## Archive for the ‘number theory’ Category

### 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.

### 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?

### Sequences 2, Padovan and Perrin numbers

Just like the golden ratio and tribonacci constant, powers of the plastic constant P can also be expressed in terms of sequences associated with it. P is a root of the equation,

$P^3=P+1$

or,

$P = \frac{1}{3}\left(\frac{27+3\sqrt{69}}{2}\right)^{1/3}+\frac{1}{3}\left(\frac{27-3\sqrt{69}}{2}\right)^{1/3}$

Define,

\begin{aligned} a & = \left(\tfrac{27+3\sqrt{69}}{2}\right)^{1/3}\\b&=\left(\tfrac{27-3\sqrt{69}}{2}\right)^{1/3}\end{aligned}

then powers of P  are,

$P^{n} = \frac{1}{9}(a^2+b^2)U_{n+1}+\frac{1}{3}(a+b)U_{n+2}+\frac{1}{3}V_n$

where U and V are the Padovan and Perrin sequences, respectively,

\begin{aligned} U_n &= 1,0,0,1,0,1,1,1,2,2,3,4,5,7,9,12,16\dots\\ V_n &=3,0,2,3,2,5,5,7,10,12,17,22,29,\dots\end{aligned}

$P = \frac{0}{9}(a^2+b^2)+\frac{1}{3}(a+b)+\frac{0}{3}$

$P^2 =\frac{1}{9}(a^2+b^2)+\frac{0}{3}(a+b)+\frac{2}{3}$

$P^3 =\frac{0}{9}(a^2+b^2)+\frac{1}{3}(a+b)+\frac{3}{3}$

and so on.  These sequences obey,

$W_n = W_{n-2} + W_{n-3}$

and their limiting ratio, of course, is P.  While the Fibonacci sequence has a nice representation as a square spiral, the Padovan is a spiral of equilateral triangles,

The Perrin sequence also has a notable feature regarding primality testing.  Let $x_1, x_2, x_3$ be the roots of,

$P^3=P+1$

then, starting with n = 0,

$V_n=x_1^n+x_2^n+x_3^n = 3,0,2,3,2,5,5,7,10,12,17,22,29,\dots$

Indexed in this manner, if n is prime, then n divides $V_n$.  For example $V_{11} = 22$.  However, there are Perrin pseudoprimes, composite numbers that pass this test, with the smallest being n = 521^2.

Lastly, like all the four limiting ratios of this family of recurrences, the plastic constant P  can be expressed in terms of the Dedekind eta function as,

\begin{aligned} P &=\frac{e^{\pi i/24}\,\eta(\tau) }{\sqrt{2}\,\eta(2\tau)}\end{aligned}

where,

$\tau=\frac{1+\sqrt{-23}}{2}$

### Sequences 1, Tribonacci numbers

In this 3-part series of posts, we’ll discuss well-known sequences with the recurrence,

$aP_{n-3} + bP_{n-2} + cP_{n-1} = P_n$

where {a, b, ccan only be zero or unity.  Aside from the Fibonacci and Lucas numbers which is a = 0, there is the Narayana sequence with b = 0, the Padovan and Perrin with c = 0, and the tribonacci has a = b = c = 1.  All four cases may then share similar properties and one of which, interestingly enough, is that their limiting ratios, a root of the following equations,

\begin{aligned} x^2 &=x+1\\ y^3 &= y^2+1\\ z^3 &= z+1\\ t^3 &= t^2+t+1\end{aligned}

can also be used to express $\zeta(3)$,  or Apery’s constant.

I. Fibonacci and Lucas numbers

Given the two roots of,

$x^2=x+1$

with $x_1 > x_2$, the larger root being the golden ratio, we get the Lucas numbers L(n) and Fibonacci numbers F(n),

\begin{aligned} L_n &= x_1^n+x_2^n = 2,1,3,4,7,11,18,29,\dots\\[2mm] F_n &= \frac{x_1^n-x_2^n}{\sqrt{5}} = 0,1,1,2,\,3,\,5,\,8,\,13,\dots\end{aligned}

(The starting index is n = 0.)  Expanding powers of the golden ratio, then for n > 0,

\begin{aligned} & {x_1}^n = \Big(\frac{1+\sqrt{5}}{2}\Big)^n = \frac{L_n+F_n\sqrt{5}}{2}\end{aligned}

We’ll see this can be generalized to powers of the tribonacci constant.

