%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Latex file % % K. Srinivasa Rao and J. Van der Jeugt % % "Stretched 9-j coefficients and summation theorems" % % J. Phys. A: Math. Gen. 27 (1994), 3083-3090 % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentstyle[12pt]{article} \headheight=0mm \headsep=0mm \oddsidemargin=1mm \evensidemargin=1mm \textheight=210mm \textwidth=150mm \addtolength{\baselineskip}{2mm} % \def\be{\begin{equation}} \def\ee{\end{equation}} \def\bea{\begin{eqnarray}} \def\eea{\end{eqnarray}} \def\beas{\begin{eqnarray*}} \def\eeas{\end{eqnarray*}} \def\nn{\nonumber} \def\ninej#1#2#3#4#5#6#7#8#9{\left\{ \begin{array}{ccc} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9 \end{array} \right\} } \def\Ga{\Gamma} % \begin{document} \centerline{\Large \bf Stretched 9$-j$ coefficients and } \centerline{\Large \bf summation theorems} \vskip 2cm \centerline{\bf K.Srinivasa Rao\footnote{Permanent address : The Institute of Mathematical Sciences, Madras - 600 113, India.} and J Van der Jeugt\footnote{Senior Research Associate of N.F.W.O (National Fund for Scientific Research of Belgium).}} \centerline{Department of Applied Mathematics and Computer Science,} \centerline{University of Gent, Krijgslaan 281-S9, B-9000 Gent, Belgium} \centerline{E-mail : rao@imsc.ernet.in and Joris.VanderJeugt@rug.ac.be} \vskip 1cm *PACS : 02.20.+b, 02.30.+g, 02.90.+p, 03.65.-w \vskip 2cm \noindent {\bf Abstract:} It is well known that the 9$-j$ coefficient has 72 symmetries and that the simplest known form for it is the triple sum series which exhibits none of these symmetries. A study of the expressions for the stretched 9$-j$ coefficients (stretched in the sense that one or more of the six triangle inequalities comprising it are stretched) via the triple sum series shows the existence of the $_2F_1(1)$, $_3F_2(1)$ and the $_4F_3(1)$ summation theorems, and reveals some new summation theorems, as well. \newpage The 9$-j$ coefficient, or the $ls-jj$ transformation coefficient which plays a crucial role in the evaluation of two-body matrix elements, arising in atomic, molecular and nuclear physics studies, exhibits 72 symmetries through its representation as a sum over the projection quantum numbers of a product of six 3$-j$ coefficients (see, for instance, Biedenharn and Louck, 1981; Wigner 1940). The simplest known form for this recoupling coefficient is the triple sum series of Jucys and Bandzaitis (1977) which however does not exhibit any of these 72 symmetries. It has been pointed out that as a consequence of the afore-said lack of symmetry, a given 9$-j$ coefficient, for large angular momenta, which has one term, could have several tens of thousands of terms when the triple sum representation for its symmetries is examined. This inherent disadvantage was converted into an advantage in the numerical computaion of the 9$-j$ coefficient (Srinivasa Rao, Rajeswari