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

%I A338671 #26 Jun 24 2021 02:27:09
%S A338671 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,
%T A338671 40,50,44,55,49,60,55,66,55,75,64,75,70,86,72,92,77,97,87,103,84,116,
%U A338671 94,114,104,126,102,135,109,138,123,143,117,164,128,153,138,171
%N A338671 a(n) is the number of distinct ways of arranging n identical square tiles into two rectangles.
%C A338671 Note that rectangles of size 0 are not accepted (i.e., the tiles may not be formed into a single rectangle).
%C A338671 Finding a(n) is equivalent to counting the positive integer solutions (x,y,z,w) of n = xy + zw such that:
%C A338671   i)   x >= y,z,w
%C A338671   ii)  z >= w
%C A338671   iii) if x = z then y >= w
%C A338671 These conditions ensure that identical pairs are not counted twice by permuting the values of the variables.
%F A338671 G.f.: (B(x)^2 + B(x^2))/2 where B(x) is the g.f. of A038548. - _Andrew Howroyd_, Apr 29 2021
%e A338671 a(5) = 3, since there are 3 ways to form 2 rectangles from 5 identical square tiles:
%e A338671 1) 2 X 2  and  1 X 1
%e A338671 2) 3 X 1  and  2 X 1
%e A338671 3) 4 X 1  and  1 X 1
%e A338671 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.
%o A338671 (Python)
%o A338671 import numpy as np
%o A338671 # This sets the number of terms:
%o A338671 nits = 20
%o A338671 # This list will contain the sequence
%o A338671 seq = []
%o A338671 # The indices of the sequence:
%o A338671 for i in range(1,nits + 1):
%o A338671     # This variable counts the pairs found for each total area i
%o A338671     count = 0
%o A338671     # The longest side of either rectangle:
%o A338671     for a in range(1,i):
%o A338671         # The other side of the same rectangle:
%o A338671         for b in np.arange(1,1 + min(a,np.floor(i/a))):
%o A338671             # Calculate the area of this rectangle and the remaining area:
%o A338671             area1    = a*b
%o A338671             rem_area = i - a*b
%o A338671             # The longer side of the second rectangle:
%o A338671             for c in np.arange(1,1 + min(a,rem_area)):
%o A338671                 # The shorter side of the second rectangle:
%o A338671                 d = rem_area / c
%o A338671                 # Check that the solution is valid and not double counted:
%o A338671                 if d != int(d) or d > min(a,c) or (a == c and d > b):
%o A338671                     continue
%o A338671                 # Count the new pair found:
%o A338671                 count += 1
%o A338671     # Add to the sequence:
%o A338671     seq.append(count)
%o A338671 for an in seq:
%o A338671     print(an)
%o A338671 (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
%Y A338671 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).
%K A338671 nonn
%O A338671 1,4
%A A338671 _Thomas Oléron Evans_, Apr 23 2021