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%I #8 Jun 25 2014 09:42:32
%S 1,0,1,0,2,2,0,12,6,9,0,108,48,36,64,0,1280,540,360,320,625,0,18750,
%T 7680,4860,3840,3750,7776,0,326592,131250,80640,60480,52500,54432,
%U 117649,0,6588344,2612736,1575000,1146880,945000,870912,941192,2097152
%N Triangle read by rows: terms T(n,k) of a binomial decomposition of n^n as Sum(k=0..n)T(n,k).
%C T(n,k)=(k)^(k-1)*(n-k)^(n-k)*binomial(n,k) for k>0, while T(n,0)=0^n by convention.
%C There are many binomial decompositions of n^n, some with all terms positive like this one (see A243203). However, for every n, the terms corresponding to k=1..n in this one are exceptionally similar in value (at least on log scale).
%H Stanislav Sykora, <a href="/A244137/b244137.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.(13), with b=-1.
%e First rows of the triangle, all summing up to n^n:
%e 1,
%e 0, 1,
%e 0, 2, 2,
%e 0, 12, 6, 9,
%e 0, 108, 48, 36, 64,
%e 0, 1280, 540, 360, 320, 625,
%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]=(-k*b)^(k-1)*(n+k*b)^(n-k)*binomial(n, k); ); );
%o return(v); }
%o a=seq(100,-1);
%Y Cf. A243203, A244116, A244117, A244118, A244119, A244120, A244121, A244122, A244123, A244124, A244125, A244126, A244127, A244128, A244129, A244130, A244131, A244132, A244133, A244134, A244135, A244136, A244138, A244139, A244140, A244141, A244142, A244143.
%K nonn,tabl
%O 0,5
%A _Stanislav Sykora_, Jun 22 2014
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