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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).
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%I #17 Sep 04 2021 20:19:29

%S 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,

%T 50,55,36,13,2,0,9,35,77,105,91,49,15,2,0,10,44,112,182,196,140,64,17,

%U 2,0,11,54,156,294,378,336,204,81,19,2,0,12,65,210,450,672,714,540,285,100

%N 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).

%C 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

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

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

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

%e T(1,1)=1 corresponds to the polynomial x.

%e T(n,1)=0 for n>=2 because no polynomial in x of degree >=2 is harmonic.

%e T(1,2)=2 because we can take, for example, x and y.

%e T(2,2)=2 because we can take, for example, x^2-y^2 and xy.

%e 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.

%e The square array starts:

%e 1, 2, 3, 4, 5, 6, 7, ...

%e 0, 2, 5, 9, 14, 20, 27, ...

%e 0, 2, 7, 16, 30, 50, 77, ...

%e 0, 2, 9, 25, 55, 105, 182, ...

%e 0, 2, 11, 36, 91, 196, 378, ...

%e 0, 2, 13, 49, 140, 336, 714, ...

%e 0, 2, 15, 64, 204, 540, 1254, ...

%e ...

%p 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

%p seq(seq(T(n,p+1-n), n=1..p), p=1..12); # yields the sequence; _Georg Fischer_, Sep 04 2021

%t 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 *)

%Y Cf. A097613 (diagonal terms), A209688.

%K nonn,tabl

%O 1,2

%A _Roger L. Bagula_, Mar 04 2006

%E Edited by _N. J. A. Sloane_, Mar 07 2006