OFFSET
1,1
COMMENTS
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.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).
FORMULA
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
a(n) = a(n-6) for n>6. - Wesley Ivan Hurt, Jun 18 2016
MAPLE
A176514:=n->[3, 1, 1, 3, 2, 1][(n mod 6)+1]: seq(A176514(n), n=0..100); # Wesley Ivan Hurt, Jun 18 2016
MATHEMATICA
LinearRecurrence[{0, 0, 0, 0, 0, 1}, {3, 1, 1, 3, 2, 1}, 96] (* Ray Chandler, Aug 26 2015 *)
PadRight[{}, 100, {3, 1, 1, 3, 2, 1}] (* Vincenzo Librandi, Jun 19 2016 *)
PROG
(Magma) &cat[[3, 1, 1, 3, 2, 1]^^20]; // Wesley Ivan Hurt, Jun 18 2016
(PARI) a(n)=[1, 3, 1, 1, 3, 2][n%6+1] \\ Charles R Greathouse IV, Jul 17 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Egil Edborg (egil.edborg(AT)ebnett.no), Apr 19 2010
STATUS
approved