A111146 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.
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, 120, 322, 359, 243, 129, 64, 0, 0, 720, 2004, 2308, 1553, 756, 321, 128, 0, 0, 5040, 14508, 17332, 11903, 5622, 2151, 769, 256, 0, 0, 40320, 119664
Offset: 0
Examples
Triangle begins: .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, 120, 322, 359, 243, 129, 64; .0, 0, 720, 2004, 2308, 1553, 756, 321, 128; .0, 0, 5040, 14508, 17332, 11903, 5622, 2151, 769, 256; .0, 0, 40320, 119664, 148232, 105048, 49840, 18066, 5756, 1793, 512; .................................................................... 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 +..) ). 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 +..) ). 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 +..) ).
Links
- Alois P. Heinz, Rows n = 0..140, flattened
Crossrefs
Programs
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Mathematica
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]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 26 2019, from PARI *)
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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)
Formula
Sum_{k, 0<=k<=n} (-1)^(n-k)*T(n, k) = A000045(n+1), Fibonacci numbers.
Sum_{k, 0<=k<=n} T(n, k) = A051295(n).
Sum_{k, 0<=k<=n} 2^(n-k)*T(n, k) = A112934(n).
T(0, 0) = 1, T(n, n) = 2^(n-1).
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
Comments