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A094791
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Triangle read by rows giving coefficients of polynomials arising in successive differences of (n!)_{n>=0}.
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5
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1, 1, 0, 1, 1, 1, 1, 3, 5, 2, 1, 6, 17, 20, 9, 1, 10, 45, 100, 109, 44, 1, 15, 100, 355, 694, 689, 265, 1, 21, 196, 1015, 3094, 5453, 5053, 1854, 1, 28, 350, 2492, 10899, 29596, 48082, 42048, 14833, 1, 36, 582, 5460, 32403, 124908, 309602, 470328, 391641, 133496
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OFFSET
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0,8
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COMMENTS
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Let D_0(n)=n! and D_{k+1}(n)=D_{k}(n+1)-D_{k}(n), then D_{k}(n)=n!*P_{k}(n) where P_{k} is a polynomial with integer coefficients of degree k.
The horizontal reversal of this triangle arises as a binomial convolution of the derangements coefficients der(n,i) (numbers of permutations of size n with i derangements = A098825(n,i) = number of permutations of size n with n-i rencontres = A008290(n,n-i), see formula section). - Olivier Gérard, Jul 31 2011
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LINKS
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FORMULA
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Sum_{k=0..n} T(n,n-k)*x^k = Sum_{i=0..n} der(n,i)*binomial( n+x, i) (an analog of Worpitzky's identity). - Olivier Gérard, Jul 31 2011
The n-th row polynomial R(n,x) = Sum _{k = 0..n} T(n,k)*x^k is P-recursive in the variable x: x*R(n,x) = (x+n+1)*R(n,x-1) - R(n,x-2). - Peter Bala, Jul 25 2021
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EXAMPLE
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D_3(n) = n!*(n^3 + 3*n^2 + 5*n + 2).
D_4(n) = n!*(n^4 + 6*n^3 + 17*n^2 + 20*n + 9).
Table begins:
1
1 0
1 1 1
1 3 5 2
1 6 17 20 9
1 10 45 100 109 44
1 15 100 355 694 689 265
...
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MAPLE
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with(LREtools): A094791_row := proc(n)
delta(x!, x, n); simplify(%/x!); seq(coeff(%, x, n-j), j=0..n) end:
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MATHEMATICA
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d[0][n_] := n!; d[k_][n_] := d[k][n] = d[k - 1][n + 1] - d[k - 1][n] // FullSimplify;
row[k_] := d[k][n]/n! // FullSimplify // CoefficientList[#, n]& // Reverse;
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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Edited and T(0,0) corrected according to the author's definition by Olivier Gérard, Jul 31 2011
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STATUS
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approved
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