and Chiu, 1989). Almost three decades ago, Bandzaitis, Karosiene and Jucys (1964) derived formulae for the stretched 9$-j$ coefficients, in which one or more of the angular momenta belonging to any of the six triads ($a,b,c$, say) corresponds to the limits of the triangular inequality : $| a-b |\leq c \leq a+b$. Sharp (1967) showed that there are in all five distinctly different doubly stretched cases and two triply stretched ones, while any singly stretched 9$-j$ coefficient can be brought to one standard form through the symmetries of the coefficient. The inherent lack of symmetry of the Jucys-Bandzaitis triple sum series for the 9$-j$ coefficient and the observation that a given 9$-j$ coefficient may have one or many terms has led us to a search of the stretched 9$-j$ coefficient formulae and the triple sum series for summation theorems. Here we show that the Vandermonde theorem (the terminating version of the famous Gauss theorem) for the $_2F_1(1)$, the $_3F_2(1)$ summation theorem of Pfaff-Saalsch\"{u}tz and the more recent Karlsson(1971)-Minton (1970) summation theorem for the $_4F_3(1)$, all occur naturally in our study. What is interesting is that besides these well-known theorems, the study opens up the scope for finding genuinely new summation theorems. After presenting the relevant details regarding the triple sum series and stretched 9$-j$ coefficients, we show their interconnection which reveals summation theorems. The triple sum series of Jucys-Bandzaitis (1977) is the simplest known form for the 9$-j$ coefficient and it is given by : \bea & \ninej{a}{b}{c}{d}{e}{f}{g}{h}{i}\,=\,K\ \ {\displaystyle \sum_{x,y,z}\, \frac {(-1)^{x+y+z}}{x!y!z!}} {\displaystyle \frac {(x1-x)!(x2+x)!(x3+x)!}{(x4-x)!(x5-x)!}} & \nn \\ & \times\,{\displaystyle \frac {(y1+y)!(y2+y)!}{(y3+y)!(y4-y)!(y5-y)!}} {\displaystyle \frac {(z1-z)!(z2+z)!}{(z3-z)!(z4-z)!(z5-z)!}} & \nn \\ & \times\,{\displaystyle \frac {(p1-y-z)!}{(p2+x+y)!(p3+x+z)!}} & \eea where $$ K\ =\ (-1)^{x5}\, {(d,a,g)(b,e,h)(i,g,h) \over(d,e,f)(b,a,c)(i,c,f)} $$ \be \begin{array}{rccclcr} 0 & \leq & x & \leq & \min(x4\,,\,x5) & = & XF \\ 0 & \leq & y & \leq & \min(y4\,,\,y5) & = & YF \\ 0 & \leq & z & \leq & \min(z4\,,\,z5) & = & ZF \end{array} \ee \be \begin{array}{lll} x1=2f & y1=-b+e+h & z1=2a \\ x2=d+e-f & y2=g+h-i & z2=-a+b+c \\ x3=c-f+i & y3=2h+1 & z3=a+d+g+1 \\ x4=-d+e+f & y4=b+e-h & z4=a+d-g \\ x5=c+f-i & y5=g-h+i & z5=a-b+c \\ p1=a+d-h+i\quad & p2=-b+d-f+h\quad & p3=-a+b-f+i\ \end{array} \ee \be (a,b,c) = \left[{(a-b+c)!(a+b-c)!(a+b+c+1)!\over (-a+b+c)!