Gasper-RahmanによるGosper予想のq類似2
Gosperによる${}_3F_2$和公式
\begin{align}
\F32{\frac 12+3x,\frac 12-3x,y}{\frac 12,3y}{\frac 34}&=\frac{2\Gamma\left(\frac 13+y\right)\Gamma\left(\frac 23+y\right)\cos\pi x}{\sqrt 3\Gamma\left(\frac 12+x+y\right)\Gamma\left(\frac 12-x+y\right)}
\end{align}
の一般化である以下の公式を
前の記事
で示した.
\begin{align}
&\F76{a,a+\frac 12,b,1-b,c,\frac{2a+1}3-c,1+\frac a2}{\frac 12,\frac{2a-b+3}3,\frac{2a+b+2}3,3c,2a+1-3c,\frac a2}{1}\\
&=\frac2{\sqrt 3}\frac{\Gamma\left(c+\frac 13\right)\Gamma\left(c+\frac 23\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(\frac{2a+b+2}3\right)\Gamma\left(\frac{2a+2}3-c\right)\Gamma\left(1+\frac{2a}3-c\right)\sin\frac{\pi(b+1)}3}{\Gamma\left(\frac{2a+2}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(c+\frac{b+1}3\right)\Gamma\left(c+\frac{2-b}3\right)\Gamma\left(\frac{2a+b+2}3-c\right)\Gamma\left(1+\frac{2a-b}3-c\right)}
\end{align}
一方,
前の記事
で示したように, Gosperの${}_3F_2$和公式にはもう一つ
\begin{align}
\F32{1+3x,1-3x,y}{\frac 32,3y-1}{\frac 34}&=\frac{2\Gamma\left(y-\frac 13\right)\Gamma\left(y+\frac 13\right)\sin\pi x}{3\sqrt 3x\Gamma(y+x)\Gamma(y-x)}
\end{align}
というものもあった. Gasper-Rahmanはこれに対しても上のような${}_7F_6$和公式の$q$類似を示している.
今回はそれについて解説したいと思う.
Gasper-Rahmanの和公式
\begin{align}
W(a;b_1,\dots,b_r;q;z):=\Q{r+3}{r+2}{a,\sqrt aq,-\sqrt aq,b_1,\dots,b_r}{\sqrt a,-\sqrt a,aq/b_1,\dots,aq/b_r}{q;z}
\end{align}
とする.
前の記事
で以下の定理を示した.
\begin{align} &\frac{(ac^2q^{3}/b,bd/ac^2;q^3)_{\infty}(dq/ab,ab;q)_{\infty}}{(cdq^{3}/ab^2,ab^2/c;q^3)_{\infty}(acq,d/ac;q)_{\infty}}\\ &\qquad\cdot\sum_{0\leq k}\frac{1-acq^{4k}}{1-ac}\frac{(a,b;q)_k(cq/b;q)_{2k}(d,a^2bc/d;q^3)_k}{(cq^3,acq^3/b;q^3)_k(ab;q)_{2k}(dq/ab,acq/d;q)_k}q^{k}\\ &=W(ac^2/b;c,d,a^2bc/d,ac/b,cq/b,cq^2/b,cq^3/b;q^3;q^3)\\ &\qquad+\frac{(ac^2q^{3}/b,bd/ac^2,d^2q^{3}/a^2bc,a^2bc/d;q^3)_{\infty}(bd/c,cq/b;q)_{\infty}}{(ac^2/bd,cdq^{3}/ab^2,ab^2/c,bd^2q^3/ac^2;q^3)_{\infty}(ac,dq/ac;q)_{\infty}}\\ &\qquad\cdot\frac{(1-c)(1-ac/b)}{(1-d/c)(1-bd/ac)}W(bd^2/ac^2;ab^2/c,d,d/c,bd/ac,dq/ac,dq^2/ac,dq^3/ac;q^3;q^3)\\ &\qquad-\frac{(q^3,d,ac^2q^{3}/b,bd/ac^2,d^2q^{3}/a^2bc,a^2bc/d;q^3)_{\infty}(a,b,cq/b;q)_{\infty}}{(cq^3,acq^3/b,d/c,bd/ac,cdq^{3}/ab^2,ab^2/c;q^3)_{\infty}(ac,dq/ac,ac/d;q)_{\infty}}\\ &\qquad\cdot\Q21{d/c,bd/ac}{d^2q^3/a^2bc}{q^3;q^3} \end{align}
定理1において, $b=q^2/a$とすると,
\begin{align}
&\frac{(a^2c^2q,dq^2/a^2c^2;q^3)_{\infty}(d/q,q^2;q)_{\infty}}{(acd/q,q^4/ac;q^3)_{\infty}(acq,d/ac;q)_{\infty}}\\
&\qquad\cdot\sum_{0\leq k}\frac{1-acq^{4k}}{1-ac}\frac{(a,q^2/a;q)_k(ac/q;q)_{2k}(d,acq^2/d;q^3)_k}{(cq^3,a^2cq;q^3)_k(q^2;q)_{2k}(d/q,acq/d;q)_k}q^{k}\\
&=W(a^2c^2/q^2;c,d,acq^2/d,a^2c/q^2,ac/q;q^3;q^3)\\
&\qquad+\frac{(a^2c^2q,dq^2/a^2c^2,d^2q/ac,acq^2/d;q^3)_{\infty}(dq^2/ac,ac/q;q)_{\infty}}{(a^2c^2/dq^2,acd/q,q^4/ac,d^2q^5/a^2c^2;q^3)_{\infty}(ac,dq/ac;q)_{\infty}}\\
&\qquad\cdot\frac{(1-c)(1-a^2c/q^2)}{(1-d/c)(1-dq^2/a^2c)}W(d^2q^2/a^2c^2;q^4/ac,d,d/c,dq^2/a^2c,dq/ac;q^3;q^3)\\
&\qquad-\frac{(q^3,d,a^2c^2q,dq^2/a^2c^2,d^2q/ac,acq^2/d;q^3)_{\infty}(a,q^2/a,ac/q;q)_{\infty}}{(cq^3,a^2cq,d/c,dq^2/a^2c,acd/q,q^4/ac;q^3)_{\infty}(ac,dq/ac,ac/d;q)_{\infty}}\\
&\qquad\cdot\Q21{d/c,dq^2/a^2c}{d^2q/ac}{q^3;q^3}
\end{align}
となる. ここで,
non-terminating Jacksonの和公式
より,
\begin{align}
&W(a^2c^2/q^2;c,d,acq^2/d,a^2c/q^2,ac/q;q^3;q^3)\\
&\qquad+\frac{(a^2c^2q,dq^2/a^2c^2,d^2q/ac,acq^2/d;q^3)_{\infty}(dq^2/ac,ac/q;q)_{\infty}}{(a^2c^2/dq^2,acd/q,q^4/ac,d^2q^5/a^2c^2;q^3)_{\infty}(ac,dq/ac;q)_{\infty}}\\
&\qquad\cdot\frac{(1-c)(1-a^2c/q^2)}{(1-d/c)(1-dq^2/a^2c)}W(d^2q^2/a^2c^2;q^4/ac,d,d/c,dq^2/a^2c,dq/ac;q^3;q^3)\\
&=\frac{(a^2c^2q,ad/q,q^3,aq^2,dq/a,d,q^4/a,dq^2/a^2c^2;q^3)_{\infty}}{(a^2cq,acd/q,cq^3,acq^2,dq^2/a^2c,q^4/ac,d/c,dq/ac;q^3)_{\infty}}
\end{align}
であるから,
\begin{align}
&\frac{(a^2c^2q,dq^2/a^2c^2;q^3)_{\infty}(d/q,q^2;q)_{\infty}}{(acd/q,q^4/ac;q^3)_{\infty}(acq,d/ac;q)_{\infty}}\\
