A338671 a(n) is the number of distinct ways of arranging n identical square tiles into two rectangles.

0, 1, 1, 2, 3, 4, 5, 7, 7, 10, 10, 13, 14, 16, 16, 21, 20, 25, 24, 29, 28, 34, 30, 40, 36, 43, 40, 50, 44, 55, 49, 60, 55, 66, 55, 75, 64, 75, 70, 86, 72, 92, 77, 97, 87, 103, 84, 116, 94, 114, 104, 126, 102, 135, 109, 138, 123, 143, 117, 164, 128, 153, 138, 171

Offset

1,4

Comments

Note that rectangles of size 0 are not accepted (i.e., the tiles may not be formed into a single rectangle).

Finding a(n) is equivalent to counting the positive integer solutions (x,y,z,w) of n = xy + zw such that:

i) x >= y,z,w

ii) z >= w

iii) if x = z then y >= w

These conditions ensure that identical pairs are not counted twice by permuting the values of the variables.

Links

Table of n, a(n) for n=1..64.

Formula

G.f.: (B(x)^2 + B(x^2))/2 where B(x) is the g.f. of A038548. - Andrew Howroyd, Apr 29 2021

Example

a(5) = 3, since there are 3 ways to form 2 rectangles from 5 identical square tiles:
1) 2 X 2  and  1 X 1
2) 3 X 1  and  2 X 1
3) 4 X 1  and  1 X 1
Note that rotation through 90 degrees and/or exchanging the order of the two rectangles in a pair naturally do not create a distinct pair. For example, the pair 2 X 1 and 1 X 3 is not distinct from pair 2 above.

Prog

(Python)
import numpy as np
# This sets the number of terms:
nits = 20
# This list will contain the sequence
seq = []
# The indices of the sequence:
for i in range(1, nits + 1):
    # This variable counts the pairs found for each total area i
    count = 0
    # The longest side of either rectangle:
    for a in range(1, i):
        # The other side of the same rectangle:
        for b in np.arange(1, 1 + min(a, np.floor(i/a))):
            # Calculate the area of this rectangle and the remaining area:
            area1    = a*b
            rem_area = i - a*b
            # The longer side of the second rectangle:
            for c in np.arange(1, 1 + min(a, rem_area)):
                # The shorter side of the second rectangle:
                d = rem_area / c
                # Check that the solution is valid and not double counted:
                if d != int(d) or d > min(a, c) or (a == c and d > b):
                    continue
                # Count the new pair found:
                count += 1
    # Add to the sequence:
    seq.append(count)
for an in seq:
    print(an)
(PARI) a(n) = {(sum(k=1, n-1, ((numdiv(k)+1)\2)*((numdiv(n-k)+1)\2)) + if(n%2==0, (numdiv(n/2)+1)\2))/2} \\ Andrew Howroyd, Apr 29 2021

Crossrefs

Cf. A038548, A338664, A055507 (where a(n) is the number of ordered ways to express n+1 as a*b+c*d with 1 <= a,b,c,d <= n).

Sequence in context: A117174 A320466 A342542 * A343246 A348531 A237824

Adjacent sequences: A338668 A338669 A338670 * A338672 A338673 A338674

Keyword

nonn

Author

Thomas Oléron Evans, Apr 23 2021

Status

approved