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A329402
Number of rectangles (w X h, w <= h) with integer side lengths w and h having area = n * perimeter.
5
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, 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, 14, 8, 5, 32, 5, 8, 23, 8, 14, 23, 5, 11, 14, 23, 5, 23, 5, 8, 23
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
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
Cf. A038548.
Sequence in context: A262507 A151679 A077664 * A339263 A179475 A373346
KEYWORD
nonn
AUTHOR
Nicolas Haverhals, Feb 28 2020
STATUS
approved