A255353 Denominators in an expansion of 3 - sqrt(5) as a sum of fractions +-1/d.
2, 3, 6, 15, 24, 40, 104, 168, 273, 714, 1155, 1870, 4895, 7920, 12816, 33552, 54288, 87841, 229970, 372099, 602070, 1576239, 2550408, 4126648, 10803704, 17480760, 28284465, 74049690, 119814915, 193864606, 507544127, 821223648, 1328767776
Offset: 1
Examples
1/(1*2) + 1/(1*3) - 1/(2*3) + 1/(3*5) + 1/(3*8) - 1/(5*8) + 1/(8*13) + 1/(8*21) - 1/(13*21) + 1/(21*34) + 1/(21*55) - 1/(34*55) + ... + = 3 - sqrt(5).
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Mohammad K. Azarian, The Value of a Series of Reciprocal Fibonacci Numbers, Problem B-1133, Fibonacci Quarterly, Vol. 51, No. 3, August 2013, p. 275; Solution published in Vol. 52, No. 3, August 2014, pp. 277-278.
- Index entries for linear recurrences with constant coefficients, signature (0,0,8,0,0,-8,0,0,1).
Programs
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Mathematica
Table[SeriesCoefficient[x (2 + 3 x + 6 x^2 - x^3 - 8 x^5 + x^8)/((1 - x) (1 + x + x^2) (1 - 7 x^3 + x^6)), {x, 0, n}], {n, 33}] (* Michael De Vlieger, Dec 17 2015 *)
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PARI
Vec(x*(2+3*x+6*x^2-x^3-8*x^5+x^8)/((1-x)*(1+x+x^2)*(1-7*x^3+x^6)) + O(x^40)) \\ Colin Barker, Dec 17 2015
Formula
3 - sqrt(5) = Sum_{n>=1} 1/(F(2*n)*F(2*n+1)) + 1/(F(2*n)*F(2*n+2)) - 1/(F(2*n+1)*F(2*n+2)), where F = A000045 (Fibonacci numbers).
From Colin Barker, Dec 17 2015: (Start)
a(n) = 8*a(n-3) - 8*a(n-6) + a(n-9) for n>9.
G.f.: x*(2+3*x+6*x^2-x^3-8*x^5+x^8) / ((1-x)*(1+x+x^2)*(1-7*x^3+x^6)).
(End)
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