## Archive for the ‘identities’ Category

### Hypergeometric formulas for Ramanujan’s continued fractions 2

(continued from yesterday’s post)

III. Icosahedral group

\begin{aligned} G(q) &= \sum_{n=0}^\infty \frac{q^{n^2}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\H(q) &= \sum_{n=0}^\infty \frac{q^{n^2+n}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\end{aligned}

I observed that,

\begin{aligned}&q^{-1/60}G(q) = j^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j}\big) = (j-1728)^{1/60}\,_2F_1\big(\tfrac{29}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{1728-j}\big)\\[2.5mm]&q^{11/60}H(q) = j^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j}\big) = (j-1728)^{-11/60}\,_2F_1\big(\tfrac{41}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{1728-j}\big)\end{aligned}

where, as in the previous post, $j=j(\tau)$ is the j-function, $q = e^{2\pi i \tau} = \exp(2\pi i \tau)$, $\tau = \sqrt{-N}$, and $N>1$.  Since it is known that,

\begin{aligned}&r(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \cfrac{q^3}{1 + \ddots}}}} = \frac{q^{11/60}H(q)}{q^{-1/60}G(q)} = \frac{q^{11/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}}{q^{-1/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(q^{5n-4})}}\end{aligned}

this implies that,

\begin{aligned}r(q) &=\frac{j^{-11/60}\,_2F_1\big(\frac{31}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{j}\big) }{j^{1/60}\,_2F_1\big(\frac{19}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{j}\big)}\\[3mm]&=\frac{(j-1728)^{-11/60}\,_2F_1\big(\frac{41}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{1728-j}\big) }{(j-1728)^{1/60}\,_2F_1\big(\frac{29}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{1728-j}\big)}\end{aligned}

Example. Let $\tau = \sqrt{-4}$, hence $j = j(\sqrt{-4}) = 66^3$. Then,

$1/r(q) -r(q) = \left(\frac{1+\sqrt{5}}{2}\right)^4+\left(\frac{1+\sqrt{5}}{2}\right)5^{3/4} = 12.2643\dots$

Furthermore, since Ramanujan established that,

$G(q^{11})H(q)-q^2G(q)H(q^{11}) = 1$

if we define the two functions,

\begin{aligned}U(\tau) &= \big(j(\tau)\big)^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j(\tau)}\big)\\V(\tau) &= \big(j(\tau)\big)^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j(\tau)}\big) \end{aligned}

then the counterpart hypergeometric identity is also beautifully simple and given by,

$U(11\tau)V(\tau)-U(\tau)V(11\tau)=1$

In the next post, we will use one of the hypergeometric formulas to solve the general quintic.

### Algebraic values of the Hypergeometric function

\begin{aligned} &\,_2F_1(a,b,c,z) = \sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!}\end{aligned}

where $(a)_n$ is a Pochhammer symbol is, for rational {a, b, c, z}, generally not a rational value.  In the link given above, equations (18) and (19) are,

$\,_2F_1\big(\frac{1}{3},\frac{2}{3},\frac{5}{6},\frac{27}{32}\big) =\frac{8}{5}$

$\,_2F_1\big(\frac{1}{4},\frac{1}{2},\frac{3}{4},\frac{80}{81}\big) =\frac{9}{5}$

There are in fact an infinite number of such equalities.  One given by M. Glasser is, let $0 < x < \frac{1}{\sqrt{3}}$, then,

$\,_2F_1\big(\frac{1}{3},\frac{2}{3},\frac{3}{2},\frac{27x^2(1-x^2)^2}{4}\big) =\frac{1}{1-x^2}$

Another, based on eq. (42) of Vidunas’ “Transformations of algebraic Gauss hypergeometric functions” is, let $0 < y < 1$, then,

$\,_2F_1\big(\frac{1}{4},\frac{-1}{12},\frac{2}{3},\frac{(9-y^4)^3(-1+y^4)}{64y^{12}}\big) =\frac{1}{y}$

A third which yields not a rational but an algebraic number is remarkable for its connection to the Rogers-Ramanujan continued fraction. Let $0 < z < v_1$ where,

$v_1 = \text{Root}[z^4+228z^3+494z^2-228z+1=0] = 0.004428\dots$

then,

$\,_2F_1\big(\frac{19}{60},\frac{-1}{60},\frac{4}{5},\frac{1728z(z^2-11z-1)^5}{(z^4+228z^3+494z^2-228z+1)^3}\big) =\frac{1}{(z^4+228z^3+494z^2-228z+1)^{1/20}}$

