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A180049
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Coefficient triangle of the numerators of the (n-th convergents to) the continued fraction 1/(w+2/(w+3/(w+4/... .
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4
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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
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OFFSET
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1,4
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COMMENTS
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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}]. Row sums equal A059480.
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LINKS
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Table of n, a(n) for n=1..78.
Hungarian discussion forum
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FORMULA
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b[0]=1; b[1]=w; b[n] = w b[n-1]+(n+1) b[n-2] (conjecture).
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EXAMPLE
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The numerator of 1/(w+2/(w+3/(w+4/(w+5/(w+6/w))))) equals 57w + 18w^3 + w^5.
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. - Philippe Deléham, Nov 06 2013
0, 6555, 0, 4680, 0, 840, 0, 52, 0, 1
10395, 0, 26685, 0, 10290, 0, 1302, 0, 63, 0, 1. - M. F. Hasler, Oct 21 2014
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MATHEMATICA
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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} ]
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PROG
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(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
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CROSSREFS
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Cf. A059480, A084950, A180047, A180048, A230698.
Sequence in context: A242451 A262964 A135481 * A244454 A238123 A128311
Adjacent sequences: A180046 A180047 A180048 * A180050 A180051 A180052
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KEYWORD
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nonn,tabl
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AUTHOR
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Wouter Meeussen, Aug 08 2010
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STATUS
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approved
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