A180049 - OEIS

%I #17 Dec 24 2021 21:54:25

%S 1,0,1,3,0,1,0,7,0,1,15,0,12,0,1,0,57,0,18,0,1,105,0,141,0,25,0,1,0,

%T 561,0,285,0,33,0,1,945,0,1830,0,510,0,42,0,1,0,6555,0,4680,0,840,0,

%U 52,0,1,10395,0,26685,0,10290,0,1302,0,63,0,1,0,89055,0,82845,0,20370,0,1926,0,75,0,1

%N Coefficient triangle of the numerators of the (n-th convergents to) the continued fraction 1/(w+2/(w+3/(w+4/... .

%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)!!). Row sums equal A059480.

%H <a href="http://www.komal.hu/forum/forum.cgi?a=to&amp;tid=94&amp;st=50&amp;dr=1&amp;sp=1233">Hungarian discussion forum</a>

%F b(0)=1; b(1)=w; b(n) = w*b(n-1) + (n+1)*b(n-2) (conjecture).

%e The numerator of 1/(w+2/(w+3/(w+4/(w+5/(w+6/w))))) equals 57w + 18w^3 + w^5.

%e From _Philippe Deléham_, Nov 06 2013: (Start)

%e Triangle begins:

%e       1;

%e       0,    1;

%e       3,    0,     1;

%e       0,    7,     0,    1;

%e      15,    0,    12,    0,     1;

%e       0,   57,     0,   18,     0,   1;

%e     105,    0,   141,    0,    25,   0,    1;

%e       0,  561,     0,  285,     0,  33,    0,  1;

%e     945,    0,  1830,    0,   510,   0,   42,  0,  1;

%e       0, 6555,     0, 4680,     0, 840,    0, 52,  0, 1;

%e   10395,    0, 26685,    0, 10290,   0, 1302,  0, 63, 0, 1;

%e   ... (End)

%e [extended by _M. F. Hasler_, Oct 21 2014]

%t Table[ CoefficientList[ Numerator[ Together[ Fold[ #2/(w+#1) &, Infinity, Reverse @ Table[ k, {k, 1, n} ] ] ] ], w ], {n, 2, 16} ] or equivalently Clear[ b ]; b[ 0 ]=1; b[ 1 ]=w; b[ n_ ]:=b[ n ] = w b[ n-1 ]+(n+1) b[ n-2 ]; Table[ CoefficientList[ b[ k ]//Expand, w ], {k, 0, 14} ]

%o (PARI) t=x-w;for(n=1,12,t=substpol(t,x,w+n/x);print(Vecrev(numerator(substpol(t,x,w))))) \\ _M. F. Hasler_, Oct 21 2014

%Y Cf. A059480, A084950, A180047, A180048, A230698.

%K nonn,tabl

%O 1,4

%A _Wouter Meeussen_, Aug 08 2010