If the polynomials are familiar, it is because they are invariants of the icosahedron.  They also appear in the j-function formula,

$j(\tau) = \frac{-(r^{20}-228r^{15}+494r^{10}+228r^5+1)^3}{r^5(r^{10}+11r^5-1)^5}$

where,

$r = r(\tau) = \cfrac{q^{1/5}}{1+ \cfrac{q}{1 + \cfrac{q^2}{1+ \cfrac{q^3}{1 + \ddots}}}}$

and

$q = e^{2\pi i \tau}$

### On Bailey and Crandall’s unusual sum

In page 20 of Bailey and Crandall’s On the Random Character of Constant Expansions, they give the wonderfully unusual sum,

\begin{aligned} x_1 = \sum_{n=0}^\infty \frac{1}{5^{5n}}\left(\frac{5}{5n+2}+\frac{1}{5n+3}\right)&=\frac{5^2}{2}\ln\left(\frac{781}{256}\left(\frac{57-5\sqrt{5}}{57+5\sqrt{5}}\right)^{\sqrt{5}}\right)\\ &= 2.833601\dots\end{aligned}

I didn’t think this was an isolated result so set about to find a generalization.  I found its counterpart,

\begin{aligned} x_2 = \sum_{n=0}^\infty \frac{1}{5^{5n}}\left(\frac{5^3}{5n+1}+\frac{1}{5n+4}\right)&=\frac{5^3}{2}\ln\left(\frac{781}{256}\left(\frac{57+5\sqrt{5}}{57-5\sqrt{5}}\right)^{\sqrt{5}}\right)\\ &= 125.256703\dots\end{aligned}

Note that,

$(57+5\sqrt{5})(57-5\sqrt{5}) = 4(781) = 5^5-1$

We can demystify the sum a bit by splitting the log function into parts. After some algebraic manipulation, we find that the first one becomes,

\begin{aligned}&x_1 = 25\Big(-5\ln(2)+\tfrac{1-\sqrt{5}}{2}\,\ln(57+5\sqrt{5})+\tfrac{1+\sqrt{5}}{2}\,\ln(57-5\sqrt{5})\Big)\end{aligned}

Thus it can be expressed in the form,

$x_1 = r\ln\big({p_1}^{q_1}\,{p_2}^{q_2}\big)$

where {$p_1, p_2$} are roots of the same equation, {$q_1, q_2$} are roots of another, and r is a rational.  The fact that,

$2\cos(\pi/5) = \frac{1+\sqrt{5}}{2}$

was my clue that trigonometric functions may be involved.  Define,

$c_k = -2\cos(2k\pi/p)$

then for p = 5,

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{5^{5n}}\left(\tfrac{5}{5n+2}+\tfrac{1}{5n+3}\right) &= 5^2\ln\big((5c_1+26)^{c_2}(5c_2+26)^{c_1}/4^2\big)\\&=2.833601\dots\end{aligned}

p = 7

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{7^{7n}}\left(\tfrac{7}{7n+3}+\tfrac{1}{7n+4}\right) &= 7^3\ln\big((7c_1+50)^{c_3}(7c_2+50)^{c_1}(7c_3+50)^{c_2}/6^2\big)\\&=2.583334\dots\end{aligned}

p = 9

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{9^{9n}}\left(\tfrac{9}{9n+4}+\tfrac{1}{9n+5}\right) &= 9^4\ln\big((9c_1+82)^{c_4}(9c_2+82)^{c_1}(9c_3+82)^{c_3}(9c_4+82)^{c_2}/8^2\big)\\&=2.450000\dots\end{aligned}

with the constants {$26, 50, 82$} easily ascertained as {$5^2+1, 7^2+1, 9^2+1$}, and so on.  On the other hand, their counterparts are easier as the exponent $c_k$ has the same subscript as the base.  Still defining,

$c_k = -2\cos(2k\pi/p)$

then for p = 5,

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{5^{5n}}\left(\tfrac{5^3}{5n+1}+\tfrac{1}{5n+4}\right) &= 5^3\ln\big((5c_1+26)^{c_1}(5c_2+26)^{c_2}/4^2\big)\\&=125.256703\dots\end{aligned}

p = 7

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{7^{7n}}\left(\tfrac{7^5}{7n+1}+\tfrac{1}{7n+6}\right) &= 7^5\ln\big((7c_1+50)^{c_1}(7c_2+50)^{c_2}(7c_3+50)^{c_3}/6^2\big)\\&=16807.169\dots\end{aligned}

p = 9

\begin{aligned}\sum_{n=0}^{\infty} \frac{1}{9^{9n}}\left(\tfrac{9^7}{9n+1}+\tfrac{1}{9n+8}\right) &= 9^7\ln\big((9c_1+82)^{c_1}(9c_2+82)^{c_2}(9c_3+82)^{c_3}(9c_4+82)^{c_4}/8^2\big)\\&\approx 4.7829\, {\rm x}\, 10^6\end{aligned}

etc.

