Archive for June, 2012

A new formula for Apery’s constant and other zeta(2n+1)?

I. Introduction

In Identities Inspired from Ramanujan’s Notebooks, Simon Plouffe recounts how, based on Ramanujan’s,

\begin{aligned}\sum_{k=1}^\infty \frac{\coth(\pi k)}{k^3} = \frac{7}{180}\pi^3\end{aligned}

he found,

\begin{aligned}\zeta(3) &= \frac{7\pi^3}{180}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{2\pi k}-1)}\\    \zeta(7) &= \frac{19\pi^7}{56700}-2\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k}-1)} \end{aligned}

and similar ones for other s = 4m+3.  On a hunch, and using Mathematica’s LatticeReduce function, I found that,

\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}\\    \frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(e^{\pi k\sqrt{2}}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(e^{2\pi k\sqrt{2}}-1)}\\    \frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k\sqrt{2}}-1)} \end{aligned}

etc.

II. Functions

If we define,

\begin{aligned} &U_a(s) = \sum_{k=1}^\infty \frac{1}{k^s(e^{a\pi k}-1)}\end{aligned}

then Plouffe discovered integer relations between,

\zeta(s), \pi^s, U_1(s),\, U_2(s),\, U_4(s)

for odd s, with s = 3 being,

\begin{aligned}    &\zeta(3) = 28U_1(3)-37U_2(3)+7U_4(3)\\    &\text{and,}\\    &\pi^3 =720U_1(3)-900U_2(3)+180U_4(3)\end{aligned}

Eliminating U_1(s),\, U_4(s) leads to the 3-term equalities in the Introduction.  See Chamberland’s and Lopatto’s Formulas for Odd Zeta Values.  On the other hand, by defining the function,

\begin{aligned} &V_b(s) = \sum_{k=1}^\infty \frac{1}{k^s(e^{b\pi k\sqrt{2}}-1)}\end{aligned}

I observed integer relations between,

\zeta(s), \pi^s \sqrt{2},\, V_1(s),\, V_2(s),\, V_3(s),\, V_6(s)

also for odd s, with s = 3 as,

\begin{aligned}    &\zeta(3)-102V_1(3)+99V_2(3)+10V_3(3)-5V_6(3) = 0\\    &\text{and,}\\    &\pi^3\,\sqrt{2}-3720V_1(3)+3540V_2(3)+360V_3(3)-180V_6(3) = 0\end{aligned}

and so on.  Eliminating V_3(s),\, V_6(s) leads to the 4-term equalities in the Introduction.

III. Conjecture

The 4-term equalities have coefficients that are simple except for one term. Recall that,

\begin{aligned}    V_1(s) &= \sum_{k=1}^\infty \frac{1}{k^s(e^{\pi k \sqrt{2}}-1)}\\    V_2(s) &= \sum_{k=1}^\infty \frac{1}{k^s(e^{2\pi k \sqrt{2}}-1)}\end{aligned}

Conjecture:

“Using the positive case of \pm 1 for s = 4m+3, and the negative for s = 4m+5, then in the equation,

\begin{aligned}&F(s)\zeta(s-1)\pi\sqrt{2} = \big(2^{(s-1)/2}\pm 1\big)\frac{\zeta(s)}{2}+2^{(s-1)/2}\,V_1(s) \pm V_2(s) \end{aligned}

F(s) is a rational number.”

The first few for s = {3, 7, 11,…} are F(s) = \frac{1}{4}, \frac{41}{40}, \frac{319}{84}, \dots while for s = {5, 9, 13,…} are F(s) = \frac{1}{3}, \frac{19}{11}, \frac{5017}{691},\dots  These rationals may have a closed-form expression in terms of Bernoulli numbers, but I do not yet know the exact formulation.

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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?

Apery-like formulas for zeta(2n)

It is well-known that,

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

D. Bailey, J. Borwein, D. Bradley gave a generalization. First define,

\begin{aligned}&A(a_0) = \sum_{k=1}^\infty \frac{1}{k^{a_0}\,\binom{2k}k}\\    &A(a_0, a_1, a_2,\dots) = \sum_{k=1}^\infty \frac{1}{k^{a_0}\,\binom{2k}k} \sum_{p=1}^{k-1}\frac{1}{p^{a_1}} \sum_{q=1}^{k-1} \frac{1}{q^{a_2}}\dots\end{aligned}

Obviously,

\zeta(2) = 3A(2)

However, a little experiment with Mathematica’s LatticeReduce command will show there are two solutions for \zeta(4),

\begin{aligned}    &a\big(\zeta(4)-3A(4)+9A(2,2)\big) =0\\    &b\big(5\zeta(4)-10A(4)-6A(2,2)\big) = 0\end{aligned}

where {a, b} are scaling variables.  Adding the two together,

(a-5b)\zeta(4)-(3a-10b)A(4)+3(3a+2b)A(2,2) = 0

hence there are an infinite number of solutions.  For appropriately chosen {a,b}, we can also eliminate one term. Thus,

\begin{aligned}    \zeta(4) &= \frac{36}{17}A(4)\\    &=\frac{108}{5}A(2,2)\end{aligned}

For \zeta(6), there are now three solutions. Given the five terms,

\zeta(6),\, A(6),\, A(4,2),\, A(2,4),\, A(2,2,2)

then the coefficients such that their sum is equal to zero are,

\begin{aligned}    &\text{1st sol:}\; (5, -9, 1, -15, 3)\\    &\text{2nd sol:}\; (2, -7, 23, 9, 15)\\    &\text{3rd sol:}\;\, (10,-17,-3,6,-36)\end{aligned}

Using the same approach above, we can eliminate two of the terms.  One solution has an interesting number pop up,

163\zeta(6) = 288A(6)+432A(2,4)

though the appearance of 163 is probably only a coincidence.   See Bailey, Borwein, Bradley’s paper for more details.