Three cubic trinomial series for $1/\pi$

Introduction

We prove the three cubic trinomial identities for $1/\pi$ in Sun’s Conjecture VI [7 Section 5]. The main step is a pair of one-parameter transformations for a common two-variable period, together with identities for its transverse derivatives. These formulas connect the three series to classical elliptic periods and modular equations.

For $b,c \in\mathbb{C}$ and an integer $n \ge0$, define

Sun recorded the discovery of the three formulas on 1 and 2 October 2011 [9 Section 4] and published them together as Conjecture VI in [7]. He still described all three type-VI series as open in 2020 [8 p. 1275]. Our main result proves the complete conjecture.

Ramanujan’s 1914 paper [5] placed series for $1/\pi$ in the setting of elliptic integrals and modular equations. The subsequent theory includes transformations between different hypergeometric descriptions of elliptic periods; the work of Berndt, Bhargava, and Garvan [1] develops the alternative bases used below. Sun’s formulas bring generalized central trinomial coefficients into this setting. Their relation to Legendre polynomials is

Here $P_n$ is the Legendre polynomial normalized by $P_n(1)=1$. For $b^2\ne4c$, the expression is independent of the square root; after expansion it is a polynomial, which also gives its value when $b^2=4c$. Thus the problem belongs to the study of generating functions of powers of Legendre polynomials.

Zudilin [11] obtained a transformation for a binomially weighted generating function of Legendre squares and used it to prove a family of Sun’s series. Van Hoeij, van Straten, and Zudilin [4 Section 6] later gave a Hadamard-product representation for the generating function of Legendre cubes and studied its fourth-order differential equation. That work provides a general framework for the cubic problem. The reductions here concern two particular rational curves: on each, the relevant period satisfies a third-order equation that is a symmetric square of a second-order equation. For the second curve, the resulting equation is related to the hypergeometric transformation for Domb numbers in Rogers [6 Theorem 3.1, equation (3.4)].

There is a further derivative to determine. If $F(w)=\sum a_nw^n$, then a weight $An+B$ corresponds to applying $Aw\,d/dw+B$ to $F$. In our two-variable generating function, this derivative holds the trinomial parameter fixed, whereas restricting to a curve gives a tangent derivative. Differentiation and Legendre’s relation have already been used systematically to obtain series for $1/\pi$ from products of elliptic periods; see Wan [10 Section 2]. Here we recover the missing direction through a first-order identity. Its constant is fixed analytically, by an endpoint estimate on the first curve and Bauer’s identity on the second. The resulting formulas apply along the entire curves; the three series follow by evaluation at the required algebraic points.

The proof has three parts. Section 2 gives the double-integral representation. Sections 3–6 establish the first period transformation, its transverse derivative, and the first and third series. Section 7 proves the second series using a signature-three transformation and an explicit continuation path. Appendices A and B contain the rational differential certificates and the polynomial sign checks for that path. All square roots and continuation branches used in the proof are specified.

An integral and a rational curve

Use the normalization

Since

uniform convergence of the geometric series gives

whenever $|w|(|h|+1)^3<1$. The integral also defines analytic continuation on real parameter intervals where its denominator remains positive.

Define

For real $s,t<1$, all square roots in this formula are positive. Expanding the last two factors and evaluating beta integrals proves

Here $(a)_n=a(a+1)\cdots(a+n-1)$ and $(a)_0=1$. The substitution $(u,v)\mapsto(1-u,1-v)$ gives $\Psi(s,t)=\Psi(t,s)$.