&\qquad\cdot\sum_{0\leq k}\frac{1-acq^{4k}}{1-ac}\frac{(a,q^2/a;q)_k(ac/q;q)_{2k}(d,acq^2/d;q^3)_k}{(cq^3,a^2cq;q^3)_k(q^2;q)_{2k}(d/q,acq/d;q)_k}q^{k}\\
&=\frac{(a^2c^2q,ad/q,q^3,aq^2,dq/a,d,q^4/a,dq^2/a^2c^2;q^3)_{\infty}}{(a^2cq,acd/q,cq^3,acq^2,dq^2/a^2c,q^4/ac,d/c,dq/ac;q^3)_{\infty}}\\
&\qquad-\frac{(q^3,d,a^2c^2q,dq^2/a^2c^2,d^2q/ac,acq^2/d;q^3)_{\infty}(a,q^2/a,ac/q;q)_{\infty}}{(cq^3,a^2cq,d/c,dq^2/a^2c,acd/q,q^4/ac;q^3)_{\infty}(ac,dq/ac,ac/d;q)_{\infty}}\\
&\qquad\cdot\Q21{d/c,dq^2/a^2c}{d^2q/ac}{q^3;q^3}
\end{align}
つまり,
\begin{align}
&\sum_{0\leq k}\frac{1-acq^{4k}}{1-ac}\frac{(a,q^2/a;q)_k(ac/q;q)_{2k}(d,acq^2/d;q^3)_k}{(cq^3,a^2cq;q^3)_k(q^2;q)_{2k}(d/q,acq/d;q)_k}q^{k}\\
&=\frac{(acd/q,q^4/ac;q^3)_{\infty}(acq,d/ac;q)_{\infty}}{(a^2c^2q,dq^2/a^2c^2;q^3)_{\infty}(d/q,q^2;q)_{\infty}}\\
&\qquad\cdot\frac{(a^2c^2q,ad/q,q^3,aq^2,dq/a,d,q^4/a,dq^2/a^2c^2;q^3)_{\infty}}{(a^2cq,acd/q,cq^3,acq^2,dq^2/a^2c,q^4/ac,d/c,dq/ac;q^3)_{\infty}}\\
&\qquad-\frac{(acd/q,q^4/ac;q^3)_{\infty}(acq,d/ac;q)_{\infty}}{(a^2c^2q,dq^2/a^2c^2;q^3)_{\infty}(d/q,q^2;q)_{\infty}}\\
&\qquad\cdot\frac{(q^3,d,a^2c^2q,dq^2/a^2c^2,d^2q/ac,acq^2/d;q^3)_{\infty}(a,q^2/a,ac/q;q)_{\infty}}{(cq^3,a^2cq,d/c,dq^2/a^2c,acd/q,q^4/ac;q^3)_{\infty}(ac,dq/ac,ac/d;q)_{\infty}}\\
&\qquad\cdot\Q21{d/c,dq^2/a^2c}{d^2q/ac}{q^3;q^3}\\
&=\frac{(acq,acq^3,d/ac,dq^2/ac,ad/q,aq^2,dq/a,q^4/a;q^3)_{\infty}}{(q^2,q^4,d/q,dq,a^2cq,cq^3,dq^2/a^2c,d/c;q^3)_{\infty}}\\
&\qquad+\frac{d}{ac}\frac{(q^3,d,d^2q/ac,acq^2/d;q^3)_{\infty}(a,q^2/a,ac/q,acq;q)_{\infty}}{(cq^3,a^2cq,d/c,dq^2/a^2c;q^3)_{\infty}(d/q,q^2,ac,acq/d;q)_{\infty}}\\
&\qquad\cdot\Q21{d/c,dq^2/a^2c}{d^2q/ac}{q^3;q^3}\\
&=\frac{(acq,acq^3,d/ac,dq^2/ac,ad/q,aq^2,dq/a,q^4/a;q^3)_{\infty}}{(q^2,q^4,d/q,dq,a^2cq,cq^3,dq^2/a^2c,d/c;q^3)_{\infty}}\\
&\qquad+\frac{d}{ac}\frac{(a,aq,aq^2,q^2/a,q^3/a,q^4/a,ac/q,acq,acq^3,d^2q/ac;q^3)_{\infty}}{(q^2,q^4,d/q,dq,acq/d,acq^3/d,cq^3,a^2cq,d/c,dq^2/a^2c;q^3)_{\infty}}\\
&\qquad\cdot\Q21{d/c,dq^2/a^2c}{d^2q/ac}{q^3;q^3}
\end{align}
つまり, 以下を得る.
