Most computationally efficient Littlewood-Richardson rule There are many, many different versions of the Littlewood-Richardson rule: the original characterization via Yamanouchi words, Remmel's version, a description via the Poirier-Reutenauer bialgebra, the Knutson-Tao hive model rule and many more. Any of these rules provides an effective algorithm for computing products of Schur functions. It is known that computing Littlewood-Richardson coefficients is, in general, #P complete.
I am interested in the most computationially efficient Littlewood-Richardson rule. Since any rule will be in #P, for which I am not aware of easy characterizations of run time (please correct me if I'm wrong on this), I suspect any answer to this question would have to be based on experimental evidence. No doubt the various computer algebra systems have coded the various rules, computed a bunch of examples and checked which one was fastest. In one of my graduate courses, we were told the Remmel rule was the easiest, though this was for by hand computations of small examples. At the end of Exercise 2.7.10 of book Symmetric Functions, Schubert Polynomials and Degeneracy Loci, Manivel asserts that the Lascoux-Schutzenberger transition equations (see here) for Schubert polynomials (and hence Stanley symmetric functions) provides the most computationally efficient means of computing the product of Schur functions.
1) What is the (experimentally) most efficient method of computing a single Littlewood-Richardson coefficient?
2) Can anyone confirm or refute Manivel's assertion that the Lascoux-Schutzenberger transition equations are the most efficient rule for computing the product of Schur functions?
 A: I can't answer your second question, but I have a pretty fast recurrence for computing a single Littlewood-Richardson coefficient for the complete flag variety. It also works for equivariant coefficients, but unlike other equivariant recurrences I've seen you can just set the equivariant part to 0 and get a recurrence for the ordinary coefficients. It does not restrict to a recurrence for Grassmannian coefficients only, so we have to expand our attention to the entire symmetric group. It is not published anywhere so I will prove it here.
The main result of my dissertation is a Leibniz formula for divided difference operators. If $p$ and $q$ are two polynomials in variables $x_1,x_2,\ldots,x_n$ and $w$ is a permutation, then
$$\partial_w(pq)=\sum_{u,v}{c_{u,v}^w(x_1,\ldots,x_n)\partial_u(p)\partial_v(q)}$$
where $c_{u,v}^w(x_1,\ldots,x_n)$ is the equivariant Littlewood-Richardson coefficient, which is the coefficient appearing in the expansion
$$S_u(x;y)S_v(x;y)=\sum_{w}{c_{u,v}^w(y_1,\ldots,y_n)S_w(x;y)}$$
of a product of two double Schubert polynomials. When $\ell(u)+\ell(v)=\ell(w)$, these coefficients are the ordinary Littlewood-Richardson coefficients (by which I mean for the complete flag variety).
For an integer $k$ I denote by $s_k$ the simple transposition $(k,k+1)$. Now, let $s_i$ be a right descent of $w$, so that
$$\partial_w=\partial_{ws_i}\partial_{s_i}$$
Then since
$$\partial_{s_i}(pq)=\partial_{s_i}(p)q+p\partial_{s_i}(q)-(x_i-x_{i+1})\partial_{s_i}(p)\partial_{s_i}(q)$$
we can write
$$\partial_w(pq)=\partial_{ws_i}(\partial_{s_i}(p)q+p\partial_{s_i}(q)-(x_i-x_{i+1})\partial_{s_i}(p)\partial_{s_i}(q))=\sum_{u,v}{c_{us_i,v}^{ws_i}\partial_{us_i}\partial_{s_i}(p)\partial_v(q)}+\sum_{u,v}{c_{u,vs_i}^{ws_i}\partial_u(p)\partial_{vs_i}\partial_{s_i}(q)}+\sum_{u',u,v}{c_{u',u,v}^{ws_i}\partial_{u'}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}$$
For the last sum,
$$\sum_{u',u,v}{c_{u',u,v}^{ws_i}\partial_{u'}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}=\sum_{u',v',u,v}{c_{u',v'}^{ws_i}c_{us_i,vs_i}^{v'}\partial_{u'}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}$$
If $v'=ws_i$, then we get a contribution of
$$\sum_{u',u,v}{c_{us_i,vs_i}^{ws_i}c_{u',ws_i}^{ws_i}\partial_{u'}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}=\sum_{u,v}{c_{us_i,vs_i}^{ws_i}(x_{w(i+1)}-x_{w(i)})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}$$
(I snuck in that last equality, it's true because acting with $ws_i$ is equivalent to summing over all $u'$ what you get by acting with $c_{u',ws_i}^{ws_i}\partial_{u'}$.) If $v'<ws_i$ (in Bruhat order) and the term is nonzero, then $\ell(u')=1$, so $u'=s_j$ for some $j$. Indeed, either $j=i-1$, $j=i$, or $j=i+1$. Thus we get
$$\sum_{s_j,v',u,v}{c_{us_i,vs_i}^{v'}c_{s_j,v'}^{ws_i}\partial_{s_j}(x_i-x_{i+1})\partial_{us_i}\partial_{s_i}(p)\partial_{vs_i}\partial_{s_i}(q)}$$
By Monk's formula, in order for this to be nonzero we must have $\ell(v')=\ell(ws_i)-1=\ell(w)-2$ and there exist $a\leq j<b$ such that
$$ws_i=v'(a,b)$$
where $(a,b)$ denotes a transposition.
Long story short, picking $i$ so that $w(i)>w(i+1)$ we have
(1) If $u(i)<u(i+1)$ and $v(i)<v(i+1)$, then
$$c_{u,v}^w=0$$
(2) If  $u(i)>u(i+1)$, and $v(i)<v(i+1)$, then
$$c_{u,v}^w=c_{us_i,v}^{ws_i}$$
(3) If $u(i)<u(i+1)$, and $v(i)>v(i+1)$, then
$$c_{u,v}^w=c_{u,vs_i}^{ws_i}$$
Now for the fun part.
(4) If $u(i)>u(i+1)$ and $v(i)>v(i+1)$, then
$$c_{u,v}^w=c_{us_i,v}^{ws_i}+c_{u,vs_i}^{ws_i}+\sum_{\stackrel{a\leq i-1<b}{\ell(ws_i(a,b))=\ell(w)-2}}{c_{us_i,vs_i}^{ws_i(a,b)}}+\sum_{\stackrel{a'\leq i+1<b'}{\ell(ws_i(a',b'))=\ell(w)-2}}{c_{us_i,vs_i}^{ws_i(a',b')}}-2\sum_{\stackrel{a''\leq i<b''}{\ell(ws_i(a'',b''))=\ell(w)-2}}{c_{us_i,vs_i}^{ws_i(a'',b'')}}-(x_{w(i+1)}-x_{w(i)})c_{us_i,vs_i}^{ws_i}$$
giving a recurrence for the equivariant coefficients in terms of $x_1,\ldots,x_n$. The formatting is a bit off, so note that the first two summations have nothing in front of them (so coefficient $+1$), and the third summation has a coefficient of $-2$. If you're not interested in the positive degree coefficients, ignore the last term (setting $x=0$). This is not so much fun for a human to use but on a computer it gets you the single coefficient fast (provided you either memoize or traverse the Bruhat graph intelligently).
The base case(s) for the recurrence is
$$c_{u,1}^u=c_{1,v}^v=1$$
or, if you're really a minimalist, it's sufficient to only assume
$$c_{1,1}^1=1$$
