A115241 Square array read by antidiagonals: T(n,p) is the number of linearly independent, homogeneous harmonic polynomials of degree n in p variables (n,p>=1).

1, 2, 0, 3, 2, 0, 4, 5, 2, 0, 5, 9, 7, 2, 0, 6, 14, 16, 9, 2, 0, 7, 20, 30, 25, 11, 2, 0, 8, 27, 50, 55, 36, 13, 2, 0, 9, 35, 77, 105, 91, 49, 15, 2, 0, 10, 44, 112, 182, 196, 140, 64, 17, 2, 0, 11, 54, 156, 294, 378, 336, 204, 81, 19, 2, 0, 12, 65, 210, 450, 672, 714, 540, 285, 100

Offset: 1

Author

Roger L. Bagula, Mar 04 2006

Keywords

• nonn
• tabl

Comments

A115241 is jointly generated with A209688 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+x*v(n-1) and v(n,x)=u(n-1,x)+v(n-1,x)+1. See the Mathematica section at A209688. - Clark Kimberling, Mar 12 2012

Clark Kimberling's recurrence in the previous comment yields an additional zero: 1, 0, 2, 0, 3, 2, ... - Georg Fischer, Sep 04 2021

Examples

			T(1,1)=1 corresponds to the polynomial x.
T(n,1)=0 for n>=2 because no polynomial in x of degree >=2 is harmonic.
T(1,2)=2 because we can take, for example, x and y.
T(2,2)=2 because we can take, for example, x^2-y^2 and xy.
T(3,3)=7 because we can take, for example, x^3-3xy^2, x^3-3xz^2, y^3-3yx^2, y^3-3yz^2, z^3-3zx^2, z^3-3zy^2 and xyz.
The square array starts:
  1, 2,  3,  4,   5,   6,    7, ...
  0, 2,  5,  9,  14,  20,   27, ...
  0, 2,  7, 16,  30,  50,   77, ...
  0, 2,  9, 25,  55, 105,  182, ...
  0, 2, 11, 36,  91, 196,  378, ...
  0, 2, 13, 49, 140, 336,  714, ...
  0, 2, 15, 64, 204, 540, 1254, ...
  ...
		

References

• Harry Hochstadt, The Functions of Mathematical Physics, Wiley, New York (1971), p. 170; also Dover, New York (1986), p. 170.

Crossrefs

Cf. A097613 (diagonal terms), A209688.

Programs

Maple

T:=(n,p)->(2*n+p-2)*binomial(n+p-3,n-1)/n: for n from 1 to 10 do seq(T(n,p),p=1..10) od; # yields the 10 by 10 upper left corner of the square array
seq(seq(T(n,p+1-n), n=1..p), p=1..12); # yields the sequence; Georg Fischer, Sep 04 2021

Mathematica

T[n_, m_] := Binomial[n + m - 3, n - 1]*(2*n + m - 2)/n; Table[Table[T[n, p+1-n],{n,1,p}], {p,1,12}] // Flatten (* amended by Georg Fischer, Sep 04 2021 *)

Formula

T(n,p) = (2n+p-2) * binomial(n+p-3,n-1)/n for n>=1, p>=1.

Extensions

Edited by N. J. A. Sloane, Mar 07 2006