cp's OEIS Frontend

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).

%I A115241 #17 Sep 04 2021 20:19:29
%S A115241 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 A115241 50,55,36,13,2,0,9,35,77,105,91,49,15,2,0,10,44,112,182,196,140,64,17,
%U A115241 2,0,11,54,156,294,378,336,204,81,19,2,0,12,65,210,450,672,714,540,285,100
%N 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).
%C A115241 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 A115241 Clark Kimberling's recurrence in the previous comment yields an additional zero: 1, 0, 2, 0, 3, 2, ... - _Georg Fischer_, Sep 04 2021
%D A115241 Harry Hochstadt, The Functions of Mathematical Physics, Wiley, New York (1971), p. 170; also Dover, New York (1986), p. 170.
%F A115241 T(n,p) = (2n+p-2) * binomial(n+p-3,n-1)/n for n>=1, p>=1.
%e A115241 T(1,1)=1 corresponds to the polynomial x.
%e A115241 T(n,1)=0 for n>=2 because no polynomial in x of degree >=2 is harmonic.
%e A115241 T(1,2)=2 because we can take, for example, x and y.
%e A115241 T(2,2)=2 because we can take, for example, x^2-y^2 and xy.
%e A115241 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 A115241 The square array starts:
%e A115241   1, 2,  3,  4,   5,   6,    7, ...
%e A115241   0, 2,  5,  9,  14,  20,   27, ...
%e A115241   0, 2,  7, 16,  30,  50,   77, ...
%e A115241   0, 2,  9, 25,  55, 105,  182, ...
%e A115241   0, 2, 11, 36,  91, 196,  378, ...
%e A115241   0, 2, 13, 49, 140, 336,  714, ...
%e A115241   0, 2, 15, 64, 204, 540, 1254, ...
%e A115241   ...
%p A115241 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 A115241 seq(seq(T(n,p+1-n), n=1..p), p=1..12); # yields the sequence; _Georg Fischer_, Sep 04 2021
%t A115241 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 A115241 Cf. A097613 (diagonal terms), A209688.
%K A115241 nonn,tabl
%O A115241 1,2
%A A115241 _Roger L. Bagula_, Mar 04 2006
%E A115241 Edited by _N. J. A. Sloane_, Mar 07 2006