}\right]^{1/2}. \ee Note that the 18 parameters $x1,x2,\cdots ,p3$ given in (3) are different for different symmetries of the 9$-j$ coefficient, since they are dependent on the positions of $a,b,\cdots ,i$ in the $3\times 3$ array. They are also not independent since there are nine relations between them, like : $ x2+x4 = y1+y4$, $x3+x5 = z2+z5$, etc. The lack of symmetry is best illustrated through the example : \be \ninej{30}{20}{10}{30}{10}{20}{60}{30}{30} = 0.00026845 \ee which has $XF = 0$, $YF = 0$ and $ZF = 0$, so that it is a single term. But its symmetry (a cyclic column permutation and an odd row permutation of the 9$-j$ treated as a $3 \times 3$ array) : \be \ninej{20}{10}{30}{30}{30}{60}{10}{20}{30} = 0.00026845 \ee has $XF = 60$, $YF = 20$, $ZF = 40$ and an actual number of 33761 terms. This number is reckoned by taking into account the constraints on the ranges of $x, y, z$, placed by $p1, p2, p3$, viz. \be y + z \leq p1,\ {\rm \ and\ if\ } \ p2,p3 < 0\, \ {\rm \ then}\ x+y \geq |p2|,\ x+z \geq |p3|. \ee If one were to use the expression due to Wigner (1940) for the 9$-j$ coefficient, where it is written as a single sum over a product of three 6$-j$ coefficients (the 6$-j$ coefficient itself being given by the single sum expression due to Racah (1942)), then the 9$-j$ coefficient in (5) would require 40 terms or 120 references to the 6$-j$ coefficients, while that in (6) would require one term or 3 references to the 6$-j$ coefficient. From this, it can be argued that the triple sum series (1) requires lesser time for its computation than the conventional single sum over a product of 6$-j$ coefficients, especially for large angular momenta. Sharp (1967) has classified the doubly stretched 9$-j$ coefficients into five distinct types : (I) - (V). These are : \be ({\rm I}) \ \ \ninej{a}{b}{c}{d}{e}{f}{a+d}{b+e}{i} {\rm \ corresponding\ to\ } y4 = 0 {\rm \ and\ } z4 = 0 \ee \be ({\rm II}) \ \ \ninej{a}{b}{c}{d}{d+f}{f}{a+d}{h}{i} {\rm \ corresponding\ to\ }e = d+f {\rm \ and\ } z4 = 0 \ee \be ({\rm III}) \ \ \ninej{a}{b}{c}{d}{e}{f}{a+d}{a+d+i}{i} {\rm \ corresponding\ to\ } y5 = 0 {\ \rm and\ } z4 = 0 \ee \be ({\rm IV}) \ \ \ninej{a}{b}{c}{d}{b+h}{f}{a+d}{h}{i} {\rm\ corresponding\ to\ } e = b+h \ {\rm and\ } z4 = 0 \ee \be ({\rm V}) \ \ \ninej{a}{b}{c}{d}{e}{f}{a+d=h+i}{h}{i} {\rm \ corresponding\ to\ } g = h+i = a+d \ee (where, in (V), $g = a+d \Leftrightarrow z4 = 0$). Any doubly stretched 9$-j$ coefficient can be brought into one of these types using the well-known symmetries (column, row permutations and transposition about the leading diagonal) of this coefficient. Explicit formulae for these five types have been derived by Sharp (1967) from the expression for the singly stretched 9$-j$ coefficient : \be \ninej{a}{b}{c}{d}{e}{f}{a+d}{h}{i} {\rm \ corresponding\ to\ } z4 = 0 \ ({\rm or\ } g = a+d) \ee The formulae of Sharp (1967) are single sum series, except for type (III) which is a single term. From the fact that $z4 = 0$ for all the five types and, in addition, $y4 = 0$ for type (I) and $y5 = 0$ for type (III), it follows that the triple sum series reduces to a double sum series for types (II), (IV) and (V) and to a single sum series for (I) and (III). However, this is not the complete