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

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

Offset

0,5

Comments

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.

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

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

Links

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

G. E. Andrews, W. Gawronski and L. L. Littlejohn, The Legendre-Stirling Numbers

M. Bruschi, F. Calogero and R. Droghei, Proof of certain Diophantine conjectures and identification of remarkable classes of orthogonal polynomials, J. Physics A, 40(2007), pp. 3815-3829.

José L. Cereceda, A refinement of Lang's formula for the sum of powers of integers, arXiv:2301.02141 [math.NT], 2023.

M. W. Coffey and M. C. Lettington, On Fibonacci Polynomial Expressions for Sums of mth Powers, their implications for Faulhaber's Formula and some Theorems of Fermat, arXiv:1510.05402 [math.NT], 2015.

Wolfdieter Lang, First ten rows and more.

Formula

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

Example

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).

Mathematica

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 *)

Prog

(Magma)
f:= func< n, k | (&+[Binomial(2*k+j, j)*StirlingFirst(2*n, 2*k+j)*n^j: j in [0..2*(n-k)]]) >;
function T(n, k) // T = A129467
  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))
def T(n, k): # T = A129467
    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

Crossrefs

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

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

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).

Cf. A008275.

Sequence in context: A268435 A039910 A352399 * A129065 A361718 A355565

Adjacent sequences: A129464 A129465 A129466 * A129468 A129469 A129470

Keyword

sign,tabl,easy

Author

Wolfdieter Lang, May 04 2007

Status

approved