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A225043
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Pascal's triangle with row n reduced modulo n+1.
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3
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0, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 1, 4, 1, 1, 5, 4, 4, 5, 1, 1, 6, 1, 6, 1, 6, 1, 1, 7, 5, 3, 3, 5, 7, 1, 1, 8, 1, 2, 7, 2, 1, 8, 1, 1, 9, 6, 4, 6, 6, 4, 6, 9, 1, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 1, 11, 7, 9, 6, 6, 6, 6, 9, 7, 11, 1, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1
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refs;
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OFFSET
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0,5
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COMMENTS
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The row sums are: {0, 2, 4, 8, 11, 20, 22, 32, 31, 52, 56, ...}.
Since row n is only defined mod n+1, it would seem better to reduce the row sums mod n+1, which gives A062173. - N. J. A. Sloane, Apr 28 2013
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LINKS
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FORMULA
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T(m,n) = binomial(m, n) mod m+1.
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EXAMPLE
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{0},
{1, 1},
{1, 2, 1},
{1, 3, 3, 1},
{1, 4, 1, 4, 1},
{1, 5, 4, 4, 5, 1},
{1, 6, 1, 6, 1, 6, 1},
{1, 7, 5, 3, 3, 5, 7, 1},
{1, 8, 1, 2, 7, 2, 1, 8, 1},
{1, 9, 6, 4, 6, 6, 4, 6, 9, 1},
{1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1},
{1, 11, 7, 9, 6, 6, 6, 6, 9, 7, 11, 1},
{1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1},...
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MATHEMATICA
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Flatten[Table[Mod[Binomial[m, n], m + 1], {m, 0, 12}, {n, 0, m}]]
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PROG
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(Haskell)
a225043 n k = a225043_tabl !! n !! k
a225043_row n = a225043_tabl !! n
a225043_tabl = zipWith (map . flip mod) [1..] a007318_tabl
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CROSSREFS
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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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