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A129334
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Triangle T(n,k) read by rows: inverse of the matrix PE = exp(P)/exp(1) given in A011971.
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1
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1, -1, 1, 0, -2, 1, 1, 0, -3, 1, 1, 4, 0, -4, 1, -2, 5, 10, 0, -5, 1, -9, -12, 15, 20, 0, -6, 1, -9, -63, -42, 35, 35, 0, -7, 1, 50, -72, -252, -112, 70, 56, 0, -8, 1, 267, 450, -324, -756, -252, 126, 84, 0, -9, 1, 413, 2670, 2250, -1080, -1890, -504, 210, 120, 0, -10, 1
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OFFSET
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0,5
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COMMENTS
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The structure of the triangle is A(r,c) = A000587(1+(r-c))*binomial(r-1,c-1) where row index r and column-index c start at 1.
Row polynomials defined recursively: P(0,x) = 1, P(n+1,x) = x*P(n,x) - P(n,x+1). The polynomials appear to be irreducible. Polynomials evaluated at x = c give sequences with e.g.f. exp(1 - cx - exp(-x)).
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LINKS
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FORMULA
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Let P be the lower-triangular Pascal-matrix, PE = exp(P-I) a matrix-exponential in exact integer arithmetic (or PE = lim exp(P)/exp(1) as limit of the exponential) then A = PE^-1 and a(n) = A(n, read sequentially). - Gottfried Helms, Apr 08 2007
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EXAMPLE
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Triangle starts:
[0] 1;
[1] -1, 1;
[2] 0, -2, 1;
[3] 1, 0, -3, 1;
[4] 1, 4, 0, -4, 1;
[5] -2, 5, 10, 0, -5, 1;
[6] -9, -12, 15, 20, 0, -6, 1;
[7] -9, -63, -42, 35, 35, 0, -7, 1;
[8] 50, -72, -252, -112, 70, 56, 0, -8, 1;
[9] 267, 450, -324, -756, -252, 126, 84, 0, -9, 1;
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MAPLE
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P := proc(n, x) option remember; if n=0 then 1 else
x*P(n-1, x) - P(n-1, x+1) fi end:
aRow := n -> seq(coeff(P(n, x), x, k), k = 0..n):
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CROSSREFS
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First column is A000587 (Uppuluri Carpenter numbers) which is also the negative of the row sums (=P(n, 1)). Polynomials evaluated at 2 are A074051, at -1 A109747.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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