A255353 - OEIS

%I #35 Dec 18 2015 00:41:35

%S 2,3,6,15,24,40,104,168,273,714,1155,1870,4895,7920,12816,33552,54288,

%T 87841,229970,372099,602070,1576239,2550408,4126648,10803704,17480760,

%U 28284465,74049690,119814915,193864606,507544127,821223648,1328767776

%N Denominators in an expansion of 3 - sqrt(5) as a sum of fractions +-1/d.

%C The minus sign in front of a fraction is considered the sign of the numerator and hence the sign of the fraction does not appear in this sequence. We note that numerators are in A131561.

%H Colin Barker, <a href="/A255353/b255353.txt">Table of n, a(n) for n = 1..1000</a>

%H Mohammad K. Azarian, <a href="http://www.fq.math.ca/Problems/ElemProbAugust2013.pdf">The Value of a Series of Reciprocal Fibonacci Numbers, Problem B-1133</a>, Fibonacci Quarterly, Vol. 51, No. 3, August 2013, p. 275; <a href="http://www.fq.math.ca/Problems/ElemProbSolnAug14.pdf">Solution</a> published in Vol. 52, No. 3, August 2014, pp. 277-278.

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,8,0,0,-8,0,0,1).

%F 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).

%F From _Colin Barker_, Dec 17 2015: (Start)

%F a(n) = 8*a(n-3) - 8*a(n-6) + a(n-9) for n>9.

%F 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)).

%F (End)

%e 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).

%t 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 *)

%o (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

%Y Cf. A131561, A187799.

%K nonn,frac,easy

%O 1,1

%A _Mohammad K. Azarian_, Feb 21 2015