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 A180048 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. 7
 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, 185, 0, 27, 0, 1, 384, 0, 975, 0, 345, 0, 35, 0, 1, 0, 2895, 0, 2640, 0, 588, 0, 44, 0, 1, 3840, 0, 12645, 0, 6090, 0, 938, 0, 54, 0, 1, 0, 35685, 0, 41685, 0, 12558, 0, 1422, 0, 65 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Equivalence to the recurrence formula needs formal proof. This C.F. converges to 0.525135276160981.. for w=1. A conjecture by Ramanujan puts this equal to -1+1/(Sqrt[E Pi/2]-Sum[1/(2k-1)!!,{k,1,Infinity}]. From Alexander Kreinin, Oct 26 2015: (Start) Let us denote the continued fraction by U(w). 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)). Indeed, R(w)=1(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(wt+w^2/2)R(t), where Q(w,t)=Sum[ Sum[q(k,m)t^m w^(k+1)/(k+1)!,{m,0,k}],{k,0,Infinity}]. Substituting t=0, we obtain R(w)=exp(w^2/2) sqrt(Pi/2) - Sum[w^(2n+1)/(2n+1)!!, {n,0,Infinity}]. If w=1 we obtain Ramanujan's formula. (End) LINKS G. C. Greubel, Table of n, a(n) for the first 75 rows, flattened Authors?, Hungarian discussion forum Alexander Kreinin, Combinatorial Properties of Mills' Ratio, arXiv:1405.5852 [math.CO], 2014. See Table 3. Alexander Kreinin, Integer Sequences and Laplace Continued Fraction, preprint, 2016. Alexander Kreinin, Integer Sequences Connected to the Laplace Continued Fraction and Ramanujan's Identity, Journal of Integer Sequences 19 (2016), Article 16.6.2. FORMULA p[0]=1; p[1]=w; p[n]= w*p[n-1] + n*p[n-2] (conjecture) 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 sum_{k=0..n} T(n,k) = A000932(n). - Philippe Deléham, Oct 28 2013 T(2n,0) = A000165(n); T(2n+1,1) = A129890(n); T(2n+2,2) = A035101(n+2). - Philippe Deléham, Oct 28 2013 EXAMPLE The denominator of 1/(w+2/(w+3/(w+4/(w+5/(w+6/w))))) equals 48 + 87w^2 + 20w^4 + w^6. From Joerg Arndt, Apr 20 2013: (Start) Triangle begins      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,   185,    0,  27,   0,  1;    384,     0,   975,     0,  345,   0,  35,  0,  1;      0,  2895,     0,  2640,    0, 588,   0, 44,  0, 1;   3840,     0, 12645,     0, 6090,   0, 938,  0, 54, 0, 1;      0, 35685,     0, 41685,    0, ... (End) MATHEMATICA 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} ] CROSSREFS Cf. A137286, A084950, A180047, A180049. Cf. A111129. Sequence in context: A238618 A132277 A137286 * A128890 A196777 A318361 Adjacent sequences:  A180045 A180046 A180047 * A180049 A180050 A180051 KEYWORD nonn,tabl AUTHOR Wouter Meeussen, Aug 08 2010 STATUS approved

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Last modified December 7 13:08 EST 2021. Contains 349581 sequences. (Running on oeis4.)