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A329402
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Number of rectangles (w X h, w <= h) with integer side lengths w and h having area = n * perimeter.
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1
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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FORMULA
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EXAMPLE
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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.
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MAPLE
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f:= n -> (numtheory:-tau(4*n^2)+1)/2;
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MATHEMATICA
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a[n_] := Ceiling[DivisorSigma[0, 4 n^2]/2]; Array[a, 75] (* Giovanni Resta, Mar 29 2020 *)
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PROG
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(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)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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