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A129467 Orthogonal polynomials with all zeros integers from 2*A000217. 12

%I #42 Feb 14 2024 06:51:03

%S 1,0,1,0,-2,1,0,12,-8,1,0,-144,108,-20,1,0,2880,-2304,508,-40,1,0,

%T -86400,72000,-17544,1708,-70,1,0,3628800,-3110400,808848,-89280,4648,

%U -112,1,0,-203212800,177811200,-48405888,5808528,-349568,10920,-168,1,0,14631321600,-13005619200,3663035136,-466619904,30977424,-1135808,23016,-240,1

%N Orthogonal polynomials with all zeros integers from 2*A000217.

%C 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.

%C The row polynomials satisfy a three-term recurrence relation which qualify them as orthogonal polynomials w.r.t. some (as yet unknown) positive measure.

%C Column sequences (without leading zeros) give A000007, A010790(n-1)*(-1)^(n-1), A084915(n-1)*(-1)^(n-2), A130033 for m=0..3.

%C 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

%H G. C. Greubel, <a href="/A129467/b129467.txt">Rows n = 0..50 of the triangle, flattened</a>

%H G. E. Andrews, W. Gawronski and L. L. Littlejohn, <a href="http://www.math.psu.edu/andrews/pdf/283.pdf">The Legendre-Stirling Numbers</a>

%H M. Bruschi, F. Calogero and R. Droghei, <a href="http://dx.doi.org/10.1088/1751-8113/40/14/005">Proof of certain Diophantine conjectures and identification of remarkable classes of orthogonal polynomials</a>, J. Physics A, 40(2007), pp. 3815-3829.

%H José L. Cereceda, <a href="https://arxiv.org/abs/2301.02141">A refinement of Lang's formula for the sum of powers of integers</a>, arXiv:2301.02141 [math.NT], 2023.

%H M. W. Coffey and M. C. Lettington, <a href="http://arxiv.org/abs/1510.05402">On Fibonacci Polynomial Expressions for Sums of mth Powers, their implications for Faulhaber's Formula and some Theorems of Fermat</a>, arXiv:1510.05402 [math.NT], 2015.

%H Wolfdieter Lang, <a href="/A129467/a129467.txt">First ten rows and more</a>.

%F Row polynomials p(n,x) = Product_{m=1..n} (x - m*(m-1)), n>=1, with p(0,x) = 1.

%F 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.

%F T(n, k) = [x^k] p(n, n, x), n >= k >= 0, otherwise 0.

%F 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

%e Triangle starts:

%e 1;

%e 0, 1;

%e 0, -2, 1;

%e 0, 12, -8, 1;

%e 0, -144, 108, -20, 1;

%e 0, 2880, -2304, 508, -40, 1;

%e ...

%e n=3: [0,12,-8,1]. p(3,x) = x*(12-8*x+x^2) = x*(x-2)*(x-6).

%e 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).

%t 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 *)

%t Table[T[n,k,n], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Feb 09 2024 *)

%o (Magma)

%o f:= func< n,k | (&+[Binomial(2*k+j,j)*StirlingFirst(2*n,2*k+j)*n^j: j in [0..2*(n-k)]]) >;

%o function T(n,k) // T = A129467

%o if k eq n then return 1;

%o else return f(n,k) - (&+[Binomial(j,2*(j-k))*T(n,j): j in [k+1..n]]);

%o end if;

%o end function;

%o [[T(n,k): k in [0..n]]: n in [0..12]]; // _G. C. Greubel_, Feb 09 2024

%o (SageMath)

%o @CachedFunction

%o 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))

%o def T(n,k): # T = A129467

%o if n==0: return 1

%o else: return - sum(binomial(j,2*j-2*k)*T(n,j) for j in range(k+1,n+1)) + f(n,k)

%o flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # _G. C. Greubel_, Feb 09 2024

%Y Cf. A129462 (v=2 member), A129065 (v=1 member), A191936 (row reversed?).

%Y Cf. A000217, A130031 (row sums), A130032 (unsigned row sums), A191936.

%Y Column sequences (without leading zeros): A000007 (k=0), (-1)^(n-1)*A010790(n-1) (k=1), (-1)^n*A084915(n-1) (k=2), A130033 (k=3).

%Y Cf. A008275.

%K sign,tabl,easy

%O 0,5

%A _Wolfdieter Lang_, May 04 2007

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