login
A180049
Coefficient triangle of the numerators of the (n-th convergents to) the continued fraction 1/(w+2/(w+3/(w+4/... .
4
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, 561, 0, 285, 0, 33, 0, 1, 945, 0, 1830, 0, 510, 0, 42, 0, 1, 0, 6555, 0, 4680, 0, 840, 0, 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
OFFSET
1,4
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)!!). Row sums equal A059480.
FORMULA
b(0)=1; b(1)=w; b(n) = w*b(n-1) + (n+1)*b(n-2) (conjecture).
EXAMPLE
The numerator of 1/(w+2/(w+3/(w+4/(w+5/(w+6/w))))) equals 57w + 18w^3 + w^5.
From Philippe Deléham, Nov 06 2013: (Start)
Triangle begins:
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, 561, 0, 285, 0, 33, 0, 1;
945, 0, 1830, 0, 510, 0, 42, 0, 1;
0, 6555, 0, 4680, 0, 840, 0, 52, 0, 1;
10395, 0, 26685, 0, 10290, 0, 1302, 0, 63, 0, 1;
... (End)
[extended by M. F. Hasler, Oct 21 2014]
MATHEMATICA
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} ]
PROG
(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
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Wouter Meeussen, Aug 08 2010
STATUS
approved