%I #15 Apr 08 2024 06:56:17

%S 0,17,424,7889,131920,2099537,32570104,498191249,7559339680,

%T 114166849937,1719485965384,25855100073809,388391603257840,

%U 5830958998038737,87510144649440664,1313063982494679569,19699665930299694400,295528344080575921937,4433225354293155251944

%N Define f(n) by e(n+1) = e(n) + 3^{n+1} - 1 + 2*f(n), where the rational numbers e(n) are defined in A134939; then a(n) is the numerator of f(n).

%H M. A. Alekseyev and T. Berger, <a href="http://arxiv.org/abs/1304.3780">Solving the Tower of Hanoi with Random Moves</a>, arXiv:1304.3780 [math.CO], 2013-204; In: J. Beineke, J. Rosenhouse (eds.) The Mathematics of Various Entertaining Subjects: Research in Recreational Math, Princeton University Press, 2016, pp. 65-79. ISBN 978-0-691-16403-8

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (32,-342,1440,-2025).

%F f(n) = (6*3^n-1)*(5^n-3^n)/(2*3^n); a(n) = (6*3^n-1)*(5^n-3^n)/2. - _Max Alekseyev_, Feb 04 2008

%F G.f.: x*(135*x^2-120*x+17) / ((3*x-1)*(5*x-1)*(9*x-1)*(15*x-1)). - _Colin Barker_, Dec 26 2012

%e The values of f(0), ..., f(3) are 0, 17/3, 424/9, 7889/27.

%Y Cf. A134939.

%K nonn,frac,easy

%O 0,2

%A Toby Berger (tb6n(AT)virginia.edu), Jan 23 2008

%E Values of f(4) onwards and a general formula found by _Max Alekseyev_, Feb 02 2008, Feb 04 2008