picture! In this article, we analyse the type (III) doubly stretched 9$-j$ coefficient through its triple sum series representation and the consequences of some of the symmetries on it. {\it Case (i) }: When we set $y5 = 0$ and $z4 = 0$ in the triple sum series we get for the type (III) doubly stretched 9$-j$ coefficient the expression : \bea \ninej{a}{b}{c}{d}{e}{f}{a+d}{a+d+i}{i}\ =\ (-1)^{x5} {\displaystyle \frac {(d,\ a,\ a+d)(b,\ e,\ a+d+i)} {(d,\ e,\ f)(b,\ a,\ c)}}\nn \\ \times {\displaystyle \frac {(i,\ a+d,\ a+d+i)}{(i,\ c,\ f)} \frac{y1!\ y2!}{y3!\ y4!} \frac{z1!\ z2!}{z3!\ z5!}}\nn \\ \times {\displaystyle \sum_x \frac{(-1)^x}{x!}\frac {(x1-x)!(x2+x)!(x3+x)!}{(x4-x)!(x5-x)!(p2+x)!(p3+x)!}} . \eea Interestingly, the triple sum series for a symmetry of the above 9$-j$ coefficient (see below) appears to be a double sum since it has $z5 = 0$ and hence $z = 0$. However, if the additional constraints on the summation indices (7) are taken into account, then, since $x \leq x5(=c-f+i)$ and $y \leq y5(=d-e+f)$, $x + y \geq -p2(=c+d-e+i)$, the apparent double sum reduces to a single term and we have the result : \bea \ninej{a+d}{a+d+i}{i}{a}{b}{c}{d}{e}{f}\ =\ (-1)^{d-e+f} {\displaystyle \frac{(a+d+i,b,e)}{(a,b,c)(d,e,f)(i,c,f)}}\nn \\ \times {\displaystyle \left[ \frac{(2a)!(2d)!(2i)!} {(2a+2d+1)\ (2a+2d+2i+1)!}\right]^{1/2} } \eea This result has been derived by Sharp (1967) from the formula for the singly stretched 9$-j$ coefficient in terms of a double sum given by Sharp and von Baeyer (1966). When the single sum series part of (14) is rearranged into a generalized hypergeometric function, we have : \be S_x\ =\ {\displaystyle \frac{x1!\ x2!\ x3!}{x4!\ x5!\ p2!\ p3!}} \ _4F_3{\left( \begin{array}{cc} 1+x2,\ 1+x3,\ -x4,\ -x5\ ;& 1\\ \ \ -x1,\ 1+p2,\ 1+p3 & \end{array}\right) } \ee where the $_4F_3(1)$ is not a Saalsch\"{u}tzian and therefore cannot be a Racah or 6$-j$ coefficient (Srinivasa Rao and Rajeswari 1993). But, it is a zero-balanced $_{p+1}F_p(1)$, i.e. its numerator and denominator parameter sums are equal : $ x1 + x2 + x3 \ =\ p2 + p3 + x4 + x5 $. A comparison of these results for the two symmetries of the doubly stretched 9$-j$ coefficient (14) and (15) gives, after simplifications, for the summation part, the result : \be S_x \ =\ (-1)^{x1-x4-x5} \ee which when expressed in terms of the $_4F_3(1)$ yields : \bea _4F_3 {\left( \begin{array}{cc} 1+x2,\ 1+x3,\ -x4,\ -x5\ ;& 1\\ \ \ -x1,\ 1+p2,\ 1+p3 & \end{array}\right) }\ =\ (-1)^{x1-x4-x5}\nn \\ \times \frac{\Ga (1+x4,\ 1+x5,\ 1+p2,\ 1+p3)}{\Ga (1+x1,\ 1+x2,\ 1+x3)} \eea where we have used the notation : $$\Ga (a,b,\cdots )\ =\ \Ga(a) \ \Ga (b)\ \cdots $$ If we now make the identifications : \be \begin{array}{ccc} \ -x1\ =\ b1, & \ -x4\ =\ b1 + m1, & (m1\ =\ x1 - x4)\\ 1 + p2\ =\ b2, & 1 + x2\ =\ b2 + m2, & (m2\ =\ x2 - p2)\\ 1 + p3\ =\ b3, & 1 + x3\ =\ b3 + m3, & (m3\ =\ x3 - p3) \end{array} \ee and rewrite our result, we have : \bea &&_4F_3 {\left( \begin{array}{cc} -(m1+m2+m3),\ b1+m1,\ b2+m2,\ b3+m3;& \ 1 \\ b1,\ b2,\ b3 & \end{array}\right) } \ =\nn \\ &&=\ (-1)^{m1+m2+m3} \frac{(m1+m2+m3)!