Oh, foo, it's basically just Pell's equation.
binom(-(sqrt(3)-sqrt(2))^n*(sqrt(6)+1)/4+(sqrt(3)+sqrt(2))^n*(sqrt(6)-1)/4+1/2,(sqrt(3)-sqrt(2))^n*(sqrt(6)+6)/24+(sqrt(3)+sqrt(2))^n*(6-sqrt(6))/24+1/2)/binom(-(sqrt(3)-sqrt(2))^n*(sqrt(6)+1)/4+(sqrt(3)+sqrt(2))^n*(sqrt(6)-1)/4-3/2,(sqrt(3)-sqrt(2))^n*(sqrt(6)+6)/24+(sqrt(3)+sqrt(2))^n*(6-sqrt(6))/24-3/2) = 6 n n (sqrt(3) - sqrt(2)) (sqrt(6) + 1) (sqrt(3) + sqrt(2)) (sqrt(6) - 1) 1 - ---------------------------------- + ---------------------------------- + - 4 4 2 ( ) n n (sqrt(3) - sqrt(2)) (sqrt(6) + 6) (sqrt(3) + sqrt(2)) (6 - sqrt(6)) 1 ---------------------------------- + ---------------------------------- + - 24 24 2 ------------------------------------------------------------------------------- = 6, n n (sqrt(3) - sqrt(2)) (sqrt(6) + 1) (sqrt(3) + sqrt(2)) (sqrt(6) - 1) 3 - ---------------------------------- + ---------------------------------- - - 4 4 2 ( ) n n (sqrt(3) - sqrt(2)) (sqrt(6) + 6) (sqrt(3) + sqrt(2)) (6 - sqrt(6)) 3 ---------------------------------- + ---------------------------------- - - 24 24 2 n = floor(n) /= 0, likewise for binom(-(sqrt(2)-1)^n*(3*sqrt(2)+2)/8+(sqrt(2)+1)^n*(3*sqrt(2)-2)/8+1/2,(3-sqrt(2))*(sqrt(2)+1)^n/12+(sqrt(2)-1)^n*(sqrt(2)+3)/12+1/2)/binom(-(sqrt(2)-1)^n*(3*sqrt(2)+2)/8+(sqrt(2)+1)^n*(3*sqrt(2)-2)/8-3/2,(3-sqrt(2))*(sqrt(2)+1)^n/12+(sqrt(2)-1)^n*(sqrt(2)+3)/12-3/2) = 9/2 n n (sqrt(2) - 1) (3 sqrt(2) + 2) (sqrt(2) + 1) (3 sqrt(2) - 2) 1 - ------------------------------ + ------------------------------ + - 8 8 2 (---------------------------------------------------------------------) n n (3 - sqrt(2)) (sqrt(2) + 1) (sqrt(2) - 1) (sqrt(2) + 3) 1 ---------------------------- + ---------------------------- + - 12 12 2 9 ----------------------------------------------------------------------- = -. n n 2 (sqrt(2) - 1) (3 sqrt(2) + 2) (sqrt(2) + 1) (3 sqrt(2) - 2) 3 - ------------------------------ + ------------------------------ - - 8 8 2 (---------------------------------------------------------------------) n n (3 - sqrt(2)) (sqrt(2) + 1) (sqrt(2) - 1) (sqrt(2) + 3) 3 ---------------------------- + ---------------------------- - - 12 12 2 Note that for n /=0 mod 4, the latter binomials are exotic. Likewise for odd n in the former. Actually, with binom(a,b) := Gamma(a+1)/(Gamma(-b+a+1)*Gamma(b+1)) a Gamma(a + 1) ( ) := -------------------------------, b Gamma(- b + a + 1) Gamma(b + 1) both identities hold for complex n /= 0. --rwg PS, the original binomial coincidence appeared during the derivation of cos(x)^odd = sum_k a_k cos(k*x). In case anyone wants them, I went ahead and derived several expansions of cos(x)^complex, and sin^complex, some of which are Fourier series valid only on an interval. One expansion for 1 - cos(pi*x)^0 gave Assuming[Element[x, Reals], Sum[Sin[2*Pi*k*x]/k, {k, 1, Infinity}]/Pi + x - 1/2] which should be (Floor[x]+Ceiling[x]-1)/2, but Mma 7.0 gives -(1/2) + x + (I*(Log[1 - E^(2*I*Pi*x)] - Log[(-1 + E^(2*I*Pi*x))/E^(2*I*Pi*x)]))/(2*Pi) which is, unlike the sum, undefined at the integers, and erroneously mimics Floor[x] if you take directional limits at the integers. An actual "analytic" Floor[x] appears to be Assuming[Element[x, Reals], FullSimplify[ArcCot[Tan[Pi*x]]/Pi + x + 1/2 == Floor[x]]] but again, no True, even Assuming[Element[x, Reals] && Not[Element[x + 1/2, Integers]],...]
Yi-Wen Liu, a former student of Julius Smith (Stanford Dept of Music) noticed, in effect,
binom(4*(n+1)^2,(n+1)*(2*n+1))/binom(2*(2*n^2+4*n+1),n*(2*n+3)) = 4
4(n+1)^2 ( ) (n+1) (2n+1) ---------------- = 4 . 2 (2n^2+4n+1) ( ) n (2n+3)
I just found
binom((2*n+1)^2,(n+1)*(2*n+1))/binom(4*n^2+4*n-1,n*(2*n+3)) = 4
(2n+1)^2 ( ) (n+1) (2n+1) --------------- = 4 . 4n^2+4n-1 ( ) n (2n+3)
Both of these are trivial to prove, but perhaps not to find, and I think at least one of them was discussed here a few years ago. There are more. Have they been characterized?
Less obvious:
binom(fib(2*n+4)-fib(2*n)-2,fib(2*n+1)-1)/binom(fib(2*n+4)-fib(2*n)-4,fib(2*n+1)-2)
fib(2n+4)-fib(2n)-2 ( ) fib(2n+1)-1 --------------------- = 5 , fib(2n+4)-fib(2n)-4 ( ) fib(2n+1)-2
which is equivalent to the identity
(fib(2*(n+2))-fib(2*n)-3)*(fib(2*(n+2))-fib(2*n)-2)/((fib(2*n+1)-1)*(fib(2*n+3)-1))=5
(fib(2 (n + 2)) - fib(2 n) - 3) (fib(2 (n + 2)) - fib(2 n) - 2) --------------------------------------------------------------- = 5 . (fib(2 n + 1) - 1) (fib(2 n + 3) - 1)
Known? --rwg ISOMETRIC EROTICISM ISOTONIC COITIONS
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