A107889 Triangular matrix T, read by rows, that satisfies: [T^-k](n,k) = -T(n,k-1) for n >= k > 0, or, equivalently, (column k of T^-k) = -SHIFT_LEFT(column k-1 of T) when zeros above the diagonal are ignored. Also, matrix inverse of triangle A107876.
1, -1, 1, 0, -1, 1, 0, -1, -1, 1, 0, -3, -2, -1, 1, 0, -15, -9, -3, -1, 1, 0, -106, -61, -18, -4, -1, 1, 0, -975, -550, -154, -30, -5, -1, 1, 0, -11100, -6195, -1689, -310, -45, -6, -1, 1, 0, -151148, -83837, -22518, -4005, -545, -63, -7, -1, 1, 0, -2401365, -1326923, -353211, -61686, -8105, -875, -84, -8, -1, 1
Offset: 0
Examples
G.f. for column 1: 1 = T(1,1)*(1-x)^-1 + T(2,1)*x*(1-x)^0 + T(3,1)*x^2*(1-x)^2 + T(4,1)*x^3*(1-x)^5 + T(5,1)*x^4*(1-x)^9 + T(6,1)*x^5*(1-x)^14 + ... = 1*(1-x)^-1 - 1*x*(1-x)^0 - 1*x^2*(1-x)^2 - 3*x^3*(1-x)^5 - 15*x^4*(1-x)^9 - 106*x^5*(1-x)^14 - 975*x^6*(1-x)^20 + ... G.f. for column 2: 1 = T(2,2)*(1-x)^-1 + T(3,2)*x*(1-x)^1 + T(4,2)*x^2*(1-x)^4 + T(5,2)*x^3*(1-x)^8 + T(6,2)*x^4*(1-x)^13 + T(7,2)*x^5*(1-x)^19 + ... = 1*(1-x)^-1 - 1*x*(1-x)^1 - 2*x^2*(1-x)^4 - 9*x^3*(1-x)^8 - 61*x^4*(1-x)^13 - 550*x^5*(1-x)^19 - 6195*x^6*(1-x)^26 + ... Triangle begins: 1; -1, 1; 0, -1, 1; 0, -1, -1, 1; 0, -3, -2, -1, 1; 0, -15, -9, -3, -1, 1; 0, -106, -61, -18, -4, -1, 1; 0, -975, -550, -154, -30, -5, -1, 1; 0, -11100, -6195, -1689, -310, -45, -6, -1, 1; ...
Programs
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Mathematica
max = 10; A107862 = Table[Binomial[If[n < k, 0, n*(n-1)/2-k*(k-1)/2 + n - k], n - k], {n, 0, max}, {k, 0, max}]; A107867 = Table[Binomial[If[n < k, 0, n*(n-1)/2-k*(k-1)/2 + n-k+1], n - k], {n, 0, max}, {k, 0, max}]; T = Inverse[Inverse[A107862].A107867]; Table[T[[n + 1, k + 1]], {n, 0, max}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 31 2024 *)
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PARI
{T(n,k)=polcoeff(1-sum(j=0,n-k-1, T(j+k,k)*x^j*(1-x+x*O(x^n))^(-1+(k+j)*(k+j-1)/2-k*(k-1)/2)),n-k)}
Formula
G.f. for column k: 1 = Sum_{j>=0} T(k+j, k)*x^j*(1-x)^(-1 + (k+j)*(k+j-1)/2 - k*(k-1)/2).
Comments