}{(b1)_{m1}\ (b2)_{m2}\ (b3)_{m3}}\qquad \qquad \eea where $(b)_m\ =\ {\Ga (b+m)}/{\Ga (b)}$, the Pochhammer symbol. This is the Karlsson(1971) - Minton (1970) summation theorem for terminating zero-balanced $_{p+1}F_p(1)$ series corresponding to $p\ =\ 3$ (cf. (1.9.3) of Gasper and Rahman, 1991). {\it Case (ii)} : For another symmetry of the 9$-j$ coefficient considered in {\it Case (i)}, the triple sum series yields the result : \bea \ninej{c}{b}{a}{f}{e}{d}{i}{a+d+i}{a+d}\ =\ \frac{(f,c,i)(b,e,a+d+i) (a+d,i,a+d+i)}{f,e,d)(b,c,a)(a+d,a,d)}\nn \\ \times \frac{(x1)!(x2)!(x3)!}{(x4)!} \frac{(y1)!(y2)!}{(y3)!(y4)!} \frac{(z1)!}{(z3)!(z5)!(p2)!}\ \ _2F_1\left( { \begin{array}{cc} -z3, -z4; & 1 \\ \ \ -z1 & \end{array} }\right) \eea where \be \begin{array}{lll} x1 = 2d & y1 = a-b+d+e+i & z1 = 2c \\ x2 = -d+e+f\quad & y2 = 2i & z2 = a+b-c \\ x3 = 2a & y3 = 2(a+d+i)+1 & z3 = c+f+i+1 \\ x4 = d+e-f & y4 = -a+b-d+e-i\quad & z4 = c+f-i \\ x5 = 0 & y5 = 0 & z5 = a-b+c \\ p1 = c+f-i & p2 =a-b+f+i & p3 = a+b-c \end{array} \ee Comparison with the column permuted result in (15) yields immediately, after simplification, using the symmetry property of the 9$-j$ coefficient, the result : \be \begin{array}{rcl} _2F_1 \left( \begin{array}{cc} -z3,\ -z5;\ & 1\\ \ \ -z1 & \end{array} \right) & = & {\displaystyle (-1)^{-a+b-c} \frac{(-a+b+c)! (a-b+f+i)!}{(-c+f-i)! (2c)!}} \\ & = & {\displaystyle \frac{(-z1 + z3)_{z5}}{(-z1)_{z5}}} \end{array} \ee which is clearly a manifestation of the Vandermonde summation theorem : \be _2F_1 \left( \begin{array}{cc} a,\ -n;\ & 1\\ \ \ c & \end{array} \right) \ =\ {\displaystyle \frac{(c - a)_{n}}{(c)_{n}}}. \ee {\it Case (iii)}: We consider another symmetry of the 9$-j$ coefficient for which the triple sum series reduces again to a single sum series : \be \begin{array}{c} \ninej{b}{c}{a}{a+d+i}{i}{a+d}{e}{f}{d}\ =\ {\displaystyle (-1)^{x5+z5} \frac{(a+d+i,b,e)}{(a+d+i,i,a+d)}} \\ \times {\displaystyle \frac{(c,i,f)(d,e,f)}{(c,b,a)(d,a,a+d)} \frac{x1!x2!x3!}{x4!x5!}\frac{(z1-z5)!(z2+z5)!} {z5!(z3-z5)!(z4-z5)!(p3+z5)!}} \\ \times {\displaystyle \sum_y \frac{(-1)^y}{y!}\frac{(y2+y)!(p1-z5-y)!} {(y3+y)!(y4-y)!(y5-y)!}}. \end{array} \ee Comparison with a symmetry (cyclic column permutation with an odd row permutation) of (15), yields, on simplification, the result : \be \begin{array}{c} _3F_2 \left( \begin{array}{cc} 1+y2,-y4,-y5; & 1 \\ 1+y3,-p1+z5 & \end{array} \right)\ =\\ =\ {\displaystyle \frac{\Ga (y3-y2+y5, 1+y3+y4+y5, 1+y3, -y2+y3+y4)} {\Ga (y3-y2,1+y3+y4, 1+y3+y5, -y2+y3+y4+y5)}} \end{array} \ee using the Saalsch\"{u}tz property satisfied by the parameters of the $_3F_2(1)$, viz. $ 1+y2-y4-y5 = y3+z5-p1$. This formula (26) can be shown to be