### The zeta function and roots of unity

In Mathworld’s entry on the Riemann zeta function, one finds in eq. 119-121 the curious evaluations,

\begin{aligned}\sum_{n=1}^\infty [\zeta(2n)-1] &= \frac{3}{4}\\ \sum_{n=1}^\infty [\zeta(3n)-1] &= \frac{1}{3}\left[-(-1)^{2/3}H_{(3-\sqrt{-3})/2}+(-1)^{1/3}H_{(3+\sqrt{-3})/2} \right]\\ \sum_{n=1}^\infty [\zeta(4n)-1] &= \frac{1}{8}\,(7-2\pi\coth(\pi))\end{aligned}

However, using the Inverse Symbolic Calculator, the first and the third, plus another one, can also be expressed as,

\begin{aligned}\sum_{n=1}^\infty [\zeta(2n)-1] &= \frac{5}{4}-\sum_{n=1}^\infty \frac{1}{2n^2+2n} = \frac{3}{4}\\ \sum_{n=1}^\infty [\zeta(4n)-1] &= \frac{5}{8}-\sum_{n=1}^\infty \frac{1}{2n^2+2}=\frac{7}{8}-\frac{1}{4}\,\pi i \cot(\pi w_4)\\ \sum_{n=1}^\infty [\zeta(6n)-1] &= \frac{5}{12}-\sum_{n=1}^\infty \frac{1}{2n^2+2n+2}=\frac{11}{12}-\frac{1}{6}\sqrt{3}\pi i\cot(\pi w_6)\end{aligned}

where $w_p = e^{2\pi i/p}$.  Interesting similar forms, isn’t it?

Unfortunately, it doesn’t seem to generalize to $\zeta(pn)$ for p = 8.  However, there is still p = 3 and, based on the even case, I assumed perhaps roots of unity are also involved.  First, given the Euler-Mascheroni constant $\gamma$, and the digamma function,

$\psi_0(z) = \psi[z]$

where we suppress the subscript for ease of notation.  Define,

$u_p = e^{\pi i/p } = (-1)^{1/p}$

and the pth root chosen such that $(-1)^{1/p} \not = -1$, then I found that p = 3 generalizes as,

\begin{aligned} 3\sum_{n=1}^\infty [\zeta(3n)-1] &= 3 + \gamma + u_3^{-1}\, \psi[u_3^{-1}]+u_3\,\psi[u_3]\\&= 0.66506\dots\\ 5\sum_{n=1}^\infty [\zeta(5n)-1] &= 6 + \gamma + \sum_{k=0}^1 \Big(u_5^{-(2k+1)}\, \psi[u_5^{-(2k+1)}]+u_5^{(2k+1)}\,\psi[u_5^{(2k+1)}]\Big)\\&=0.18976\dots\\ 7\sum_{n=1}^\infty [\zeta(7n)-1] &= 9 + \gamma + \sum_{k=0}^2 \Big(u_7^{-(2k+1)}\, \psi[u_7^{-(2k+1)}]+u_7^{(2k+1)}\,\psi[u_7^{(2k+1)}]\Big)\\&=0.05887\dots\end{aligned}

and so on, though a rigorous proof is needed that it holds true for all odd numbers p.

P.S. Going back to even p, note that p = 2, 4, 6 can also be expressed by the digamma function since,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{an^2+bn+c} = \frac{1}{\sqrt{b^2-4ac}}\Big(\psi[\tfrac{2a+b+\sqrt{b^2-4ac}}{2a}]-\psi[\tfrac{2a+b-\sqrt{b^2-4ac}}{2a}]\Big)\end{aligned}

for $a \not=0$.

