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Triangle T(n,k) = Sum_{i=0..k} (-1)^(i+k)*binomial(k,i)*Sum_{j=0..n} (i+1)^j*(3n-3j+1) read by rows.
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%I #9 Feb 13 2022 23:16:16

%S 1,5,1,12,7,2,22,26,20,6,35,74,112,84,24,51,183,484,672,456,120,70,

%T 417,1818,4140,4968,3000,720,92,904,6288,22014,41400,42840,23040,5040,

%U 117,1900,20672,106920,295056,464040,418320,201600,40320,145,3917,65816,489696,1902960

%N Triangle T(n,k) = Sum_{i=0..k} (-1)^(i+k)*binomial(k,i)*Sum_{j=0..n} (i+1)^j*(3n-3j+1) read by rows.

%C Start from an array A(n,i) = Sum_{j=0..n} (i+1)^j*(3n-3j+1) which contains A000012

%C in row n=0, A000027 starting from 5 in row n=1, A117950 starting from 12 in row n=2 etc.

%C T(n,.) is obtained by computing the inverse binomial transform of row n of A(.,.).

%C Since A(n,i) is a polynomial in i, the inverse binomial transform is essentially finite; all trailing zeros are discarded while building the triangle.

%F T(n,0) = A000326(n+1).

%F T(n,n) = n! = A000142(n).

%F T(n,k) = Sum_{j=0..n} A028246(j+1,k+1)*(3*n-3*j+1). - _R. J. Mathar_, Mar 27 2010

%e First few rows of the array A(.,.):

%e 1, 1, 1, 1, 1, ...

%e 5, 6, 7, 8, 9, ...

%e 12, 19, 28, 39, 52, ...

%e ...

%e such that for example the inverse binomial transform of 12, 19, 28, ... becomes row n=2 of the triangle: 12, 7, 2, (0, 0, 0, 0, ...).

%e First few rows of the triangle T(n,k):

%e 1,

%e 5, 1;

%e 12, 7, 2;

%e 22, 26, 20, 6;

%e 35, 74, 112, 84, 24;

%e 51, 183, 484, 672, 456, 120;

%e ...

%p A := proc(n,i) add( (i+1)^j*(3*n-3*j+1),j=0..n) ; end proc:

%p A116923 := proc(n,m) add((-1)^(i+m)*binomial(m,i)*A(n,i),i=0..m) ; end proc:

%p seq(seq(A116923(n,k),k=0..n),n=0..15) ; # _R. J. Mathar_, Mar 27 2010

%Y Cf. A000326.

%K nonn,tabl

%O 0,2

%A _Gary W. Adamson_, Feb 26 2006

%E Offset set to 0 and precise definition added by _R. J. Mathar_, Mar 27 2010