The silver ratio and a continued fraction for log(2)

Define the three sequences,

C_n = \sum_{k=0}^n {\binom n k} {\binom {n+k}k} = 1, 3, 13, 63, 321,\dots

B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251,\dots

A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, \dots

The last two are Apery numbers and have been discussed previously. The first are the central Delannoy numbers which obeys the limit,

\lim_{n \to \infty} \frac{C_{n+1}}{C_n} = \left(1+\sqrt{2}\right)^2 = 5.8284\dots

which is the square of the silver ratio.  (The ratios for the others have already been mentioned.)  These have the recurrence relations,

n C_n = 3(2n-1)C_{n-1}-(n-1)C_{n-2}

n^2 B_n = (11n^2-11n+3)B_{n-1}+(n-1)^2B_{n-2}

n^3 A_n = (34n^3-51n^2+27n-5)A_{n-1}-(n-1)^3A_{n-2}

To recall, the polynomials P(n) =11n^2-11n+3 and P(n) = 34n^3-51n^2+27n-5  generated numbers for the continued fractions of \zeta(2), \zeta(3), so I was curious if P(n) = 3(2n-1) could be used in an analogous manner.  It turns out, depending on what sign to use, it gives either \log(2) or \arctan(\frac{1}{3}),

\frac{1}{2}\log(2) = \cfrac{1}{v_1 - \cfrac{1^2}{v_2 - \cfrac{2^2}{v_3 - \cfrac{3^2}{v_4-\ddots}}}}


\arctan(\frac{1}{3}) = \cfrac{1}{v_1 + \cfrac{1^2}{v_2 + \cfrac{2^2}{v_3 + \cfrac{3^2}{v_4 + \ddots}}}}

where, starting with n = 1,

v_n = 3(2n-1)

This can be partly demystified since one continued fraction for the natural logarithm, and arctan, is,

\frac{1}{2}\log\left(1+\frac{2x}{y}\right)= \cfrac{x}{1(x+y) - \cfrac{(1x)^2}{3(x+y) - \cfrac{(2x)^2}{5(x+y) - \cfrac{(3x)^2}{7(x+y) - \ddots}}}}


\arctan(\frac{x}{z}) = \cfrac{x}{1z + \cfrac{(1x)^2}{3z + \cfrac{(2x)^2}{5z + \cfrac{(3x)^2}{7z + \ddots}}}}

Hence, if x+y = z, as with the case {x, y, z} = {1, 2, 3}, then there will be identical-looking continued fractions that differ only in the signs.  But it remains interesting how the recurrence relations of these three binomial sums are involved in the continued fractions of \log(2), \zeta(2), \zeta(3).  Later, we shall see there is a recurrence relation for the cfrac of \zeta(4) as well.


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