\begin{align} &\sum_{0\leq k}\frac{1-acq^{4k}}{1-ac}\frac{(a,q^2/a;q)_k(ac/q;q)_{2k}(d,acq^2/d;q^3)_k}{(cq^3,a^2cq;q^3)_k(q^2;q)_{2k}(d/q,acq/d;q)_k}q^{k}\\ &=\frac{(acq,acq^3,d/ac,dq^2/ac,ad/q,dq/a,aq^2,q^4/a;q^3)_{\infty}}{(q^2,q^4,d/q,dq,a^2cq,cq^3,dq^2/a^2c,d/c;q^3)_{\infty}}\\ &\qquad+\frac{d}{ac}\frac{(a,aq,aq^2,q^2/a,q^3/a,q^4/a,ac/q,acq,acq^3,d^2q/ac;q^3)_{\infty}}{(q^2,q^4,d/q,dq,acq/d,acq^3/d,cq^3,a^2cq,d/c,dq^2/a^2c;q^3)_{\infty}}\\ &\qquad\cdot\Q21{d/c,dq^2/a^2c}{d^2q/ac}{q^3;q^3} \end{align}
Gosper予想の類似
定理2において, $a,c,d$を$b,a^2/b,c^3$と置き換えると
\begin{align}
&\sum_{0\leq k}\frac{1-a^2q^{4k}}{1-a^2}\frac{(b,q^2/b;q)_k(a^2/q;q)_{2k}(c^3,a^2q^2/c^3;q^3)_k}{(a^2q^3/b,a^2bq;q^3)_k(q^2;q)_{2k}(c^3/q,a^2q/c^3;q)_k}q^{k}\\
&=\frac{(a^2q,a^2q^3,c^3/a^2,c^3q^2/a^2,bc^3/q,c^3q/b,bq^2,q^4/b;q^3)_{\infty}}{(q^2,q^4,c^3/q,c^3q,a^2bq,a^2q^3/b,c^3q^2/a^2b,bc^3/a^2;q^3)_{\infty}}\\
&\qquad+\frac{c^3}{a^2}\frac{(b,bq,bq^2,q^2/b,q^3/b,q^4/b,a^2/q,a^2q,a^2q^3,c^6q/a^2;q^3)_{\infty}}{(q^2,q^4,c^3/q,c^3q,a^2q/c^3,a^2q^3/c^3,a^2q^3/b,a^2bq,b^2c^3/a,c^3q^2/a^2b;q^3)_{\infty}}\\
&\qquad\cdot\Q21{bc^3/a^2,c^3q^2/a^2b}{c^6q/a^2}{q^3;q^3}
\end{align}
を得る. ここで, $a\mapsto q^a,b\mapsto q^b,c\mapsto q^c$として$q\to 1$として, Gaussの超幾何定理を用いると,
\begin{align}
&\F76{a-\frac 12,a,b,2-b,c,\frac{2a+2}3-c,\frac a2+1}{\frac 32,1+\frac{2a-b}3,\frac{2a+b+1}3,3c-1,2a+1-3c,\frac a2}1\\
&=\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(c+\frac{2-2a-b}3\right)\Gamma\left(c+\frac{b-2a}3\right)}{\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(c-\frac{2a}3\right)\Gamma\left(c+\frac{2-2a}3\right)\Gamma\left(c+\frac{b-1}3\right)\Gamma\left(c+\frac{1-b}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{4-b}3\right)}\\
&\qquad+\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+1}3-c\right)\Gamma\left(1+\frac{2a}3-c\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(c+\frac{2b-a}3\right)\Gamma\left(c+\frac{2-2a-b}3\right)}{\Gamma\left(\frac b3\right)\Gamma\left(\frac{b+1}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{2-b}3\right)\Gamma\left(\frac{3-b}3\right)\Gamma\left(\frac{4-b}3\right)\Gamma\left(\frac{2a-1}3\right)\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(2c+\frac{1-2a}3\right)}\\
&\qquad\cdot\F21{c+\frac{b-2a}3,c+\frac{2-2a-b}3}{2c+\frac{1-2a}3}1\\
