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%I #22 Jun 12 2013 16:36:46
%S 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,
%T 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,
%U 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
%N Pascal's triangle with row n reduced modulo n+1.
%C The row sums are: {0, 2, 4, 8, 11, 20, 22, 32, 31, 52, 56, ...}.
%C 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
%H Reinhard Zumkeller, <a href="/A225043/b225043.txt">Rows n = 0..120 of triangle, flattened</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BinomialDistribution.html">Binomial Distribution</a>
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F T(m,n) = binomial(m, n) mod m+1.
%e {0},
%e {1, 1},
%e {1, 2, 1},
%e {1, 3, 3, 1},
%e {1, 4, 1, 4, 1},
%e {1, 5, 4, 4, 5, 1},
%e {1, 6, 1, 6, 1, 6, 1},
%e {1, 7, 5, 3, 3, 5, 7, 1},
%e {1, 8, 1, 2, 7, 2, 1, 8, 1},
%e {1, 9, 6, 4, 6, 6, 4, 6, 9, 1},
%e {1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1},
%e {1, 11, 7, 9, 6, 6, 6, 6, 9, 7, 11, 1},
%e {1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1},...
%t Flatten[Table[Mod[Binomial[m, n], m + 1], {m, 0, 12}, {n, 0, m}]]
%o (PARI) T(m,n)=binomial(m,n)%(m+1) \\ _Charles R Greathouse IV_, Apr 25 2013
%o (Haskell)
%o a225043 n k = a225043_tabl !! n !! k
%o a225043_row n = a225043_tabl !! n
%o a225043_tabl = zipWith (map . flip mod) [1..] a007318_tabl
%o -- _Reinhard Zumkeller_, Jun 12 2013
%Y Cf. A007318, A062173, A001803, A086117.
%K nonn,tabl
%O 0,5
%A _Roger L. Bagula_, Apr 25 2013
%E Definition edited by _N. J. A. Sloane_, Apr 28 2013
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