II. Tribonacci numbers

These are a generalization of the Fibonacci numbers, being,

$t_n = t_{n-1}+t_{n-2}+t_{n-3}$

Pin-Yen Lin has a nice paper involving these numbers.  First, define the following three sequences with this recurrence, but with different initial values,

\begin{aligned}S_n &=0,0,1,1,2,4,7,13,24,\dots\\U_n &=0,3,2,5,10,17,32,49,\dots\\V_n &=3,1,3,7,11,21,39,71,\dots \end{aligned}

(The starting index as usual is n = 0.)  The first and the third are recognized by the OEIS, with the first being the tribonacci numbers.  The limiting ratio for all three is the tribonacci constant, T, the real root of,

$x^3=x^2+x+1$

or,

$T = \frac{1}{3}+\frac{1}{3}(19+3\sqrt{33})^{1/3}+\frac{1}{3}(19-3\sqrt{33})^{1/3}$

I’ve already written about the tribonacci constant before.  But I want to include how Lin found that powers of x can be expressed in terms of those three sequences. Define,

$a =\sqrt[3]{19+3\sqrt{33}}$

$b =\sqrt[3]{19-3\sqrt{33}}$

then, similar to the golden ratio,

$T^n = \frac{1}{9}(a^2+b^2)\,S_n+\frac{1}{9}(a+b)\,U_n+\frac{1}{3}\,V_n$

Hence, starting with = 1,

$T = \frac{0}{9}(a^2+b^2)+\frac{3}{9}(a+b)+\frac{1}{3}$

$T^2 = \frac{1}{9}(a^2+b^2)+\frac{2}{9}(a+b)+\frac{3}{3}$

$T^3 = \frac{1}{9}(a^2+b^2)+\frac{5}{9}(a+b)+\frac{7}{3}$

and so on.  Interesting, isn’t it, that powers of the tribonacci constant can be expressed in this manner.

There is a primality test regarding Lucas numbers: if n is a prime then $L_n-1$ is divisible by n.  For example $L_5 = 11$, minus 1, is divisible by 5.  However there are Lucas pseudoprimes, composite numbers that pass this test, with the smallest being n = 705.

The third tribonacci sequence can be formed analogously to the Lucas numbers.  Given the three roots $x_1, x_2, x_3$ of,

$x^3=x^2+x+1$

then, starting with n = 0,

$V_n = x_1^n+x_2^n+x_3^n = 3,1,3,7,11,21,39,71,\dots$

I notice that likewise, if n is prime, then $V_n-1$ is divisible by n.  But there are also tribonacci-like pseudoprimes.  The smallest is n = 182.  Steven Stadnicki was nice enough to compute the first 36.  It turns out they are relatively rarer, as there are only 21 less than $10^8$, while there are  852 Lucas pseudoprimes in the same range.

### Even powers of Fibonacci numbers

In a previous post, it was pointed out that powers of Fibonacci numbers also obey recurrence relations.  For example, it is the case that,

$F_n^2-2F_{n+1}^2-2F_{n+2}^2+F_{n+3}^2 = 0$

In general, for even kth powers, it takes k+2 consecutive Fibonacci numbers to sum up to zero.  However, using Mathematica’s LatticeReduce function which has an integer relations algorithm, I found that if reduced to k+1 terms, then it can still sum up to a constant, though it is now non-zero.  Thus,

\begin{aligned} &F_n^2-3F_{n+1}^2+F_{n+2}^2 = (-1)^{n+1}\,2\\[1.5mm] &F_n^4-4F_{n+1}^4-19F_{n+2}^4-4F_{n+3}^4+F_{n+4}^4 = 6\\[1.5mm] &F_n^6-14F_{n+1}^6-90F_{n+2}^6+350F_{n+3}^6-90F_{n+4}^6-14F_{n+5}^6+F_{n+6}^6 = (-1)^n\, 80\\[1.5mm] &F_n^8-33F_{n+1}^8-747F_{n+2}^8+3894F_{n+3}^8+16270F_{n+4}^8+3894F_{n+5}^8-747F_{n+6}^8\\&\;\;-33F_{n+7}^8+F_{n+8}^8 = 2520\end{aligned}

and so on, with k = 10 summing to $(-1)^{n+1}\,226800$.  Notice the formulas are palindromic, the same read forwards or backwards.

I was curious if this sequence of constants,

$C(2p) = 2, 6, 80, 2520, 226800, \dots$

had a generating function. Unfortunately, OEIS didn’t recognize it, so that question is unanswered for now.

Update, May 26, 2012:  Jim Cullen found a recurrence relation which is equivalent to the formula,

\begin{aligned}C(2p)&=\prod_{n=1}^p \frac{2(2n-1)(F(n))^2}{n}=\frac{(2p)!}{p!^2}\prod_{n=1}^p (F(n))^2\\&=2,6,80, 2520, 226800,53222400,\dots\end{aligned}

hence the next constant is(12) = 53222400.  The product of the first p Fibonacci numbers (n) is called a fibonorial.

