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A051632
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Rows of triangle formed using Pascal's rule except we begin and end the n-th row with n-2.
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1
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-2, -1, -1, 0, -2, 0, 1, -2, -2, 1, 2, -1, -4, -1, 2, 3, 1, -5, -5, 1, 3, 4, 4, -4, -10, -4, 4, 4, 5, 8, 0, -14, -14, 0, 8, 5, 6, 13, 8, -14, -28, -14, 8, 13, 6, 7, 19, 21, -6, -42, -42, -6, 21, 19, 7, 8, 26, 40, 15, -48, -84, -48, 15, 40, 26, 8, 9, 34, 66, 55, -33, -132, -132
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OFFSET
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0,1
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COMMENTS
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Row sums are -2.
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LINKS
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FORMULA
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-t(n,k)=((2*k + 1 - n)/(k + 1))*Binomial[n, k] + ((1 - n + 2 (-k + n))/(1 - k + n)) Binomial[n, -k + n]. [From Roger L. Bagula, Feb 17 2009]
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EXAMPLE
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Contribution from Roger L. Bagula, Feb 17 2009: (Start)
The rows of the triangle, negated, are:
{2},
{1, 1},
{0, 2, 0},
{-1, 2, 2, -1},
{-2, 1, 4, 1, -2},
{-3, -1, 5,5, -1, -3},
{-4, -4, 4, 10, 4, -4, -4},
{-5, -8, 0, 14, 14, 0, -8, -5},
{-6, -13, -8, 14, 28, 14, -8, -13, -6},
{-7, -19, -21, 6, 42,42, 6, -21, -19, -7},
{-8, -26, -40, -15, 48, 84, 48, -15, -40, -26, -8} (End)
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MATHEMATICA
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t[n_, k_] = ((2*k + 1 - n)/(k + 1))*Binomial[n, k] + ((1 - n + 2 (-k + n))/(1 - k + n)) Binomial[n, -k + n];
Table[Table[t[n, k], {k, 0, n}], {n, 0, 10}];
Flatten[%] (* Roger L. Bagula , Feb 17 2009 *)
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PROG
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(Haskell)
a051632 n k = a051632_tabl !! n !! k
a051632_list = concat a051632_tabl
a051632_tabl = iterate (\rs -> zipWith (+) ([1] ++ rs) (rs ++ [1])) [-2]
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CROSSREFS
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A156644 [From Roger L. Bagula, Feb 17 2009]
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KEYWORD
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AUTHOR
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STATUS
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approved
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