nothing but the Pfaff-Saalsch\"{u}tz summation theorem : \be _3F_2 \left( \begin{array}{cc} a,b, -n; & 1 \\ c,1+a+b-c-n & \end{array} \right)\ =\ \frac{(c-a)_n (c-b)_n} {(c)_n(c-a-b)_n}. \ee (cf. Gasper and Rahman, 1991). {\it Case (iv) }: Next, we consider a symmetry of the type (III) 9$-j$ coefficient for which the triple sum series reduces to a double sum series : \bea \ninej{f}{i}{c}{e}{a+d+i}{b}{d}{a+d}{a}\ =\ \frac{(e,d,f)(i,a+d+i,a+d)(a,d,a+d)}{(e,a+d+i,b)(i,f,c)(a,c,b)} \nn \\ \times {\displaystyle \frac{y1! y2!}{ y3!y4!y5!} \sum_{x,z} \frac{(-1)^{x+z}}{x! z!} \frac{(x1-x)!(x2+x)!(x3+x)!}{(x4-x)!(x5-x)!}} \nn \\ \times {\displaystyle \frac{(z1-z)! (z2+z)!} {(z3-z)!(z5-z)!(p3+z)!(p2+x+z)!} }.\qquad \eea Comparison of this result, with a symmetry (an odd column permutation, a cyclic row permutation and a reflection about the diagonal) of the single term expression (15), after simplifications, yields the result for the double sum : \bea {\displaystyle \sum_{x,z} \frac{(-1)^{x+z}}{x! z!} \frac{(x1-x)!(x2+x)!(x3+x)!}{(x4-x)!(x5-x)!(p2+x)!}} \qquad \qquad \nn \\ \times {\displaystyle \frac{(z1-z)!(z2+z)!}{(z3-z)!(z5-z)!(p3+x+z)!}} \ = \qquad \qquad \nn \\ \ = \ {\displaystyle \frac{(-1)^{z2+z5} (x1+x2+1)!x2! (z1+z2-z5)!} {(x2+x4+1)!(x2-p3+1)! (x2-p3-z1)!} },\qquad \eea where \be \begin{array}{lcl} x1 = 2b & & z1 = 2f \\ x2 = a-b+d+e+i & & z2 = c-f+i\\ x3 = a-b+c& & z3 = d+e+f+1\\ x4 = a+b+d-e+i & & \ z5 = c+f-i \\ x5 = -a+b+c & & p3 = a-b-f+i \\ p2 = a-b+d+e-i\quad & & \end{array} \ee This sum depends on seven independent variables, e.g. $x1, x2, x4$, $z1, z2, z5$ and $p3$; the remaining variables are then given by : $x3 = z5+p3$, $x5 = z2-p3$, $z5 = x2-p3+1$ and $p2 = x2-z1-z2+z5$. Furthermore, there are inequality conditions governed by the requirement that the factorials should be non-negative. The factor $(p3+x+z)!$ in this sum makes it a genuine double sum series. Numerically, the validity of this summation theorem has been checked with Macsyma(1985). Also, the summation in $x$ or $z$ separately cannot be performed with the help of other summation theorems (like the Minton, or Karlson-Minton summation theorems for the $_4F_3(1)$). If we consider the symmetry of the 9$-j$ coefficient $$\ninej{b}{c}{a}{d}{e}{f}{a+d+i}{i}{a+d} $$ for which the triple sum series also reduces to a double sum, it can be summed using first the Minton theorem for a $_4F_3(1)$ $-$ though a numerator parameter contains the other summation index, the characteristic of the Minton theorem is the presence of a numerator and a denominator parameter which differ by 1 $-$ and the resulting series can be summed using the $_2F_1(1)$ summation theorem of Vandermonde. {\it Case (v)} : Here we consider a symmetry of the 9$-j$ coefficient which does not reduce the triple sum series : \bea \ninej{d}{e}{f}{a+d}{a+d+i}{i}{a}{b}{c}\ =\ (-1)^{x5} {\displaystyle \frac{(a+d,d,a)(e,a+d+i,b)(c,a,b)} {(a+d,a+d+i,i)(e,d,f)(c,f,i)}}\nn \\ \times \sum_{x,y,z}{\displaystyle \frac{(-1)^{x+y+z}}{x!y!z!