### Odd powers of Fibonacci numbers

The Fibonacci numbers $F_n$,

$F_n = 0, 1, 1, 2, 3, 5, 8, 13, \dots$

obey the following recurrence relations,

\begin{aligned}&F_n-F_{n-1}-F_{n-2} = 0\\[1.5mm]&F_n^2-2F_{n-1}^2-2F_{n-2}^2+F_{n-3}^2 = 0\\[1.5mm]&F_n^3-3F_{n-1}^3-6F_{n-2}^3+3F_{n-3}^3+F_{n-4}^3 = 0\\[1.5mm]&F_n^4-5F_{n-1}^4-15F_{n-2}^4+15F_{n-3}^4+5F_{n-4}^4-F_{n-5}^4 = 0\\[1.5mm]&F_n^5-8F_{n-1}^5-40F_{n-2}^5+60F_{n-3}^5+40F_{n-4}^5-8F_{n-5}^5-F_{n-6}^5 = 0\end{aligned}

and so on.  As a number triangle, the coefficients are,

\begin{aligned} &1;\; \bold{1, -1,-1}=0\\ &2;\; 1, -2, -2, \;1=0\\ &3;\; \bold{1, -3, -6, \;3, \;1}=0\\ &4;\; 1, -5, -15, 15, \;5, \;-1=0\\ &5;\; \bold{1, -8, -40, 60, 40, -8, -1}=0 \end{aligned}

See Ron Knott’s article on the fibonomials, so-called since the above is reminiscent of the binomial triangle.  However, I found another set of recurrence relations can be given as,

\begin{aligned} &F_{n-1}^2-F_{n+1}^2 = -F_{2n}\\[1.5mm] &F_{n-1}^3-F_{n}^3-F_{n+1}^3 = -F_{3n}\\[1.5mm] &F_{n-2}^4+3F_{n-1}^4-3F_{n+1}^4-F_{n+2}^4 = -6F_{4n}\\[1.5mm] &F_{n-2}^5-3F_{n-1}^5-6F_{n}^5+3F_{n+1}^5+F_{n+2}^5 = 6F_{5n}\\[1.5mm] &F_{n-3}^6+4F_{n-2}^6-20F_{n-1}^6+20F_{n+1}^6-4F_{n+2}^6-F_{n+3}^6 = -120F_{6n}\\[1.5mm] &F_{n-3}^7-8F_{n-2}^7-40F_{n-1}^7+60F_{n}^7+40F_{n+1}^7-8F_{n+2}^7-F_{n+3}^7 = -240F_{7n}\\[1.5mm]\end{aligned}

etc.  As a number triangle,

\begin{aligned} &2;\; 1, -1 =-1\\ &3;\; \bold{1, -1,-1}=-1\\ &4;\; 1, \;\;\;3, -3, \;-1=-6\\ &5;\; \bold{1, -3, -6, \;3, \;\;1}\;=\;6\\ &6;\; 1, \;\;\;4, -20, 20, -4, -1=-120\\ &7;\; \bold{1, -8, -40, 60, 40, -8, -1}=-240 \end{aligned}

Compare the two triangles.  Notice how, for odd powers, the same coefficients appear, though moved up by one odd power.  I have no explanation for the phenomenon, other than the fact that I’ve seen several instances already of a “recycled” polynomial appearing in many contexts.

### Binomial sums and the zeta function

Recall the three sequences,

$C_n = \sum_{k=0}^n {\binom n k} {\binom {n+k}k} = 1, 3, 13, 63, 321, 1683\dots$

$B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251, 11253\dots$

$A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, 819005\dots$

Equivalently,

$C_n = \,_2F_1(-n,n+1;\,1;\,-1)$

$B_n = \,_3F_2(-n,-n,n+1;\,1,1;\,1)$

$A_n = \,_4F_3(-n,-n,n+1,n+1;\,1,1,1;\,1)$

where $_pF_q$ is the generalized hypergeometric function.  Then it is known that,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\, C_n C_{n+1}} = \frac{1}{2}\log(2)\\ &\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2 B_n B_{n+1}} = \frac{1}{5}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3 A_n A_{n+1}} = \frac{1}{6}\, \zeta(3)\end{aligned}

Beautiful, aren’t they? Since the numbers increase at a near-geometric rate (for example, $\frac{A_{n+1}}{A_n} \to (1+\sqrt{2}\,)^4 \approx 33.97$  as n  goes to infinity), then the convergence is very fast.