### Roots of unity and binomial sums

Thanks to Robert Israel who answered my question in mathstackexchange, we have a generalization of the binomial sums of the previous post.  Interestingly, it turns out roots of unity are involved.  Given,

$w = e^{2\pi\, {\rm i}/k}$

where k is an even integer then,

\begin{aligned}&\sum_{n=0}^\infty \frac{(2z)^{kn}}{\binom{kn}{kn/2}} = \frac{1}{1-z^k}+\frac{1}{k}\sum_{j=0}^{k-1}\frac{w^j z \arcsin(w^j z)}{(1-w^{2j}z^2)^{3/2}}\end{aligned}

for appropriate z such that the sum converges.  For the special case when,

$2z =w^{1/2}=e^{\pi\,{\rm i}/k}$

then,

\begin{aligned}&\sum_{n=0}^\infty\frac{(-1)^n}{\binom{kn}{kn/2}}=\frac{2^k}{2^k+1}+\frac{1}{k}\sum_{j=0}^{k-1}\frac{w^{j+1/2} \arcsin(\tfrac{1}{2}\,w^{j+1/2})}{2(1-\tfrac{1}{4}\,w^{2j+1})^{3/2}}\end{aligned}

Note that the terms are complex, but the sum is a real number so they must come in conjugate pairs. The arcsin of a complex root of unity can be given as,

$\arcsin(\frac{1}{2}\,e^{\pi\,{\rm i}/k}) = -\arcsin\big(\frac{-a+b}{4}\big)+{\rm i}\ln\Big(\frac{a+b+\sqrt{-6+2ab}}{4}\Big)$

where,

\begin{aligned}a &= \sqrt{5+4\cos(\pi/k)}\\ b &= \sqrt{5-4\cos(\pi/k)}\end{aligned}

With this transformation, it is now possible to have an expression all in real terms.  The case k = 2, 4 was given in the previous post.  For k = 6, we have the counterpart to Sprugnoli’s equality as,

\begin{aligned}\sum_{n=0}^\infty \frac{(-1)^n}{\binom{6n}{3n}}&=\tfrac{64}{65}-\tfrac{2\sqrt{26}\,(1+7\sqrt{13})}{3\cdot13^2\,\sqrt{1+\sqrt{13}}}\arcsin\left(\tfrac{-\sqrt{6}}{2\sqrt{5+\sqrt{13}}}\right)\\&-\tfrac{\sqrt{26}\,(-1+7\sqrt{13})}{3\cdot13^2\,\sqrt{-1+\sqrt{13}}}\ln\left(\tfrac{\sqrt{7+2\sqrt{13}}\,+1}{\sqrt{7+2\sqrt{13}}\,-1}\right)-\tfrac{4\sqrt{5}}{3\cdot5^2}\ln\left(\tfrac{1+\sqrt{5}}{2}\right) = 0.95106\dots\end{aligned}

Note that the prime factors of 65 are 5 and 13, and the square root of both appear above. However, for k = 8, while the expression contains the fraction $\frac{256}{257}$ as expected, the argument of the log and arcsin do not factor over the quadratic extension $\sqrt{257}$, but rather only over $\sqrt{2}$.  Furthermore, the argument of the log for both k = 6, 8 are no longer simply expressible in terms of the Dedekind eta function, so observations for lower k do not generalize to higher ones.

### Fermat primes and Binomial sums

We have,

\begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n}{\binom n{n/2}} &= \frac{4}{3}-\frac{4\pi\sqrt{3}}{27}\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} &= \frac{4}{5} - \frac{4\sqrt{5}}{25}\ln\left(\frac{1+\sqrt{5}}{2}\right) \end{aligned}

For the next step, Renzo Sprugnoli gave the Ramanujan-like identity,

\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}+\frac{4\sqrt{34}\,(-2+\sqrt{17}\,)}{17^2\,\sqrt{-1+\sqrt{17}}}\arctan\left(\frac{\sqrt{2}}{\sqrt{-1+\sqrt{17}}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{17^2\,\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right) = 0.846609\dots \end{aligned}

(The sign of the third term has been changed by this author.)  However, to make it more symmetrical, we can express the arctan in terms of the log function.  Since,

\begin{aligned}&\arctan(z) = \frac{i}{2}\ln\left(\frac{1-i z}{1+i z}\right)\end{aligned}

then,

\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}-\frac{2\sqrt{-34}\,(-2+\sqrt{17}\,)}{17^2\,\sqrt{-1+\sqrt{17}}}\ln\left(\frac{\sqrt{-1+\sqrt{17}}+\sqrt{-2}}{\sqrt{-1+\sqrt{17}}-\sqrt{-2}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{17^2\,\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right) =0.846609\dots \end{aligned}

