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crystal lattice

Chinese Remainder Theorem

$x=bm^2 + a(m+1)$ solves $x\equiv a \pmod m \land x\equiv b \pmod{m+1}$.
$x=2k^2(a+b) + k(a-b)$ solves $x\equiv a \pmod{2k-1} \land x\equiv b \pmod{2k+1}$.
$x=a(3k+r+3)((3-r)k+1)+b(3k+r)(rk+2r-1)$ solves $x\equiv a \pmod{3k+r} \land x\equiv b \pmod{3k+r+3}$.

Consider 3k+r, 3k+r+3 where r=1 or 2. Then:
(3k+r+3)((3-r)k+1) gives 50: 1mod7, 0mod10, and 33: 1mod8, 0mod11
(3k+r)(rk+2r-1) gives 21: 0mod7, 1mod10, and 56: 0mod8, 1mod11

Binomial sums

This polynomial sums the first 7 entries of a row in Pascal's triangle:

$\dfrac{720+444n+304n^2-75n^3+55n^4-9n^5+n^6}{720}$

See A190782 for more.

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