&=\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(c+\frac{2-2a-b}3\right)\Gamma\left(c+\frac{b-2a}3\right)}{\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(c-\frac{2a}3\right)\Gamma\left(c+\frac{2-2a}3\right)\Gamma\left(c+\frac{b-1}3\right)\Gamma\left(c+\frac{1-b}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{4-b}3\right)}\\
&\qquad+\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+1}3-c\right)\Gamma\left(1+\frac{2a}3-c\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(c+\frac{2b-a}3\right)\Gamma\left(c+\frac{2-2a-b}3\right)}{\Gamma\left(\frac b3\right)\Gamma\left(\frac{b+1}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{2-b}3\right)\Gamma\left(\frac{3-b}3\right)\Gamma\left(\frac{4-b}3\right)\Gamma\left(\frac{2a-1}3\right)\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(2c+\frac{1-2a}3\right)}\\
&\qquad\cdot\frac{\Gamma\left(2c+\frac{1-2a}3\right)\Gamma\left(\frac{2a-1}3\right)}{\Gamma\left(c+\frac{1-b}3\right)\Gamma\left(c+\frac{b-1}3\right)}\\
&=\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(c+\frac{2-2a-b}3\right)\Gamma\left(c+\frac{b-2a}3\right)}{\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(c-\frac{2a}3\right)\Gamma\left(c+\frac{2-2a}3\right)\Gamma\left(c+\frac{b-1}3\right)\Gamma\left(c+\frac{1-b}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{4-b}3\right)}\\
&\qquad+\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+1}3-c\right)\Gamma\left(1+\frac{2a}3-c\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(c+\frac{2b-a}3\right)\Gamma\left(c+\frac{2-2a-b}3\right)}{\Gamma\left(\frac b3\right)\Gamma\left(\frac{b+1}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{2-b}3\right)\Gamma\left(\frac{3-b}3\right)\Gamma\left(\frac{4-b}3\right)\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(c+\frac{1-b}3\right)\Gamma\left(c+\frac{b-1}3\right)}\\
&=\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(c+\frac{2-2a-b}3\right)\Gamma\left(c+\frac{b-2a}3\right)}{\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(c-\frac{2a}3\right)\Gamma\left(c+\frac{2-2a}3\right)\Gamma\left(c+\frac{b-1}3\right)\Gamma\left(c+\frac{1-b}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{4-b}3\right)}\\
&\qquad\cdot\left(1+\frac{\sin\frac{\pi b}3\sin\frac{\pi(b+1)}3}{\sin\pi\left(c-\frac{2a}3\right)\sin\pi\left(c+\frac{2-2a}3\right)}\right)
\end{align}
ここで,
\begin{align}
&1+\frac{\sin\frac{\pi b}3\sin\frac{\pi(b+1)}3}{\sin\pi\left(c-\frac{2a}3\right)\sin\pi\left(c+\frac{2-2a}3\right)}\\
