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%I #44 Dec 25 2021 21:25:59
%S 1,0,1,2,0,1,0,5,0,1,8,0,9,0,1,0,33,0,14,0,1,48,0,87,0,20,0,1,0,279,0,
%T 185,0,27,0,1,384,0,975,0,345,0,35,0,1,0,2895,0,2640,0,588,0,44,0,1,
%U 3840,0,12645,0,6090,0,938,0,54,0,1,0,35685,0,41685,0,12558,0,1422,0,65
%N Coefficient triangle of the denominators of the (n-th convergents to) the continued fraction 1/(w+2/(w+3/(w+4/... . Conjectured to equal unsigned version of A137286.
%C Equivalence to the recurrence formula needs formal proof. This continued fraction converges to 0.525135276160981... for w=1. A conjecture by Ramanujan puts this equal to -1 + 1/(sqrt(e*Pi/2) - Sum_{k>=1} 1/(2k-1)!!).
%C From _Alexander Kreinin_, Oct 26 2015: (Start)
%C Let us denote the continued fraction by U(w).
%C Then it is easy to show that Mill's ratio, R(w) = (1 - Phi(w))/f(w), where Phi is the standard normal distribution function and f is the standard normal density function, satisfies R(w) = 1/(w + U(w)).
%C Indeed, R(w) = 1/(w+1/(w+2/(w+3/(w+... Then we find U(w) = 1/R(w) - w. It was proved in Alexander Kreinin (arXiv:1405.5852) that R(w+t) + Q(w, t) = exp(w*t + w^2/2)*R(t), where Q(w,t) = Sum_{k>=0} Sum_{m=0..k} q(k,m) * t^m * w^(k+1)/(k+1)!.
%C Substituting t=0, we obtain R(w) = exp(w^2/2)*sqrt(Pi/2) - Sum_{n>=0} w^(2n+1)/(2n+1)!!. If w=1 we obtain Ramanujan's formula. (End)
%H G. C. Greubel, <a href="/A180048/b180048.txt">Table of n, a(n) for the first 75 rows, flattened</a>
%H Authors?, <a href="http://www.komal.hu/forum/forum.cgi?a=to&tid=94&st=50&dr=1&sp=1233">Hungarian discussion forum</a>
%H Alexander Kreinin, <a href="http://arxiv.org/abs/1405.5852">Combinatorial Properties of Mills' Ratio</a>, arXiv:1405.5852 [math.CO], 2014. See Table 3.
%H Alexander Kreinin, <a href="https://www.researchgate.net/profile/Alexander_Kreinin/publication/294260037_Integer_Sequences_and_Laplace_Continued_Fraction/links/56bf5f5308ae2f498ef7effa.pdf">Integer Sequences and Laplace Continued Fraction</a>, preprint, 2016.
%H Alexander Kreinin, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Kreinin/kreinin4.html">Integer Sequences Connected to the Laplace Continued Fraction and Ramanujan's Identity</a>, Journal of Integer Sequences 19 (2016), Article 16.6.2.
%F p(0)=1; p(1)=w; p(n) = w*p(n-1) + n*p(n-2) (conjecture).
%F T(n,k) = T(n-1,k-1) + n*T(n-2,k), T(0,0) = 1, T(1,0) = 0, T(1,1) = 1. - _Philippe Deléham_, Oct 28 2013
%F sum_{k=0..n} T(n,k) = A000932(n). - _Philippe Deléham_, Oct 28 2013
%F T(2n,0) = A000165(n); T(2n+1,1) = A129890(n); T(2n+2,2) = A035101(n+2). - _Philippe Deléham_, Oct 28 2013
%e The denominator of 1/(w+2/(w+3/(w+4/(w+5/(w+6/w))))) equals 48 + 87w^2 + 20w^4 + w^6.
%e From _Joerg Arndt_, Apr 20 2013: (Start)
%e Triangle begins
%e 1;
%e 0, 1;
%e 2, 0, 1;
%e 0, 5, 0, 1;
%e 8, 0, 9, 0, 1;
%e 0, 33, 0, 14, 0, 1;
%e 48, 0, 87, 0, 20, 0, 1;
%e 0, 279, 0, 185, 0, 27, 0, 1;
%e 384, 0, 975, 0, 345, 0, 35, 0, 1;
%e 0, 2895, 0, 2640, 0, 588, 0, 44, 0, 1;
%e 3840, 0, 12645, 0, 6090, 0, 938, 0, 54, 0, 1;
%e 0, 35685, 0, 41685, 0, ... (End)
%t Table[ CoefficientList[ Denominator[ Together[ Fold[ #2/(w+#1) &, Infinity, Reverse @ Table[ k, {k, 1, n} ] ] ] ], w ], {n, 16} ] (* or equivalently *) Clear[ p ];p[ 0 ]=1; p[ 1 ]=w; p[ n_ ]:=p[ n ]= w*p[ n-1 ] + n*p[ n-2 ]; Table[ CoefficientList[ p[ k ]//Expand, w ], {k,0,15} ]
%Y Cf. A137286, A084950, A180047, A180049.
%Y Cf. A111129.
%K nonn,tabl
%O 0,4
%A _Wouter Meeussen_, Aug 08 2010
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