%I #36 Aug 26 2023 22:05:32
%S 2,3,5,4,5,8,5,5,8,8,5,11,5,8,14,6,5,13,5,11,14,8,5,14,8,8,11,11,5,23,
%T 5,7,14,8,14,18,5,8,14,14,5,23,5,11,23,8,5,17,8,13,14,11,5,18,14,14,
%U 14,8,5,32,5,8,23,8,14,23,5,11,14,23,5,23,5,8,23
%N Number of rectangles (w X h, w <= h) with integer side lengths w and h having area = n * perimeter.
%C This sequence was inspired by a generalization of the Isis Problem.
%C Number of ways of writing 1/n as a sum of inverses of consecutive triangular numbers (possibly infinite). For example a(2) = 3 because there are 3 ways of writing 1/2 as a sum of inverses of consecutive triangular numbers: 1/2 = 1/C(4,2) + ... + 1/C(12,2) = 1/C(3,2) + 1/C(4,2) = 1/C(5,2) + 1/C(6,2) + ... - _Ludovic Schwob_, Jul 25 2023
%H Robert Israel, <a href="/A329402/b329402.txt">Table of n, a(n) for n = 1..10000</a>
%H B. Greer, D. De Bock and W. Van Dooren, <a href="https://doi.org/10.1016/j.jmathb.2009.10.004">The Isis Problem as an Experimental probe and teaching resource</a>, Journal of Mathematical Behavior, 28, (2009), 237-246.
%F a(n) = A038548(4*n^2). - _Peter Bala_, Mar 03 2020
%e a(1) = 2 because there are two rectangles which have area = perimeter: 4 X 4 and 3 X 6.
%e a(2) = 3 because there are 3 rectangles for which area = 2 * perimeter: 8 X 8, 6 X 12, and 5 X 20.
%e a(3) = 5 because there are 5 rectangles for which area = 3 * perimeter: 12 X 12, 10 X 15, 9 X 18, 8 X 24, and 7 X 42.
%p f:= n -> (numtheory:-tau(4*n^2)+1)/2;
%p map(f, [$1..100]); # _Robert Israel_, Mar 31 2020
%t a[n_] := Ceiling[DivisorSigma[0, 4 n^2]/2]; Array[a, 75] (* _Giovanni Resta_, Mar 29 2020 *)
%o (Python 3.5.3)
%o numbers=[]
%o for n in range(500):
%o c=int(0)
%o n=int(n+1)
%o for x in range(2*n+1,4*n+1):
%o y=(2*n*x)/(x-2*n)
%o if y==y//1:
%o y=int(y)
%o c=c+1
%o numbers.append(c)
%o print(numbers)
%Y Cf. A038548.
%K nonn
%O 1,1
%A _Nicolas Haverhals_, Feb 28 2020