&=\frac{\sin\pi\left(c-\frac{2a}3\right)\sin\pi\left(c+\frac{2-2a}3\right)+\sin\frac{\pi b}3\sin\frac{\pi(b+1)}3}{\sin\pi\left(c-\frac{2a}3\right)\sin\pi\left(c+\frac{2-2a}3\right)}\\
&=-\frac 12\frac{\cos\pi\left(2c+\frac{2-4a}3\right)-\cos\frac{\pi(2-2b)}3}{\sin\pi\left(c-\frac{2a}3\right)\sin\pi\left(c+\frac{2-2a}3\right)}\\
&=\frac{\sin\pi\left(c+\frac{2-2a-b}3\right)\sin\pi\left(c+\frac{b-2a}3\right)}{\sin\pi\left(c-\frac{2a}3\right)\sin\pi\left(c+\frac{2-2a}3\right)}
\end{align}
であるから,
\begin{align}
&\F76{a-\frac 12,a,b,2-b,c,\frac{2a+2}3-c,\frac a2+1}{\frac 32,1+\frac{2a-b}3,\frac{2a+b+1}3,3c-1,2a+1-3c,\frac a2}1\\
&=\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(c+\frac{2-2a-b}3\right)\Gamma\left(c+\frac{b-2a}3\right)}{\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(c-\frac{2a}3\right)\Gamma\left(c+\frac{2-2a}3\right)\Gamma\left(c+\frac{b-1}3\right)\Gamma\left(c+\frac{1-b}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{4-b}3\right)}\\
&\qquad\cdot\frac{\sin\pi\left(c+\frac{2-2a-b}3\right)\sin\pi\left(c+\frac{b-2a}3\right)}{\sin\pi\left(c-\frac{2a}3\right)\sin\pi\left(c+\frac{2-2a}3\right)}\\
&=\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(\frac{2a+1}3-c\right)\Gamma\left(1+\frac{2a}3-c\right)}{\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(c+\frac{b-1}3\right)\Gamma\left(c+\frac{1-b}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{4-b}3\right)\Gamma\left(\frac{2a+b+1}3-c\right)\Gamma\left(1+\frac{2a-b}3-c\right)}
\end{align}
を得る. つまり, 以下が得られた.
\begin{align} &\F76{a-\frac 12,a,b,2-b,c,\frac{2a+2}3-c,\frac a2+1}{\frac 32,1+\frac{2a-b}3,\frac{2a+b+1}3,3c-1,2a+1-3c,\frac a2}1\\ &=\frac{\Gamma\left(\frac 23\right)\Gamma\left(\frac 43\right)\Gamma\left(c-\frac 13\right)\Gamma\left(c+\frac 13\right)\Gamma\left(\frac{2a+b+1}3\right)\Gamma\left(1+\frac{2a-b}3\right)\Gamma\left(\frac{2a+1}3-c\right)\Gamma\left(1+\frac{2a}3-c\right)}{\Gamma\left(\frac{2a+1}3\right)\Gamma\left(1+\frac{2a}3\right)\Gamma\left(c+\frac{b-1}3\right)\Gamma\left(c+\frac{1-b}3\right)\Gamma\left(\frac{b+2}3\right)\Gamma\left(\frac{4-b}3\right)\Gamma\left(\frac{2a+b+1}3-c\right)\Gamma\left(1+\frac{2a-b}3-c\right)} \end{align}
$a\to i\infty$としてから$b\mapsto 1+3x,c\mapsto y$とすると, 冒頭に紹介したGosperの${}_3F_2$和公式
\begin{align}
\F32{1+3x,1-3x,y}{\frac 32,3y-1}{\frac 34}&=\frac{2\Gamma\left(y-\frac 13\right)\Gamma\left(y+\frac 13\right)\sin\pi x}{3\sqrt 3x\Gamma(y+x)\Gamma(y-x)}
\end{align}
を得ることができる.
