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%I #18 Jan 29 2023 10:53:27
%S 1,0,1,0,-2,3,0,3,-18,16,0,-4,72,-192,125,0,5,-240,1440,-2500,1296,0,
%T -6,720,-8640,30000,-38880,16807,0,7,-2016,45360,-280000,680400,
%U -705894,262144,0,-8,5376,-217728,2240000,-9072000,16941456,-14680064,4782969
%N Triangle read by rows: terms of a binomial decomposition of 1 as Sum(k=0..n)T(n,k).
%C T(n,k)=(1+k)^(k-1)*(-k)^(n-k)*binomial(n,k) for k>0, while T(n,0)=0^n by convention.
%C Sequence A161628, arising from a different context, appears to be the same, but with opposite signs of odd rows.
%H Stanislav Sykora, <a href="/A244119/b244119.txt">Table of n, a(n) for rows 0..100</a>
%H S. Sykora, <a href="http://dx.doi.org/10.3247/SL5Math14.004">An Abel's Identity and its Corollaries</a>, Stan's Library, Volume V, 2014, DOI 10.3247/SL5Math14.004. See eq.(4) with b=-1.
%e First rows of the triangle, all summing up to 1:
%e 1
%e 0 1
%e 0 -2 3
%e 0 3 -18 16
%e 0 -4 72 -192 125
%e 0 5 -240 1440 -2500 1296
%p A244119 := (n, k) -> (1+k)^(k-1)*(-k)^(n-k)*binomial(n,k):
%p seq(seq(A244119(n, k), k = 0..n), n = 0..8); # _Peter Luschny_, Jan 29 2023
%o (PARI) seq(nmax,b)={my(v,n,k,irow);
%o v = vector((nmax+1)*(nmax+2)/2);v[1]=1;
%o for(n=1,nmax,irow=1+n*(n+1)/2;v[irow]=0;
%o for(k=1,n,v[irow+k]=(1-k*b)^(k-1)*(k*b)^(n-k)*binomial(n,k););
%o );return(v);}
%o a=seq(100,-1);
%Y Cf. A161628, A244116, A244117, A244118, A244120, A244121, A244122, A244123, A244124, A244125, A244126, A244127, A244128, A244129, A244130, A244131, A244132, A244133, A244134, A244135, A244136, A244137, A244138, A244139, A244140, A244141, A244142, A244143.
%Y Cf. A273954.
%K sign,tabl
%O 0,5
%A _Stanislav Sykora_, Jun 21 2014
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