Proof. Write $d\nu(X)=dX/(\pi\sqrt{1-X^2})$ on $(-1,1)$. For $|r|<1$, the increasing fractional linear map

satisfies

In the three cosine variables of (6), use respectively $M_a(X)$, $M_a(Y)$, $M_b(Z)$. The definitions of $h,w$ give the polynomial identity

The Jacobian factors therefore cancel the three linear factors from the transformed denominator, leaving

Put $A=1-ab+a(X+Y)$ and $B=b+aXY$. After $X=1-2u$, $Y=1-2v$, direct factorization gives

Both factors are positive under the stated hypotheses, so $A>|B|$. Thus the integral in $Z$ equals $(A^2-B^2)^{-1/2}$. Moreover $d\nu(X)$ and $d\nu(Y)$ become the beta measures $du/(\pi\sqrt{u(1-u)})$ and $dv/(\pi\sqrt{v(1-v)})$. The constant extracted from $\sqrt{A^2-B^2}$ is $\sqrt{(1-a^2)(1-b^2)}$, which proves precisely the prefactor in (9). This argument also proves positivity of the original denominator, since all the linear factors in the substitution are positive. For (10), substitution gives $s=2a/(3a+1)<0$ and $t=(a+1)/(1-a)\in(1/2,1)$. Also $|a|,|b|<1$ there. Hence the same direct integral argument proves (11) on the entire stated interval. □

Set

The curve in the $(s,t)$-plane is $s+t-2st=1$, parametrized by $s=(1-U)/2$ and $t=(1+U^{-1})/2$.

The elliptic transformation

Write

Thus $f(x)=2K(x)/\pi$, where $K(x)=\int_0^{\pi/2}(1-x\sin^2v)^{-1/2}\,\mathrm{d}v$ uses the parameter convention. Our hypergeometric normalization is

initially in its disk of convergence, and continued analytically where stated. We use Clausen’s identity, Pfaff’s transformation, and Legendre’s relation in the forms recorded in [2 16.12.2, 15.8.1, 19.7.1]; the quadratic forms used below also follow by substitution in the Gauss equation.

Proof. The equation follows by exact elimination from the Legendre recurrence. The derivation and explicit certificate are given in Appendix A. All identities used in that elimination are over a rational-function field. □

Proof of Theorem 3.1. Let $f_0=f(1/2)$ and $f_1=f'(1/2)$. The Gauss equation and Legendre’s relation give

Put $J(t)={}_3F_2(1/2,1/2,1/2;2,2;t)$. Coefficient comparison in the defining series gives the contiguous identity

The apparent singularity at $t=0$ is removable. The quadratic Clausen identity is

It follows from (15) that, with $r=\sqrt{1-t}$,

Substitution in (16) cancels the terms in $r^{-1}$ and gives

From (8), uniformly for $0\leq t\leq1$ and small real $s$,

For the uniform bound, the coefficient in (8) is at most the product of the corresponding coefficients of $H(s)H(t)$, and $H(1)$ is finite. Moreover $J'(1)$ is finite by absolute convergence of its differentiated series, so $J(t)=J(1)+O(1-t)$. Consequently, as $\delta\downarrow0$,

To identify a local basis, put

The Gauss equation, followed by the change of variable $4m(1-m)=U^{-2}$, shows that both functions satisfy

Its symmetric square is exactly (14), after division by its leading coefficient. Hence $u^2,uv,v^2$ solve that equation. As $\delta=U-1\downarrow0$, the three combinations

have leading terms

respectively, all with nonzero coefficients. They therefore form a local basis. The right side of (13) is $(3u^2+v^2)/4$ and has the three coefficients in (19). Comparison in this basis proves the identity near $U=1$. Since the leading coefficient of (14) does not vanish on $1<U<\infty$, the identity continues along that interval. □

The transverse derivative

The weight $An+B$ requires a derivative in $w$ with $h$ fixed. The period transformation gives instead a derivative along the curve $a\mapsto(h(a),w(a))$. The next identity supplies the additional derivative needed to pass between these two directions.

On $-1/3<a<0$ define

In particular $R(a)>0$.

