;A%w?0:(F=(l,n)=&gt;n?S(w-n,x=&gt;S(A/w-n,y=&gt;l.some(([X,Y,W])=&gt;X&lt;x+n&amp;X+W&gt;x&amp;Y&lt;y+n&amp;Y+W&gt;y)?0:F([...l,[x,y,n]],n-1))):1)([],n))?A:f(n,-~A)


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Commented

Helper function

We first define a helper function \$S\$ which invokes a callback function \$c\$ for \$n\$ to \$0\$ (both included) and stops as soon as a call returns a truthy value.

S = (n, c) =&gt;               // n = integer, c = callback function
  n &gt;= 0 ?                  // if n is greater than or equal to 0:
    c(n) ||                 //   invoke c with n; stop if it's truthy
    S(n - 1, c)             //   or go on with n - 1 if it's falsy
  :                         // else:
    0                       //   stop recursion and return 0


Main function

We start with \$A=1\$.

For each pair \$(w,h)\$ such that \$w\times h = A\$, we try to insert all squares of size \$1\times1\$ to \$n\times n\$ (actually starting with the largest one) in the corresponding area, in such a way that they don't overlap with each other.

We keep track of the list of squares with their position \$(X,Y)\$ and their width \$W\$ in \$l[\text{ }]\$.

We either return \$A\$ if a valid arrangement was found, or try again with \$A+1\$.

f = ( n,                    // n = input
      A ) =&gt;                // A = candidate area (initially undefined)
S(A, w =&gt;                   // for w = A to w = 0:
  A % w ?                   //   if w is not a divisor of A:
    0                       //     do nothing
  : (                       //   else:
    F = (l, n) =&gt;           //     F = recursive function taking a list l[] and a size n
      n ?                   //       if n is not equal to 0:
        S(w - n, x =&gt;       //         for x = w - n to x = 0
          S(A / w - n, y =&gt; //           for y = A / w - n to y = 0:
            l.some(         //             for each square in l[]
            ([X, Y, W]) =&gt;  //             located at (X, Y) and of width W:
              X &lt; x + n &amp;   //               test whether this square is overlapping
              X + W &gt; x &amp;   //               with the new square of width n that we're
              Y &lt; y + n &amp;   //               trying to insert at (x, y)
              Y + W &gt; y     //
            ) ?             //             if some existing square does overlap:
              0             //               abort
            :               //             else:
              F([ ...l,     //               recursive call to F:
                  [x, y, n] //                 append the new square to l[]
                ],          //
                n - 1       //                 and decrement n
              )             //               end of recursive call
          )                 //           end of iteration over y
        )                   //         end of iteration over x
      :                     //       else (n = 0):
        1                   //         success: stop recursion and return 1
    )([], n)                //     initial call to F with an empty list of squares
) ?                         // end of iteration over w; if it was successful:
  A                         //   return A
:                           // else:
  f(n, -~A)                 //   try again with A + 1