We also have the nice evaluations,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} = -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right) \\ &\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} = \frac{17}{36}\,\zeta(4)\\ &\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} = -\frac{19}{3}\,\zeta(5) +\frac{2}{3}\,\zeta(2)\zeta(3)+\frac{\pi\sqrt{3}}{2^3\cdot 3^2}\left(\zeta(4, \tfrac{1}{3})-\zeta(4,\tfrac{2}{3}) \right)\\ &\sum_{n=1}^\infty \frac{1}{n^6\,\binom{2n}n} = \;\;?\\ &\sum_{n=1}^\infty \frac{1}{n^7\,\binom{2n}n} = -\frac{493}{24}\zeta(7)+2\zeta(2)\zeta(5)+\frac{17}{18}\zeta(3)\zeta(4)+\frac{11\pi\sqrt{3}}{2^5\cdot 3^4}\left(\zeta(6,\tfrac{1}{3})-\zeta(6,\tfrac{2}{3})\right)\\ \end{aligned}

with the Riemann zeta function $\zeta(s)$ and the more general Hurwitz zeta function $\zeta(s,a)$,

\begin{aligned} &\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}\\&\zeta(s,a) = \sum_{n=0}^\infty \frac{1}{(n+a)^s} \end{aligned}

respectively.  (Note that for $a = 1$, the Hurwitz reduces into the Riemann.)  The expression for p = 5 in the paper here used Dirichlet L-functions, but a poster from mathstackexchange gave it in terms of the Hurwitz zeta.  The one for p = 7 is from Mathworld’s article on central binomial coefficients.

However, none are known for p > 7 (as well as p = 6).  Based on odd p, it is easy to assume that the next has the form,

\begin{aligned} & a_0\sum_{n=1}^\infty \frac{1}{n^9\,\binom {2n}n} = a_1\,\zeta(9) +a_2\,\zeta(2)\zeta(7)+a_3\, \zeta(3)\zeta(6)+a_4\, \zeta(4)\zeta(5)+a_5\,\pi\sqrt{3}\,H_8\\ \end{aligned}

where,

$H_8 = \left(\zeta(8,\tfrac{1}{3})-\zeta(8,\tfrac{2}{3})\right)$

and the six $a_i$ are integers.  One can use Mathematica’s LatticeReduce function (which employs an integer relations algorithm) to find them, if any exists. Unfortunately, it didn’t find any exact relation, nor for analogous forms for prime p = 11 or 13.  Either my old version of Mathematica is just not powerful enough, or odd p > 7 do not have analogous forms to the ones above.

Can you find the next in the family?

### Ramanujan’s continued fraction for Catalan’s constant

Ramanujan was a goldmine when it came to continued fractions (and many others).  In this post, two families will be given: they involve pi and Catalan’s constant as special cases.  However, versions will be given that roughly double the rate of convergence.

Recall that Catalan’s constant C is given by,

$C = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 0.915965\dots$

Ramanujan gave the beautiful pair of continued fractions.  Let $|x| > 1$, then,

$f(x) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(x+2n+1)^2} = \cfrac{1}{x^2-1 + \cfrac{2^2}{1 + \cfrac{2^2}{x^2-1 + \cfrac{4^2}{1 + \cfrac{4^2}{x^2-1 +\ddots}}}}}$

$g(x) \; = \;\frac{\Gamma^4 (\frac{x+1}{4})}{\Gamma^4 (\frac{x+3}{4})}\; =\; \cfrac{16}{x^2-1 + \cfrac{1^2}{1 + \cfrac{1^2}{x^2-1 + \cfrac{3^2}{1 + \cfrac{3^2}{x^2-1 +\ddots}}}}}$

One can just admire Ramanujan’s artistry — one continued fraction uses even numerators, the other, odd numerators.  For even x > 0, it is easily seen the first one involves Catalan’s constant.  For example,

$f(2) = 2(1-C)$

On the other hand, for odd x > 1, both involve pi,

$f(3) = \frac{1}{24}(12-\pi^2)$

$g(3) = \pi^2$

Notice though that numerators are repeated. Thanks to the insight of J.M. from a mathstackexchange post about Apery’s constant, we can speed up the rate of convergence of this particular form by getting rid of every other level and extracting even convergents.  After some slightly tedious algebraic manipulation, given,

$y = \cfrac{a_1}{b_1 + \cfrac{a_2}{1 + \cfrac{a_3}{b_2 + \cfrac{a_4}{1 +\ddots}}}}$

then,

$y_{even} = \cfrac{a_1}{b_1+a_2 - \cfrac{a_2\, a_3}{b_2+a_3+a_4 - \cfrac{a_4\, a_5}{b_3+a_5+a_6 - \cfrac{a_6\, a_7}{b_4+a_7+a_8 -\ddots}}}}$

