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A211229
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Matrix inverse of lower triangular array A211226.
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1
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1, -1, 1, 0, -1, 1, 0, 0, -1, 1, 1, 0, 0, -2, 1, -1, 1, 0, 0, -1, 1, 2, -3, 3, 0, 0, -3, 1, -2, 2, -3, 3, 0, 0, -1, 1, 9, -8, 8, -12, 6, 0, 0, -4, 1, -9, 9, -8, 8, -6, 6, 0, 0, -1, 1, 44, -45, 45, -40, 20, -30, 10, 0, 0, -5, 1, -44, 44, -45, 45, -20, 20, -10
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OFFSET
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0,14
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COMMENTS
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This triangle is connected to the derangement numbers. The subtriangles (T(2*n,2*k))n,k>=0, -(T(2*n+1,2*k))n,k>=0, and (T(2*n+1,2*k+1))n,k>=0 are all equal to A008290, whilst the subtriangle (T(2*n,2*k+1))n,k>=0 equals -A180188 (with an extra initial row of zeros).
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LINKS
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Table of n, a(n) for n=0..72.
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FORMULA
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T(2*n,2*k) = T(2*n+1,2*k+1) = -T(2*n+1,2*k) = binomial(n,k)*A000166(n-k) = n!/k!*sum {i = 0..n-k} (-1)^i/i!;
T(2*n,2*k+1) = -n*binomial(n-1,k)*A000166(n-k-1) = -n!/k!*sum {i = 0..n-k-1} (-1)^i/i!.
T(n,k) = T(n-k,0)*A211226(n,k).
Column entries:
T(2*n,0) = A000166(n), T(2*n,2) = A000240(n), T(2*n,4) = A000387(n), T(2*n,6) = A000449(n), T(2*n,8) = A000475(n).
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EXAMPLE
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Triangle begins
.n\k.|....0....1....2....3....4....5....6....7....8....9
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
..0..|....1
..1..|...-1....1
..2..|....0...-1....1
..3..|....0....0...-1....1
..4..|....1....0....0...-2....1
..5..|...-1....1....0....0...-1....1
..6..|....2...-3....3....0....0...-3....1
..7..|...-2....2...-3....3....0....0...-1....1
..8..|....9...-8....8..-12....6....0....0...-4....1
..9..|...-9....9...-8....8...-6....6....0....0...-1....1
...
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CROSSREFS
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Cf. A000166, A000240, A000387, A000449, A000475, A008290, A180188, A211226.
Sequence in context: A005091 A086831 A191340 * A111405 A089053 A214979
Adjacent sequences: A211226 A211227 A211228 * A211230 A211231 A211232
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KEYWORD
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sign,easy,tabl
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AUTHOR
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Peter Bala, Apr 05 2012
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STATUS
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approved
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