The elementary endpoint estimate in the single cosine integral is

uniformly for $h$ in a sufficiently small neighborhood of 1. The analogous differentiated estimate follows by inserting $n(h+\cos v)^{n-1}$ in the same integral. Expanding $\cos v=1-v^2/2+O(v^4)$ near zero and bounding the remaining interval geometrically proves both estimates. Cubing and summing gives

Here the error in the differentiated sum is bounded by a convergent multiple of $\sum n^{-3/2}$, and $\sum_{n\geq1}q^n n^{-1/2}\sim\sqrt{\pi}/\sqrt{1-q}$ follows by a rescaled Riemann sum. Since $h'/h\to4$ and $w'/w\to-4$, we have $g'\sim2N$. Finally,

so $E/R\to3\sqrt{2}/(8\pi)$. This proves (20) on a neighborhood of zero and hence throughout the interval by its first-order equation and analytic continuation of the positive integral. □

Classical modular evaluations

For $0<\lambda<1$, put

The quadratic transformation and Clausen’s identity give

The modular calculation will be used for $\lambda=1/2$ and $\lambda=1/3$; its midpoint identity also treats $\lambda=1/4$.

Proof. Both $F_\lambda(u)$ and $F_\lambda(1-u)$ solve the Gauss equation

Abel’s identity makes their Wronskian a constant multiple of $1/(u(1-u))$. Euler’s integral, and the same integral differentiated in $u$, give

Since $F_\lambda(u)=1+O(u)$, the Wronskian constant is $-\sigma_\lambda$. This proves (22); its midpoint specialization gives (23). It also gives

Together with $R_\lambda(1/2)=1$ and $R_\lambda(u)\to\infty$ as $u\downarrow0$, this proves existence and uniqueness of $m$.

At the complementary pair, $F_\lambda(1-m)=\sqrt{N}F_\lambda(m)$, so (25) gives $\beta'(m)=1$. Set

(22) and logarithmic differentiation of the defining equation for $\beta$ give, respectively,

Finally, differentiation of (21) gives

Eliminating $A,B$ proves (24).

For $N=3$ or $7$, put $\gamma=1/(N+1)$. The two modular equations needed are

Here $\alpha,\beta\in(0,1)$ satisfy $R_{1/2}(\beta)=NR_{1/2}(\alpha)$, and all roots are positive. For $N=3$ this is [3 (1.2), p. 31]; for $N=7$ it is [3 Example 2, p. 39]. Their convention for degree is stated in their equation (1.1).

Evaluation of the first and third series

Proof of (2) and (4). On the curve put

Theorem 3.1 says $g=W(3f(m)^2+f(1-m)^2)$. At the two points needed here, direct substitution gives the following exact data:

Let $V=f(m)^2$ at the relevant point. Differentiation, using Lemma 5.2 and the complementary derivative identities in the proof of Lemma 5.1, yields

The derivatives are taken in the period identity before specializing the parameter. For any constants $A,B$, Proposition 4.1 gives

At $a=-1/7$, insert $(A,B)=(66,17)$. The coefficient of $V$ is zero, and the right side is $540\sqrt{2}/(11\pi)$. At $a=-7/25$, insert $(A,B)=(3990,1147)$. Again the coefficient of $V$ is zero, and the result is

To identify the original series, use the scaling identity $T_n(b,c)=(2\sqrt{c})^nT_n(b/(2\sqrt{c}),1/4)$. At the first point, $h=5/11$ and $w=1331/6912=2^2 3^3/55296$. At the second point, $h=-31/95$ and $w=857375/2985984$; the parity identity $T_n(-h,1/4)=(-1)^nT_n(h,1/4)$ supplies the negative denominator in (4). Finally, $|T_n(b,c)|\leq(|b|+2\sqrt{c})^n$ proves absolute convergence: the respective geometric bounds are $16/27$ and $343/512$. It also justifies termwise application of the linear weight. □

The second identity

The second target lies on $s+t=-1$, rather than on the curve of Theorem 3.1. We determine its period and its transverse derivative separately. Write

Thus $f_3=F_{1/3}$ and $H_3=H_{1/3}$ in the notation of Section 5. The branch of each hypergeometric function is the one continued from its power series at zero to the plane cut along $[1,\infty)$.

