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, Discrete Mathematics, Volume 311, Issue 14, 28 July 2011, Pages 1255-1272.
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.
José L. Cereceda, Sums of powers of integers and the sequence A304330, arXiv:2405.05268 [math.GM], 2024. See p. 14.
Mark W. Coffey and Matthew 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
KEYWORD
AUTHOR
Wolfdieter Lang, May 04 2007
STATUS
approved