cp's OEIS Frontend

A329402 Number of rectangles (w X h, w <= h) with integer side lengths w and h having area = n * perimeter.

%I A329402 #40 Mar 23 2025 17:40:47
%S A329402 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 A329402 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 A329402 14,8,5,32,5,8,23,8,14,23,5,11,14,23,5,23,5,8,23
%N A329402 Number of rectangles (w X h, w <= h) with integer side lengths w and h having area = n * perimeter.
%C A329402 This sequence was inspired by a generalization of the Isis Problem.
%C A329402 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 A329402 Robert Israel, <a href="/A329402/b329402.txt">Table of n, a(n) for n = 1..10000</a>
%H A329402 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 A329402 a(n) = A038548(4*n^2). - _Peter Bala_, Mar 03 2020
%e A329402 a(1) = 2 because there are two rectangles which have area = perimeter: 4 X 4 and 3 X 6.
%e A329402 a(2) = 3 because there are 3 rectangles for which area = 2 * perimeter: 8 X 8, 6 X 12, and 5 X 20.
%e A329402 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 A329402 f:= n -> (numtheory:-tau(4*n^2)+1)/2;
%p A329402 map(f, [$1..100]); # _Robert Israel_, Mar 31 2020
%t A329402 a[n_] := Ceiling[DivisorSigma[0, 4 n^2]/2]; Array[a, 75] (* _Giovanni Resta_, Mar 29 2020 *)
%o A329402 (Python)
%o A329402 numbers=[]
%o A329402 for n in range(500):
%o A329402     c=int(0)
%o A329402     n=int(n+1)
%o A329402     for x in range(2*n+1,4*n+1):
%o A329402         y=(2*n*x)/(x-2*n)
%o A329402         if y==y//1:
%o A329402             y=int(y)
%o A329402             c=c+1
%o A329402     numbers.append(c)
%o A329402 print(numbers)
%Y A329402 Cf. A038548.
%K A329402 nonn
%O A329402 1,1
%A A329402 _Nicolas Haverhals_, Feb 28 2020