Proof. The rational certificate in Appendix A shows that $P(s)=\Psi(s,-1-s)$ satisfies the symmetric square of

With $x=s(s+1)$, this equation becomes

Substitute $x=1-4z-\frac{1}{16z}$ and $Y=z^{1/4}V$. Direct differentiation gives

Its symmetric square is the Domb equation:

Clausen’s identity gives $H_3(v)={}_2F_1(1/6,1/3;1;v)^2$. Substitution of $v=108z^2/(1-4z)^3$ into the Gauss equation directly verifies that $V=(1-4z)^{-1/2}{}_2F_1(1/6,1/3;1;v)$ satisfies (32). In particular $D=(1-4z)^{-1}H_3(v)$ satisfies its symmetric square. Hence a local square root $q$ of $A$ satisfies (31). Its quadratic products with its conjugate satisfy the same third-order equation as $P$. The hypergeometric pullback above is the one in Rogers’ generating function for Domb numbers [6 Theorem 3.1, equation (3.4)].

It remains to determine the linear combination, including its branches. The coefficient expansion (8), analytically continued in $t$ to a neighborhood of $-1$ using the positive beta integral, and the contiguous identity (16) give

Put

Clausen’s identity and the differential equation for $f$ therefore yield

There is no term in $s^{1/2}$.

We use the signature-two/signature-three transformation

which holds near $p=0$ with the square root equal to one; see [1 Theorem 5.6, equation (5.17)]. At $p_0=(\sqrt{2}-1+\mathrm{i})/2$ its two arguments are

and its squared multiplier is $1/\sqrt{2}+\mathrm{i}/2$. The path verification in Lemma B.1 identifies this continuation with the required upper boundary value. By the quadratic form of Clausen’s identity, $H_3(4u(1-u))=f_3(u)^2$, so

Write $q=q_0+q_1s^{1/2}+q_2s+O(s^{3/2})$, choosing $q^2=A$. Differentiation of (34) gives

For an explicit check, the arguments in this differentiation satisfy

The coefficient of $s^{-1}$ in (31) gives $q_2=-q_0/16$. Using $|\kappa|=\sqrt{6}/4$, substitution into

gives exactly the coefficients in (33), including the zero coefficient of $s^{1/2}$. The second-order equation has exponents $0,1/2$ at zero and two convergent Frobenius series without logarithms. Their quadratic products have exponents $0,1/2,1$, so these three coefficients uniquely determine a solution of the symmetric square. This proves (30) near zero.

For completeness, there is no hidden cut or zero in continuing the formula through $0<s<1$. Write $8z=e^{\mathrm{i}\phi}$, $0<\phi<\pi$. The sign of $\operatorname{Im}v$ is the sign of $\sin\phi(16\cos\phi-11)$. Its only zero inside the interval is at $x=5/16$, where $v=-4$. Thus $v$ never meets $[1,\infty)$. Euler’s integral for $f$ gives $0<f_1/f_0<1/2$, and consequently

Thus $q$ and $\overline{q}$ are independent. Their Wronskian never vanishes on this interval, so $q$ cannot vanish there. In particular $|A|=q\overline{q}$ is analytic on the interval. Uniqueness for the ordinary differential equation completes the continuation. □

Classical values needed for the second identity

Equation (21) with $\lambda=1/3$ gives

and the same identity holds by analytic continuation along negative real $m$. We first record a boundary value for the function $H$ defined above.

Proof. We give the modular differentiation argument to specify the needed classical input. Set $R(m)=f_3(1-m)/f_3(m)$. The degree-five modular equation in signature three is

when $R(\beta)=5R(\alpha)$; see [3 (1.5)]. The unique $m\in(0,1/2)$ with $R(m)=\sqrt{5}$ forms a complementary degree-five pair with $1-m$. Substitution in (39) gives

Consider the inverse local branch $\beta(\alpha)$ through $(m,1-m)$, for which $R(\beta)=R(\alpha)/5$. The modular equation is symmetric, so it holds on this branch as well. Twice differentiating it gives, at this pair,

For an explicit check, its partial derivative in $\beta$ is $-5cp^{-2/3}/6$, while its second derivative in direction $(1,1)$ is $p^{-5/3}(-13/9+10p/3)$. Apply Lemma 5.1 with $\lambda=1/3$ and $N=5$. Its formula gives

By (40), the coefficient in parentheses is $(2/15-2p)/(1-4p)=4/33$. The right side is $5\sqrt{3}/(22\pi)$, which proves (38).

