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%I #16 May 26 2019 14:55:03
%S 1,0,1,0,0,2,0,0,1,4,0,0,2,5,8,0,0,6,15,17,16,0,0,24,62,68,49,32,0,0,
%T 120,322,359,243,129,64,0,0,720,2004,2308,1553,756,321,128,0,0,5040,
%U 14508,17332,11903,5622,2151,769,256,0,0,40320,119664
%N Triangle T(n,k), read by rows, given by [0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] DELTA [1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938.
%C Let R(m,n,k), 0<=k<=n, the Riordan array (1, x*g(x)) where g(x) is g.f. of the m-fold factorials . Then Sum_{k, 0<=k<=n} = R(m,n,k) = Sum_{k, 0<=k<=n} T(n,k)*m^(n-k).
%C For m = -1, R(-1,n,k) is A026729(n,k).
%C For m = 0, R(0,n,k) is A097805(n,k).
%C For m = 1, R(1,n,k) is A084938(n,k).
%C For m = 2, R(2,n,k) is A111106(n,k).
%H Alois P. Heinz, <a href="/A111146/b111146.txt">Rows n = 0..140, flattened</a>
%F Sum_{k, 0<=k<=n} (-1)^(n-k)*T(n, k) = A000045(n+1), Fibonacci numbers.
%F Sum_{k, 0<=k<=n} T(n, k) = A051295(n).
%F Sum_{k, 0<=k<=n} 2^(n-k)*T(n, k) = A112934(n).
%F T(0, 0) = 1, T(n, n) = 2^(n-1).
%F G.f.: A(x, y) = 1/(1 - x*y*Sum_{j>=0} (y-1+j)!/(y-1)!*x^j ). - _Paul D. Hanna_, Oct 26 2005
%e Triangle begins:
%e .1;
%e .0, 1;
%e .0, 0, 2;
%e .0, 0, 1, 4;
%e .0, 0, 2, 5, 8;
%e .0, 0, 6, 15, 17, 16;
%e .0, 0, 24, 62, 68, 49, 32;
%e .0, 0, 120, 322, 359, 243, 129, 64;
%e .0, 0, 720, 2004, 2308, 1553, 756, 321, 128;
%e .0, 0, 5040, 14508, 17332, 11903, 5622, 2151, 769, 256;
%e .0, 0, 40320, 119664, 148232, 105048, 49840, 18066, 5756, 1793, 512;
%e ....................................................................
%e At y=2: Sum_{k=0..n} 2^k*T(n,k) = A113327(n) where (1 + 2*x + 8*x^2 + 36*x^3 +...+ A113327(n)*x^n +..) = 1/(1 - 2/1!*x*(1! + 2!*x + 3!*x^2 + 4!*x^3 +..) ).
%e At y=3: Sum_{k=0..n} 3^k*T(n,k) = A113328(n) where (1 + 3*x + 18*x^2 + 117*x^3 +...+ A113328(n)*x^n +..) = 1/(1 - 3/2!*x*(2! + 3!*x + 4!*x^2 + 5!*x^3 +..) ).
%e At y=4: Sum_{k=0..n} 4^k*T(n,k) = A113329(n) where (1 + 4*x + 32*x^2 + 272*x^3 +...+ A113329(n)*x^n +..) = 1/(1 - 4/3!*x*(3! + 4!*x + 5!*x^2 + 6!*x^3 +..) ).
%t T[n_, k_] := Module[{x = X + X*O[X]^n, y = Y + Y*O[Y]^k}, A = 1/(1 - x*y*Sum[x^j*Product[y + i, {i, 0, j - 1}], {j, 0, n}]); Coefficient[ Coefficient[A, X, n], Y, k]];
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, May 26 2019, from PARI *)
%o (PARI) {T(n,k)=local(x=X+X*O(X^n),y=Y+Y*O(Y^k)); A=1/(1-x*y*sum(j=0,n,x^j*prod(i=0,j-1,y+i))); return(polcoeff(polcoeff(A,n,X),k,Y))} (Hanna)
%Y Cf. A000045, A026729, A051295, A084938, A097805, A111106, A112934.
%Y Cf. m-fold factorials : A000142, A001147, A007559, A007696, A008548, A008542.
%Y Cf. A113326, A113327 (y=2), A113328 (y=3), A113329 (y=4), A113330 (y=5), A113331 (y=6).
%K easy,nonn,tabl
%O 0,6
%A _Philippe Deléham_, Oct 19 2005
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