|
%I #10 Nov 17 2017 19:58:22
%S 1,3,1,15,6,1,108,45,9,1,1029,432,90,12,1,12288,5145,1080,150,15,1,
%T 177147,73728,15435,2160,225,18,1,3000000,1240029,258048,36015,3780,
%U 315,21,1,58461513,24000000,4960116,688128,72030,6048,420,24,1,1289945088
%N Triangle, read by rows, of coefficients for the third iteration of the hyperbinomial transform.
%C Equals the matrix cube of A088956 when treated as a lower triangular matrix. The 3rd hyperbinomial transform of a sequence {b} is defined to be the sequence {d} given by d(n) = sum(k=0..n, T(n,k)*b(k)), where T(n,k) = 3*(n-k+3)^(n-k-1)*C(n,k). Given a table in which the n-th row is the n-th binomial transform of the first row, then the 3rd hyperbinomial transform of any diagonal results in the 3rd diagonal lower in the table.
%H G. C. Greubel, <a href="/A089463/b089463.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F T(n, k) = 3*(n-k+3)^(n-k-1)*C(n, k).
%F E.g.f.: exp(x*y)*(-LambertW(-y)/y)^3.
%F Note: (-LambertW(-y)/y)^3 = sum(n>=0, 3*(n+3)^(n-1)*y^n/n!).
%e Rows begin:
%e {1},
%e {3,1},
%e {15,6,1},
%e {108,45,9,1},
%e {1029,432,90,12,1},
%e {12288,5145,1080,150,15,1},
%e {177147,73728,15435,2160,225,18,1},
%e {3000000,1240029,258048,36015,3780,315,21,1},..
%t Flatten[Table[3(n-k+3)^(n-k-1) Binomial[n,k],{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, Jun 26 2013 *)
%o (PARI) for(n=0,10, for(k=0,n, print1(3*(n-k+3)^(n-k-1)*binomial(n,k), ", "))) \\ _G. C. Greubel_, Nov 17 2017
%Y Cf. A089464(row sums), A089465(diagonal), A089460, A088956.
%K nonn,tabl
%O 0,2
%A _Paul D. Hanna_, Nov 05 2003
|
