It is well known that $$ (x \pm 1)^n = \sum_{k=0}^n (-1)^{n-k}{n \choose k} x^k $$ In other terms $n \choose k$ are expansion coefficients of $(x+1)^n$ over $x^k$ basis.
I'm looking for similar expansion, but for Legendre polynomials $$ P_n(x\pm 1) = \sum_{k=0}^n c_{n,k}^\pm P_k(x). $$ Sure, I can obtain $c_k$ for concrete $n$ using repeated polynomial division, but I suppose that the answer is already known.
Update. Here's a (reversed) table of $c_{n,k}^+$ for $n \leq 10$: $$ \begin{array}{c|ccccccccccc} n & c_0 & c_1 & c_2 & c_3 & c_4 & c_5 & c_6 & c_7 & c_8 & c_9 & c_{10}\\\hline 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 3 & \frac{3}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 3 & 1 & 5 & \frac{15}{2} & \frac{7}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & 1 & 7 & \frac{35}{2} & \frac{41}{2} & \frac{75}{8} & 0 & 0 & 0 & 0 & 0 & 0 \\ 5 & 1 & 9 & \frac{63}{2} & \frac{115}{2} & \frac{483}{8} & \frac{211}{8} & 0 & 0 & 0 & 0 & 0 \\ 6 & 1 & 11 & \frac{99}{2} & \frac{245}{2} & \frac{1515}{8} & \frac{1473}{8} & \frac{1239}{16} & 0 & 0 & 0 & 0 \\ 7 & 1 & 13 & \frac{143}{2} & \frac{447}{2} & \frac{3619}{8} & \frac{5023}{8} & \frac{9195}{16} & \frac{3763}{16} & 0 & 0 & 0 \\ 8 & 1 & 15 & \frac{195}{2} & \frac{737}{2} & \frac{7371}{8} & \frac{13069}{8} & \frac{33495}{16} & \frac{29253}{16} & \frac{93747}{128} & 0 & 0 \\ 9 & 1 & 17 & \frac{255}{2} & \frac{1131}{2} & \frac{13475}{8} & \frac{28971}{8} & \frac{93275}{16} & \frac{112335}{16} & \frac{756195}{128} & \frac{297667}{128} & 0 \\ 10 & 1 & 19 & \frac{323}{2} & \frac{1645}{2} & \frac{22763}{8} & \frac{57497}{8} & \frac{221367}{16} & \frac{330689}{16} & \frac{3031795}{128} & \frac{2474713}{128} & \frac{1920237}{256} \\ \end{array} $$ and scaled $2^kc_{n,k}^+$ $$ \begin{array}{c|ccccccccccc} n & c_0 & 2c_1 & 2^2c_2 & 2^3c_3 & 2^4c_4 & 2^5c_5 & 2^6c_6 & 2^7c_7 & 2^8c_8 & 2^9c_9 & 2^{10}c_{10}\\\hline 1 & 1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 6 & 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 3 & 1 & 10 & 30 & 28 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & 1 & 14 & 70 & 164 & 150 & 0 & 0 & 0 & 0 & 0 & 0 \\ 5 & 1 & 18 & 126 & 460 & 966 & 844 & 0 & 0 & 0 & 0 & 0 \\ 6 & 1 & 22 & 198 & 980 & 3030 & 5892 & 4956 & 0 & 0 & 0 & 0 \\ 7 & 1 & 26 & 286 & 1788 & 7238 & 20092 & 36780 & 30104 & 0 & 0 & 0 \\ 8 & 1 & 30 & 390 & 2948 & 14742 & 52276 & 133980 & 234024 & 187494 & 0 & 0 \\ 9 & 1 & 34 & 510 & 4524 & 26950 & 115884 & 373100 & 898680 & 1512390 & 1190668 & 0 \\ 10 & 1 & 38 & 646 & 6580 & 45526 & 229988 & 885468 & 2645512 & 6063590 & 9898852 & 7680948 \\ \end{array} $$
I will just deal with the "positive" case, the "negative" case is similar.
Since Legendre polynomials give an orthogonal base of $L^2(-1,1)$ with respect to the usual inner product, we have:
$$ c_{n,k}^+ = \frac{2k+1}{2}\int_{-1}^{1} P_n(x+1)P_k(x)\,dx, \tag{1} $$ so by using Rodrigues' formula and repeated integration by parts we have that the computation of $c_{n,k}^+$ boils down to the computation of: $$c_{n,k}^+ = \frac{2k+1}{2^{n+k+1}n!k!}\int_{-1}^{1}(1-x^2)^k \cdot \frac{d^{n+k}}{dx^{n+k}}\left(x^n(x+2)^n\right)\,dx \tag{2} $$ that is straightforward to carry on through the binomial theorem and the Euler beta function. By $(2)$ it is not difficult to estimate the magnitude of such coefficients. However, they do not seem to have a compact/nice closed form. They also have quite large prime factors even for small values of $k,n$.