} \frac{(x2+x)!(x3+x)!}{(x5-x)!} \frac{(y1+y)!(y2+y)!}{(y3+y)!(y4-y)!(y5-y)!}} \nn \\ \times {\displaystyle \frac{(z2+z)!}{(z3-z)!(z5-z)!} \frac{(p1-y-z)!}{(p2+x+y)!(p3+x+z)!}}\qquad \qquad \eea On comparing this with an appropriate symmetry of the 9$-j$ coefficient in (15), we get : \bea \sum_{x,y,z}{\displaystyle \frac{(-1)^{x+y+z}}{x!y!z!} \frac{(x2+x)!(x3+x)!}{(x5-x)!} \frac{(y1+y)!(y2+y)!}{(y3+y)!(y4-y)!(y5-y)!}}\nn \\ \times {\displaystyle \frac{(z2+z)!}{(z3-z)!(z5-z)!} \frac{(p1-y-z)!}{(p2+x+y)!(p3+x+z)!}}\ = \nn \\ =\ {\displaystyle (-1)^{x5+z5} \frac{x3!y1!z2!}{x5!y5!z5!} \frac{1} {z3 (y3+y5)!(-y1+y3-1)!}} \qquad \qquad \eea where we choose $x3, x5, y1, y3, y4, z5$ and $p2$ to be the independent variables and the dependent variables are then related to these through : $x2 = y4+p2$, $y2 = x5-z5+p2$, $y5 = x3+y1-y3-z5+1$, $z2 = x3+x5-z5$, $z3 = y4+p2+1$, $p1 = -x5+y4+z5$ and $p3 = x3-z5$. The numerical validity of this summation theorem has also been verified with the help of Macsyma. This triple sum cannot be summed with the help of the Minton theorem. In fine, we have shown that the comparison of a stretched 9$-j$ coefficient formula with the Jucys-Bandzaitis triple sum series, in conjunction with the symmetries of the 9$-j$ coefficient, reveals the Vandermonde, Pfaff-Saalsch\"{u}tz, Karlsson-Minton summation theorems, respectively, for the $_2F_1(1)$, $_3F_2(1)$ and $_4F_3(1)$ series, as well as {\bf new} summation theorems for double and triple sum series. This is a direct consequence of the highly asymmetric nature of the triple sum series. A complete classification of the summation theorems from a study of all the 72 symmetries of the 9$-j$ coefficient on the triple sum series and their relation to the other stretched formulae will be reported elsewhere. It is a pleasure to thank Professor R.Askey and Professor R.P.Agarwal for fruitful exchanges regarding the summation theorems and Professors G.Vanden Berghe and Hans de Meyer for stimulating discussions. This work was partly supported by the E.E.C. (contract No. CI1*-CT92-0101). \vskip 1cm \centerline{\bf References} \begin{itemize} \item[]Bandzaitis A A, Karosienne A and Jucys A P (1964) Liet. Fiz. Rin. {\bf 4} 457. \item[]Biedenharn L C and Louck J D (1981) {\it Angular Momentum in Quantum Physics}, Encyc. of Maths. and its Applns., Vol. {\bf 8} (Academic Press). \item[]Biedenharn L C and Van Dam H (1965) (Eds.) {\it Quantum Theory of Angular Momentum : A collection of Reprints and original Papers} (Academic Press). \item[]Gasper G and Rahman M (1991) {\it Basic Hypergeometric series}, Encyc. of Maths. and its applns. Vol. {\bf 35} (Academic Press). \item[]Jucys A P and Bandzaitis A A (1977) {\it Angular Momentum in Quantum Physics} (Mokslas, Vilnius). 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