:::

Proof. The algebraic transformation of Theorem 6.1 of [1] is

Although stated there for $0\leq v<1$, it extends from a neighborhood of zero along $v_0\leq v\leq0$, where $v_0=2/\sqrt{5}-1$. On this interval $\alpha(v)\leq0$, $0\leq\beta(v)<1/2$, and $1+v>0$; thus no hypergeometric branch cut is crossed. With $A(v)=4\alpha(v)(1-\alpha(v))$ and $B(v)=4\beta(v)(1-\beta(v))$, squaring and using (37) gives

At $v=v_0$, exact differentiation gives

These quantities give (42) and (43) by the chain rule. Consequently,

and Lemma 7.3 proves the result.

The transverse derivative and the weighted evaluation

On $0 \le a < 1/3$, define

Applying Lemma 2.1 with $b=(3a-1)/(a+1)$ gives, for $0<a<1/3$,

Proof. Set $E=(\theta_wG)(h_2,w_2)+U_0g_2+U_1g_2'$. The normal row identity (50) gives $E'=(R_2'/R_2)E$, since

Thus $E/R_2$ is constant. At $a=0$, one has $h_2=1$, $w_2=-1/8$, $U_0=1/4$, $U_1=0$, and $R_2=1$. The triple integral is analytic near this point, because its denominator is $1+\prod_j(1+\cos v_j)/8\ge1$ there. In particular $g_2'$ stays bounded. Since $T_n(1,1/4)=2^{-n}\binom{2n}{n}$, Bauer’s identity (Lemma 7.2) gives

This proves (45). □

We can now prove the second formula in Conjecture VI. At $a=1/9$,

Let $S_2$ denote the left side of (3). Then $S_2=126\theta_wG+31g_2$. At this point

Therefore (45) and (44) yield

At $s=1/4$ the quantities in Theorem 7.1 satisfy

Differentiation of their defining rational functions gives

Since $H_3(-4)$ is real and positive, differentiating (30) now gives, with $\theta_v=v\,d/dv$,

Substitution into (46) yields

The classical value in Lemma 7.4 is $(4+15\vartheta)H_3(-4)=3\sqrt{3}/\pi$. Consequently

The coefficient bound $T_n(22,441)\leq64^n$ gives geometric ratio $64/125<1$ after cubing and dividing by $512000^n$. This proves absolute convergence, including the linear weight, and completes the second identity.

Exact differential certificates

This appendix gives finite certificates for Lemma 3.2, Proposition 4.1, and the corresponding second-curve equations. All operations are differentiation and arithmetic of rational functions.

Put $\theta=z\partial_z$ and

The value assigned to $P_{-1}$ does not affect $W$; take $P_{-1}=1$. For compactness define

Multiplication by a coefficient in these expressions is always on the left. The Legendre recurrence gives the exact identities

For completeness, put $V_j=\sum_{n\geq0}P_n(y)^{3-j}P_{n-1}(y)^jz^n$. Raising $(n+1)P_{n+1}=(2n+1)yP_n-nP_{n-1}$ to powers one, two, and three, respectively, gives

Since $V_3=1+zF$, elimination proves the first two equations in (47). The third follows from $(y^2-1)P'_n=n(yP_n-P_{n-1})$.

Here is a short specification of every coefficient of the resulting four-dimensional connection. Define the operator

where

Thus $W=\mathcal{R}F$. Write $C=\sum_{j=0}^{2}c_j\theta^j$, set $r_{-1}=c_3=0$, and put

Let $M$ have its first three rows $e_1,e_2,e_3$ and its last row $(m_0,m_1,m_2,m_3)$, where $e_j$ is the row vector with its $j$th coordinate equal to one, indexed from zero. Set

and let $N$ have rows $N_0,\ldots,N_3$. For $V=(F,\theta F,\theta^2F,\theta^3F)^{t}$, equations (47) give

The differential identities are understood after clearing denominators at apparent poles; these poles need not be singularities of $F$.

