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A176514
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Period 6: repeat [3, 1, 1, 3, 2, 1].
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1
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3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 1, 3, 2, 1
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OFFSET
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1,1
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COMMENTS
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3/n expressed as Egyptian fractions with 3 unit fractions: 3/n = 1/x + 1/y + 1/z, all integers, n > 0 and x < y < z.
I have introduced a variable t, an integer that varies with n in accordance with a periodic sequence of 6 terms: [3, 1, 1, 3, 2, 1] starting at n = 0, although 0 is outside the defined interval.
x, the denominator of the first unit fraction, varies with n in accordance with the sequence:
[(1), 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5,..]; x_0 = x_1 = x_2 = 1. Then x_3 = x_4 = x_5 = 2 and so on, increasing the value by 1 for every 3 terms.
y is calculated by:
y = (nx + t)/2 for n = 1, 4, 7, 10,.. etc.
y = (nx + t)/1 = nx + t for n = 2, 5, 8, 11,.. etc.
y = (nx + t)/t for n = 3, 6, 9, 12,.. etc.
z is calculated by: z = nxy/t for all n > 0
This algorithm produces the "first" Egyptian fraction of each type that has 3 unit fractions.
By "first" I indicate the Egyptian fraction that otherwise would be arrived at by employing Fibonacci's greedy algorithm.
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LINKS
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FORMULA
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G.f.: ( 3+x+x^2+3*x^3+2*x^4+x^5 ) / ( (1-x)*(1+x)*(1+x+x^2)*(x^2-x+1) ). - R. J. Mathar, Oct 08 2011
a(n) = 11/6 -cos(Pi*n/3)/6 -sqrt(3)*sin(Pi*n/3)/6 +7*cos(2*Pi*n/3)/6 +sqrt(3)*sin(2*Pi*n/3)/6 +(-1)^n/6. - R. J. Mathar, Oct 08 2011
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MAPLE
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MATHEMATICA
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LinearRecurrence[{0, 0, 0, 0, 0, 1}, {3, 1, 1, 3, 2, 1}, 96] (* Ray Chandler, Aug 26 2015 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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Egil Edborg (egil.edborg(AT)ebnett.no), Apr 19 2010
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STATUS
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approved
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