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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 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)!!).
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+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)!.
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)
LINKS
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. A111129.
Sequence in context: A238618 A132277 A137286 * A128890 A196777 A318361
KEYWORD
nonn,tabl
AUTHOR
Wouter Meeussen, Aug 08 2010
STATUS
approved

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Last modified March 19 03:33 EDT 2024. Contains 370952 sequences. (Running on oeis4.)