OFFSET
1,1
COMMENTS
This sequence was inspired by a generalization of the Isis Problem.
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
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
B. Greer, D. De Bock and W. Van Dooren, The Isis Problem as an Experimental probe and teaching resource, Journal of Mathematical Behavior, 28, (2009), 237-246.
FORMULA
a(n) = A038548(4*n^2). - Peter Bala, Mar 03 2020
EXAMPLE
a(1) = 2 because there are two rectangles which have area = perimeter: 4 X 4 and 3 X 6.
a(2) = 3 because there are 3 rectangles for which area = 2 * perimeter: 8 X 8, 6 X 12, and 5 X 20.
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.
MAPLE
f:= n -> (numtheory:-tau(4*n^2)+1)/2;
map(f, [$1..100]); # Robert Israel, Mar 31 2020
MATHEMATICA
a[n_] := Ceiling[DivisorSigma[0, 4 n^2]/2]; Array[a, 75] (* Giovanni Resta, Mar 29 2020 *)
PROG
(Python 3.5.3)
numbers=[]
for n in range(500):
c=int(0)
n=int(n+1)
for x in range(2*n+1, 4*n+1):
y=(2*n*x)/(x-2*n)
if y==y//1:
y=int(y)
c=c+1
numbers.append(c)
print(numbers)
CROSSREFS
KEYWORD
nonn
AUTHOR
Nicolas Haverhals, Feb 28 2020
STATUS
approved