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.


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):


Try it online!

Original answer,  209 183 178  174 bytes

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


Try it online!


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) =>               // n = integer, c = callback function
  n >= 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 ) =>                // A = candidate area (initially undefined)
S(A, w =>                   // for w = A to w = 0:
  A % w ?                   //   if w is not a divisor of A:
    0                       //     do nothing
  : (                       //   else:
    F = (l, n) =>           //     F = recursive function taking a list l[] and a size n
      n ?                   //       if n is not equal to 0:
        S(w - n, x =>       //         for x = w - n to x = 0
          S(A / w - n, y => //           for y = A / w - n to y = 0:
            l.some(         //             for each square in l[]
            ([X, Y, W]) =>  //             located at (X, Y) and of width W:
              X < x + n &   //               test whether this square is overlapping
              X + W > x &   //               with the new square of width n that we're
              Y < y + n &   //               trying to insert at (x, y)
              Y + W > 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
  • 2
    \$\begingroup\$ Save 6* by not defining h and moving the test for a%w<1 to the tail of the recursion TIO. Of course it's much slower. (* at least - I'm no JavaScript expert!) \$\endgroup\$ – Jonathan Allan May 6 at 10:52
  • \$\begingroup\$ @JonathanAllan Thanks. :) Actually, I wonder if a%w<1 could be replaced with just 1. I'll have to double-check that later. \$\endgroup\$ – Arnauld May 6 at 11:12

Python 2 (PyPy), 250 236 bytes

-14 bytes thanks to msh210's suggestions.

Outputs the 1-indexed nth term of the sequence.

for Q in r(m):
 for X in r(n,0,-1):P|=([x for x in[{(x+a,y+b)for a in r(X)for b in r(X)}for x in r(Q%k-X+1)for y in r(Q/k-X+1)]if not x&P]+[{0}])[0]
 if len(P)>k:m=min(Q%k*(Q/k),m)
print m

Try it online! For n>4, this takes lot of time. I have verified the result up to n=7 locally.

  • \$\begingroup\$ Would you mind including an explanation of how it works? Also, I imagine you can shave bytes by indenting one space at a time instead of seven (for the second indentation). (In fact, I think maybe the two for lines can be on one line, and you only need to indent once.) \$\endgroup\$ – msh210 May 7 at 18:16
  • 1
    \$\begingroup\$ @msh210 the "7 spaces" are in fact a tab, as in Python 2 you can indent first with a space, then with a tab. Putting the two for loops on one line would be invalid syntax unfortunately. \$\endgroup\$ – ArBo May 7 at 18:24
  • 1
    \$\begingroup\$ @msh210 I found a different way to combine those for loops. Those 7 spaces where only on line, thanks for the catch. I will try to write an explanation tomorrow \$\endgroup\$ – ovs May 7 at 18:40

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