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A129467
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Orthogonal polynomials with all zeros integers from 2*A000217.
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12
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1, 0, 1, 0, -2, 1, 0, 12, -8, 1, 0, -144, 108, -20, 1, 0, 2880, -2304, 508, -40, 1, 0, -86400, 72000, -17544, 1708, -70, 1, 0, 3628800, -3110400, 808848, -89280, 4648, -112, 1, 0, -203212800, 177811200, -48405888, 5808528, -349568, 10920, -168, 1, 0, 14631321600, -13005619200, 3663035136, -466619904, 30977424, -1135808, 23016, -240, 1
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OFFSET
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0,5
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COMMENTS
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The row polynomials p(n,x) = Sum_{k=0..n} T(n,k)*x^k have the n integer zeros 2*A000217(j), j=0..n-1.
The row polynomials satisfy a three-term recurrence relation which qualify them as orthogonal polynomials w.r.t. some (as yet unknown) positive measure.
Apparently this is the triangle read by rows of Legendre-Stirling numbers of the first kind. See the Andrews-Gawronski-Littlejohn paper, table 2. The mirror version is the triangle A191936. - Omar E. Pol, Jan 10 2012
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LINKS
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FORMULA
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Row polynomials p(n,x) = Product_{m=1..n} (x - m*(m-1)), n>=1, with p(0,x) = 1.
Row polynomials p(n,x) = p(n, v=n, x) with the recurrence: p(n,v,x) = (x + 2*(n-1)^2 - 2*(v-1)*(n-1) - v + 1)*p(n-1,v,x) - (n-1)^2*(n-1-v)^2*p(n-2,v,x)) with p(-1,v,x) = 0 and p(0,v,x) = 1.
T(n, k) = [x^k] p(n, n, x), n >= k >= 0, otherwise 0.
T(n, k) = Sum_{j=0..2*(n-k)} ( binomial(2*k+j, j)*s(n,k)*n^j ) - Sum_{j=k+1..n} binomial(j, 2*(j-k))*T(n, j) (See Coffey and Lettington formula (4.7)). - G. C. Greubel, Feb 09 2024
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EXAMPLE
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Triangle starts:
1;
0, 1;
0, -2, 1;
0, 12, -8, 1;
0, -144, 108, -20, 1;
0, 2880, -2304, 508, -40, 1;
...
n=3: [0,12,-8,1]. p(3,x) = x*(12-8*x+x^2) = x*(x-2)*(x-6).
n=5: [0,2880,-2304,508,-40,1]. p(5,x) = x*(2880-2304*x+508*x^2-40*x^3 +x^4) = x*(x-2)*(x-6)*(x-12)*(x-20).
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MATHEMATICA
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T[n_, k_, m_]:= T[n, k, m]= If[k<0 || k>n, 0, If[n==0, 1, (2*(n-1)*(n-m) -(m-1))*T[n-1, k, m] -((n-1)*(n-m-1))^2*T[n-2, k, m] +T[n-1, k-1, m]]]; (* T=A129467 *)
Table[T[n, k, n], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 09 2024 *)
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PROG
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(Magma)
f:= func< n, k | (&+[Binomial(2*k+j, j)*StirlingFirst(2*n, 2*k+j)*n^j: j in [0..2*(n-k)]]) >;
if k eq n then return 1;
else return f(n, k) - (&+[Binomial(j, 2*(j-k))*T(n, j): j in [k+1..n]]);
end if;
end function;
[[T(n, k): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 09 2024
(SageMath)
@CachedFunction
def f(n, k): return sum(binomial(2*k+j, j)*(-1)^j*stirling_number1(2*n, 2*k+j)*n^j for j in range(2*n-2*k+1))
if n==0: return 1
else: return - sum(binomial(j, 2*j-2*k)*T(n, j) for j in range(k+1, n+1)) + f(n, k)
flatten([[T(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 09 2024
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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