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A134431
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Triangle read by rows: T(n,k) is the number of arrangements of the set {1,2,...,n} in which the sum of the entries is equal to k (n >= 0, k >= 0; to n=0 there corresponds the empty set).
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2
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1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 6, 1, 1, 1, 3, 3, 4, 8, 8, 6, 6, 24, 1, 1, 1, 3, 3, 5, 10, 10, 14, 14, 36, 30, 30, 24, 24, 120, 1, 1, 1, 3, 3, 5, 11, 12, 16, 22, 44, 44, 66, 60, 78, 174, 168, 144, 144, 120, 120, 720, 1, 1, 1, 3, 3, 5, 11, 13, 18, 24, 52, 52, 80, 98, 120, 234
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OFFSET
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0,7
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COMMENTS
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Row n has 1 + n(n+1)/2 terms (n >= 0). Row sums yield the arrangement numbers (A000522). T(n, n(n+1)/2) = n!. Sum_{k=0..n(n+1)/2} k*T(n,k) = A134432(n).
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LINKS
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FORMULA
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The row generating polynomials P[n](t) are equal to Q[n](t,1), where the polynomials Q[n](t,x) are defined by Q[0]=1 and Q[n]=Q[n-1] + xt^n (d/dx)xQ[n-1]. [Q[n](t,x) is the bivariate generating polynomial of the arrangements of {1,2,...,n}, where t (x) marks the sum (number) of the entries; for example, Q[2](t,x)=1+tx + t^2*x + 2t^3*x^2, corresponding to: empty, 1, 2, 12 and 21, respectively.]
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EXAMPLE
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T(4,7)=8 because we have 34,43 and the six permutations of {1,2,4}.
Triangle starts:
1;
1, 1;
1, 1, 1, 2;
1, 1, 1, 3, 2, 2, 6;
1, 1, 1, 3, 3, 4, 8, 8, 6, 6, 24;
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MAPLE
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Q[0]:=1: for n to 7 do Q[n]:=sort(simplify(Q[n-1]+t^n*x*(diff(x*Q[n-1], x))), t) end do: for n from 0 to 7 do P[n]:=sort(subs(x=1, Q[n])) end do: for n from 0 to 7 do seq(coeff(P[n], t, j), j=0..(1/2)*n*(n+1)) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n, s, t) option remember;
`if`(n=0, t!*x^s, b(n-1, s, t)+b(n-1, s+n, t+1))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
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MATHEMATICA
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b[n_, s_, t_] := b[n, s, t] = If[n == 0, t!*x^s, b[n - 1, s, t] + b[n - 1, s + n, t + 1]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]] @ b[n, 0, 0];
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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