In this manner, it reduces to the concise,

\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}-\frac{x_1}{17}\ln(y_1)-\frac{x_2}{17}\ln(y_2)\end{aligned}

where, $x_1,x_2$ and $y_1,y_2$ are the appropriate roots of,

\begin{aligned} &289x^4-799x^2-676 = 0\\ &y^4-5y^3+4y^2-5y+1 = 0\end{aligned}

I found that, curiously, the argument of the log can be expressed in terms of the Dedekind eta function, $\eta(z)$.  Let,

\begin{aligned} t_1 &=\frac{1+\sqrt{-5}}{2}\\ t_2 &= \frac{1+\sqrt{-17}}{2}\\ \zeta_{48} &=\exp(\pi i/24)\end{aligned}

then,

\begin{aligned} &\frac{1}{2}\left(\frac{\zeta_{48}\, \eta(t_1)}{\eta(2t_1)}\right)^4 = \frac{1+\sqrt{5}}{2}\\ &\frac{1}{2}\left(\frac{\zeta_{48}\, \eta(t_2)}{\eta(2t_2)}\right)^4 = \frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\end{aligned}

Is this coincidence?  Furthermore, using these as the argument of the polylogarithm,

\begin{aligned} &L_s(z) = \text{Li}_s (z) = \sum_{k=1}^\infty\frac{z^k}{k^s}\end{aligned}

one can find a polylogarithm ladder to express Apery’s constant.  For example, getting the square root and reciprocal of $y_2$ so that z < 1,

\begin{aligned} z &= \sqrt{\frac{\sqrt{1+\sqrt{17}}-\sqrt{2}}{\sqrt{1+\sqrt{17}}+\sqrt{2}}} = 0.480533\dots\end{aligned}

then,

$-12L_3(z)+75L_3(z^2)-68L_3(z^3)-33L_3(z^4)+43L_3(z^6)+12L_3(z^8)-7L_3(z^{12})+2\log^3(1/z) = 3\zeta(3)$

A simpler one exists for the other argument. The next step, of course, is,

\begin{aligned}&\sum_{n=0}^\infty \frac{(-1)^n}{\binom{8n}{2n}}=\,?\end{aligned}

Since the first three Fermat primes 3, 5, 17 have already appeared, it should be interesting to conjecture if 257 will be next.

### A missing binomial sum identity?

D.Bailey, J. Borwein, and D.Bradley found the beautiful pair involving binomial sums.  In Theorem 1 of this paper (2008), let x $\not=$ non-zero integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^2-x^2} &= 3\,\sum_{k=1}^\infty \frac{1}{k^2 \binom{2k}k (1-x^2/k^2)} \prod_{m=1}^{k-1} \left(\frac{1-4x^2/m^2}{1-x^2/m^2}\right)\\[2.5mm] \sum_{k=1}^\infty \frac{1}{k^3(1-x^4/k^4)} &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3 \binom{2k}k (1-x^4/k^4)} \prod_{m=1}^{k-1} \left(\frac{1+4x^4/m^4}{1-x^4/m^4}\right)\end{aligned}

When x = 0, they reduce into,

\begin{aligned} \zeta(2) &= 3\,\sum_{k=1}^\infty \frac{1}{k^2\binom{2k}k }\\ \zeta(3) &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\binom{2k}k }\end{aligned}

However, there is a third single-term equality,

\begin{aligned} \frac{17}{36}\,\zeta(4) &= \sum_{k=1}^\infty \frac{1}{k^4\binom{2k}k }\end{aligned}

so there might be a third identity that reduces to this as the special case x = 0.

To compare, there are three identities such that as $x \to 0$, then those zeta values are the respective limit.  For x $\not=$ integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\,\frac{3k^2+x^2}{k^2-x^2}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{\pi x \csc(\pi x)-1}{x^2}\\[2.5mm] \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\,\binom{2k}k}\,\frac{5k^2-x^2}{2(k^2-x^2)}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{-\psi^{(0)}(1-x)-\psi^{(0)}(1+x)-2\gamma}{2x^2}\\[2.5mm] \sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\,\frac{1}{k^2-x^2}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{\pi x \csc(\pi x)+3\cos(\pi x/3)-4}{4x^4}\end{aligned}

The first two were found by Leshchiner and Koecher, respectively, while the third is Theorem 2 in the same paper by Bailey, Borwein, and Bradley. The function $\psi^{(0)}$ is given in Mathematica as,

$\psi^{(0)}(z) = \text{PolyGamma[0,z]}$

while $\gamma$ is the Euler-Mascheroni constant.  So are the Bailey-Borwein-Bradley pair of binomial sum identities in fact a triplet?