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%I #23 Jul 21 2021 10:01:59
%S 1,0,-1,-1,-1,1,-1,1,2,-1,0,3,0,-3,1,1,2,-5,-2,4,-1,1,-2,-7,6,5,-5,1,
%T 0,-5,0,15,-5,-9,6,-1,-1,-3,12,9,-25,1,14,-7,1,-1,3,15,-18,-29,35,7,
%U -20,8,-1,0,7,0,-42,14,63,-42,-20,27,-9,1,1,4,-22,-24,85,14,-112,42
%N Triangle T(n,k) read by rows: difference between A098489 and A098490 at triangular rows.
%C Also, coefficients of polynomials that have values in A098495 and A094954.
%H Alex Fink, Richard K. Guy, and Mark Krusemeyer, <a href="https://doi.org/10.11575/cdm.v3i2.61940">Partitions with parts occurring at most thrice</a>, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114.
%F T(n, k) = A098489[n(n+1)/2, k] - A098490[n(n+1)/2, k].
%F Recurrence: T(n, k) = T(n-1, k)-T(n-1, k-1)-T(n-2, k); T(n, k)=0 for n<0, k>n, k<0; T(n, n)=(-1)^n; T(n, n-1)=(-1)^n*(1-n).
%F G.f.: (1-x)/(1+(y-1)*x+x^2). [_Vladeta Jovovic_, Dec 14 2009]
%F From _Peter Bala_, Jul 13 2021: (Start)
%F Riordan array ( (1 - x)/(1 - x + x^2), -x/(1 - x + x^2) ).
%F T(n,k) = (-1)^k * the (n,k)-th entry of Q^(-1)*P = Sum_{j = k..n} (-1)^(k+binomial(n-j+1,2))*binomial(floor((1/2)*n+(1/2)*j),j)* binomial(j,k), where P denotes Pascal's triangle A007318 and Q denotes triangle A061554 (formed from P by sorting the rows into descending order). (End)
%e Triangle begins:
%e 1;
%e 0, -1;
%e -1, -1, 1;
%e -1, 1, 2, -1;
%e 0, 3, 0, -3, 1;
%e ...
%p A098493 := proc (n, k)
%p add((-1)^(k+binomial(n-j+1,2))*binomial(floor((1/2)*n+(1/2)*j),j)* binomial(j,k), j = k..n);
%p end proc:
%p seq(seq(A098493(n, k), k = 0..n), n = 0..10); # _Peter Bala_, Jul 13 2021
%o (PARI) T(n,k)=if(k>n||k<0||n<0,0,if(k>=n-1,(-1)^n*if(k==n,1,-k),if(n==1,0,if(k==0,T(n-1,0)-T(n-2,0),T(n-1,k)-T(n-2,k)-T(n-1,k-1)))))
%Y Columns include A010892, -A076118. Diagonals include A033999, A038608, (-1)^n*A000096. Row sums are in A057077.
%Y Cf. A098494 (diagonal polynomials).
%K sign,tabl
%O 0,9
%A _Ralf Stephan_, Sep 12 2004
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