A145142 Triangle T(n,k), n>=1, 0<=k<=n-1, read by rows: T(n,k)/(n-1)! is the coefficient of x^k in polynomial p_n for the n-th row sequence of A145153.
1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 24, 6, 11, 6, 1, 120, 144, 50, 35, 10, 1, 720, 1200, 634, 225, 85, 15, 1, 5040, 9960, 6804, 2464, 735, 175, 21, 1, 80640, 89040, 71868, 29932, 8449, 1960, 322, 28, 1, 1088640, 1231776, 789984, 375164, 112644, 25473, 4536, 546, 36, 1
Offset: 1
Examples
Triangle begins: 1; 0, 1; 0, 1, 1; 0, 2, 3, 1; 24, 6, 11, 6, 1; 120, 144, 50, 35, 10, 1;
Links
- Alois P. Heinz, Rows n = 1..45, flattened
Crossrefs
Row sums are in A052593.
Programs
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Maple
row:= proc(n) option remember; local f,i,x; f:= unapply(simplify(sum('cat(a||i) *x^i', 'i'=0..n-1) ), x); unapply(subs(solve({seq(f(i+1)= coeftayl(x/ (1-x-x^4)/ (1-x)^i, x=0, n), i=0..n-1)}, {seq(cat(a||i), i=0..n-1)}), sum('cat(a||i) *x^i', 'i'=0..n-1) ), x); end: T:= (n,k)-> `if`(k<0 or k>=n,0, coeff(row(n)(x),x,k)*(n-1)!): seq(seq(T(n,k), k=0..n-1), n=1..12);
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Mathematica
row[n_] := Module[{f, eq}, f = Function[x, Sum[a[k]*x^k, {k, 0, n-1}]]; eq = Table[f[k+1] == SeriesCoefficient[x/(1-x-x^4)/(1-x)^k, {x, 0, n}], {k, 0, n-1}]; Table[a[k], {k, 0, n-1}] /. Solve[eq] // First]; Table[row[n]*(n-1)!, {n, 1, 12}] // Flatten (* Jean-François Alcover, Feb 04 2014, after Alois P. Heinz *)
Formula
See program.