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A145142
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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.
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18
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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
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OFFSET
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1,8
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LINKS
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FORMULA
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See program.
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EXAMPLE
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Triangle begins:
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
24, 6, 11, 6, 1;
120, 144, 50, 35, 10, 1;
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MAPLE
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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
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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 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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