So Ramanujan’s continued fractions are now the more compact,

$f(x) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(x+2n+1)^2} = \cfrac{1}{u_1 - \cfrac{2^4}{u_2 - \cfrac{4^4}{u_3 - \cfrac{6^4}{u_4 -\ddots}}}}$

where, starting with n = 1,

$u_n = (2n-2)^2 + (2n)^2 + x^2-1$

and,

$g(x) \; = \; \frac{\Gamma^4 (\frac{x+1}{4})}{\Gamma^4(\frac{x+3}{4})}\; =\; \cfrac{16}{-1+v_1 - \cfrac{1^4}{v_2 - \cfrac{3^4}{v_3 - \cfrac{5^4}{v_4 -\ddots}}}}$

$v_n = (2n-3)^2 + (2n-1)^2 + x^2-1$

Thus, we have the slightly faster continued fraction for Catalan’s constant C,

$f(2) = 2(1-C) = \cfrac{1}{u_1 - \cfrac{2^4}{u_2 - \cfrac{4^4}{u_3 - \cfrac{6^4}{u_4 -\ddots}}}}$

where,

$u_n = 8n^2-8n+7$

(There is an even faster one by Zudilin given in An Apery-like difference equation for Catalan’s constant though he states this still does not prove C is irrational.)

### Zudilin’s continued fraction for Zeta(4)

Euler proved the following general continued fraction formula,

$\frac{1}{u_1}+\frac{1}{u_2}+\frac{1}{u_3} = \cfrac{1}{u_1 - \cfrac{u_1^2}{u_1+u_2 - \cfrac{u_2^2}{u_2+u_3 - \cfrac{u_3^2}{u_3+u_4 - \ddots}}}}$

which automatically gives a representation for the Riemann zeta function $\zeta(s)$.  However, the convergence is rather slow.  Apery found a much faster version for $\zeta(3)$ and proved its irrationality.  The status for other odd s, however, remains open.  While it has already been proved irrational for all even s, Wadim Zudilin nonetheless found an interesting faster version for $\zeta(4)$.

First, consider the following known binomial sums,

$\sum_{n=1}^\infty \frac{1}{n\,\binom{2n}n} = \frac{1}{3\sqrt{3}}\,\pi$

$\sum_{n=1}^\infty \frac{1}{n^2\,\binom{2n}n} = \frac{1}{3}\,\zeta(2)$

$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3\,\binom{2n}n} = \frac{2}{5}\,\zeta(3)$

$\sum_{n=1}^\infty \frac{1}{n^4\,\binom{2n}n} = \frac{17}{36}\,\zeta(4)$

This nice pattern stops at s = 4.  (There are sums for other s, but they do not have this succinct form.)  It is suggestive then that $\zeta(4)$ belongs to the same family and may share certain similarities.

Zudilim found that,

$\zeta(4) = \cfrac{13}{p_0 + \cfrac{1^8\, q_1}{p_1+ \cfrac{2^8\,q_2}{p_2+\cfrac{3^8\,q_3}{p_3 + \ddots}}}}$

where, starting with n = 0,

$p_n = 3(2n+1)(3n^2+3n+1)(15n^2+15n+4) = 12, 2142, 26790, 142968,\dots$

$q_n = 3(3n-1)(3n+1) = -3, 24, 105, 240, 429\dots$

(Note that the cfrac does not use $q_0$.)  Like Apery’s accelerated version for $\zeta(2)$, the partial denominators use addition, in contrast to Euler’s form which has subtraction.  Also, like its siblings, it obeys a recurrence relation,

$(n+1)^5U_{n+1}=p_n U_n+n^3q_nU_{n-1}$

Starting with $U_0=1, U_1=12$, one gets the sequence,

$U_n = 1, 12, 804, 88680, \dots$

which is not yet in the OEIS.  In Theorem 4, Zudilin mentions they are positive rationals, but some Mathematica experimentation will show that apparently all the $U_i$ are integral.  (Proof anyone?)  Furthermore, they have the limit,

$\lim_{n \to \infty} \frac{U_{n+1}}{U_n} = (3+2\sqrt{3})^3$

For more details, refer to Zudilin’s An Apery-like difference equation for Catalan’s constant.  Now if only someone will do something more about $\zeta(5)$

### 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 $\arctan(\frac{1}{3})$,

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

or,

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

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,

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

and,

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

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.