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A273631
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a(n) = Sum_{k = 0..n} (-1)^k*binomial(k,2)^3*binomial(n,k)^3.
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2
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0, 0, 1, 0, -1296, 0, 303750, 0, -36879360, 0, 3157481250, 0, -217564322976, 0, 12926105848656, 0, -689598074880000, 0, 33901459248661290, 0, -1562983866658500000, 0, 68423756889802253940, 0, -2870422192164339671040, 0, 116191495035298068750000
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OFFSET
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0,5
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COMMENTS
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Let d(n) = Sum_{k = 0..n} (-1)^k*binomial(n,k)^3. Clearly, by symmetry of the binomial coefficients we have d(2*n + 1) = 0. Dixon's identity is the result d(2*n) = (-1)^n*(3*n)!/n!^3. A generalization is: for r a nonnegative integer there holds Sum_{k = 0..n} (-1)^k*binomial(k,r)^3*binomial(n,k)^3 = (-1)^r*binomial(n,r)^3*d(n - r). This is the case r = 2. See A273630 (case r = 1) and A245086 (case r = 0).
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LINKS
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FORMULA
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a(0) = 0 and a(2*n + 2) = (-1)^n*binomial(2*n + 2,2)^3*(3*n)!/n!^3 for n >= 0. a(2*n + 1) = 0.
a(2*n + 2) = (-1)^n*(n + 1)^3*(2*n + 1)^3 * A006480(n) for n >= 0.
a(n) = Sum_{k = 2..n} (-1)^k*multinomial(n, 2, k - 2, n - k)^3.
Recurrence: a(n) = -3*n^3*(n - 1)^3*(3*n - 8)*(3*n - 10)/((n - 2)^5*(n - 3)^3) * a(n-2).
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MAPLE
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seq(add((-1)^k*binomial(k, 2)^3*binomial(n, k)^3, k = 0..n), n = 0..30);
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MATHEMATICA
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Table[Sum[(-1)^k*Binomial[k, 2]^3 Binomial[n, k]^3, {k, 0, n}], {n, 0, 27}] (* Michael De Vlieger, Jul 22 2016 *)
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PROG
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(PARI) a(n) = sum(k=0, n, (-1)^k*binomial(k, 2)^3*binomial(n, k)^3) \\ Felix Fröhlich, Jul 22 2016
(Magma) [&+[(-1)^k*Binomial(k, 2)^3*Binomial(n, k)^3: k in [0..n]]: n in [0..70]]; // Vincenzo Librandi, Jul 23 2016
(Python)
from math import factorial
def A273631(n): return 0 if n&1 or n == 0 else (-1 if (m:=n-1>>1)&1 else 1)*((m+1)*(n-1))**3*factorial(3*m)//factorial(m)**3 # Chai Wah Wu, Oct 04 2022
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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