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A144431
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A sub-Pascal recursion triangle sequence: m=-1; A(n,k) := (m*n - m*k + 1)A(n - 1, k - 1) + (m*k - (m - 1))A(n - 1, k).
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0
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1, 1, 1, 1, 0, 1, 1, -1, -1, 1, 1, -2, 2, -2, 1, 1, -3, 2, 2, -3, 1, 1, -4, 7, -8, 7, -4, 1, 1, -5, 9, -5, -5, 9, -5, 1, 1, -6, 16, -26, 30, -26, 16, -6, 1, 1, -7, 20, -28, 14, 14, -28, 20, -7, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,12
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COMMENTS
| Row sums are:{1, 2, 2, 0, 0, 0, 0, 0, 0, 0}.
m=0 is the Pascal sequence, so m=-1 is sub-Pascal.
The triangle starts off like A098593, but is different further on.
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FORMULA
| m=-1; A(n,k) := (m*n - m*k + 1)A(n - 1, k - 1) + (m*k - (m - 1))A(n - 1, k).
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EXAMPLE
| {1},
{1, 1},
{1, 0, 1},
{1, -1, -1, 1},
{1, -2, 2, -2, 1},
{1, -3, 2, 2, -3, 1},
{1, -4, 7, -8, 7, -4, 1},
{1, -5, 9, -5, -5, 9, -5, 1},
{1, -6, 16, -26, 30, -26, 16, -6, 1},
{1, -7, 20, -28, 14, 14, -28, 20, -7, 1
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MATHEMATICA
| m=-1; A[n_, 1] := 1; A[n_, n_] := 1; A[n_, k_] := (m*n - m*k + 1)A[n - 1, k - 1] + (m*k - (m - 1))A[n - 1, k]; a = Table[A[n, k], {n, 10}, {k, n}]; Flatten[a]
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CROSSREFS
| Cf. A098593.
Sequence in context: A104754 A206827 A098593 * A053821 A076545 A162246
Adjacent sequences: A144428 A144429 A144430 * A144432 A144433 A144434
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KEYWORD
| uned,sign
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AUTHOR
| Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Oct 04 2008
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