We now give the substitution certificate. In either column of the following table, let $g(a)=G(h(a),w(a))$, where $G(h,w)=\sum T_n(h,1/4)^3w^n$.

	I	II
$h$	$\dfrac{a^2-4a-1}{5a^2-1}$	$-\dfrac{a^2+4a-1}{3a^2+1}$
$w$	$-\dfrac{(5a^2-1)^3}{8(a-1)^2(a+1)^3(3a+1)}$	$\dfrac{(3a^2+1)^3}{8(a-1)^3(a+1)^3}$
$b$	$-\dfrac{5a+1}{a+1}$	$\dfrac{3a-1}{a+1}$
$s$	$\dfrac{2a}{3a+1}$	$\dfrac{2a}{1-a}$
$t$	$\dfrac{a+1}{1-a}$	$-1-s$

In column I, $t=(1-s)/(1-2s)$. Put

These substitutions give $g=F(y,z)=k\Psi(s,t)$ and $V'=TV$. The square roots are taken consistently; every entry of $T$ is rational in $a$ after cancellation.

Put $\ell=k'/k$ and define rational rows

Then $d^j\Psi(s,t(s))/ds^j=k^{-1}B_jV$. The first certificate is the rational row identity

with

	$p(s)$	$q(s)$
I	$\dfrac{6s^2-6s+1}{2s(s-1)(2s-1)}$	$\dfrac{1}{4(2s-1)^2}$
II	$\dfrac{(2s+1)(s^2+s-1)}{s(s-1)(s+1)(s+2)}$	$\dfrac{4s^2+4s-1}{16s(s-1)(s+1)(s+2)}$

Consequently each restricted function $\Psi(s,t(s))$ satisfies the symmetric square of $Y''+pY'+qY=0$. For column I, putting $u=1-2s$ gives, equivalently,

For the normal derivative certificate define

The second row identity is

where

Both (49) and (50) are checked by inserting (48), differentiating, and clearing denominators: all four resulting numerators vanish identically. The displayed definitions specify these checks completely and avoid large expanded connection matrices. Since $EV=\theta G+U^{\mathrm{I}}_0g+U^{\mathrm{I}}_1g'$, the second identity proves the asserted first-order equation for the normal expression. Its multiplicative constant is then fixed by the analytic boundary arguments in the text.

The continuation path

Recall that $p_0=(\sqrt{2}-1+\mathrm{i})/2$ and $m_0=(1-\sqrt{2})/2$.

Proof. Denote the two arguments in (34) by $u(r)$ and $m(r)$. The denominators do not vanish: $\operatorname{Re}(1+2p)>0$ and

The following real polynomial checks specify the continuation explicitly. Put

Then $D(r)\le-6+3\sqrt{2}<0$ on $[0,1]$, and

Also $\operatorname{Im}u(r)=27r^2N(r)/(4D(r)^3)$, where

Its four coefficients of degrees five through eight are positive and all other nonzero coefficients are negative. Hence $N(r)\le40-29\sqrt{2}<0$, proving $\operatorname{Im}u(r)>0$. Finally

where

Only its coefficients of degrees seven through ten are negative; therefore $J(r) \ge2113 - 1489\sqrt{2} > 0$. Thus $\operatorname{Re} u(r) < 1/2$ for $r < 1$. It follows that $4u(r)(1-u(r))$ remains in the upper half-plane and approaches $27/2$ from above. These signs prove all asserted branch choices. $\square$

AI assistance

OpenAI ChatGPT/Codex (GPT-6-astra) assisted with symbolic calculations, literature checks. The main ideas were from the author. ChatGPT Astra pro assistance also included the developing of transverse derivative method. The October 2026 revisions used GPT-6 Astra Pro. Responsibility for the mathematical claims, references, and final text rests with the author.