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%I #2 Mar 30 2012 17:34:41
%S 1,1,1,1,5,1,1,18,18,1,1,52,144,52,1,1,131,766,766,131,1,1,303,3273,
%T 6743,3273,303,1,1,664,12312,45422,45422,12312,664,1,1,1406,42844,
%U 261230,463348,261230,42844,1406,1,1,2913,141936,1358100,3915312
%N A symmetrical triangle between the Eulerian numbers, A008292, and MacMahon numbers, A060187, by linear combination with Pascal(A007318), Eulerian numbers ,and Narayana numbers,A001263,:k=3;t(n,m,k)=Binomial[n, m] - k*(Binomial[n, m]*Binomial[n + 1, m]/(m + 1)) + k*Eulerian[n + 1, m]
%C Row sums are:
%C {1, 2, 7, 38, 250, 1796, 13897, 116798, 1074310, 10836524, 119575066,...}.
%C By solving the linear combination of Pascal,Narayana and Eulerian to give
%C the MacMahon at k=4, I got this functional set of symmetrical triangles.
%C By modulo two pattern the k=3 set is Narayana numbers like and the k=6 ,{1.8.1} level,
%C is a new even Sierpinski type.
%F k=3;
%F t(n,m,k)=Binomial[n, m] - k*(Binomial[n, m]*Binomial[n + 1, m]/(m + 1)) + k*Eulerian[n + 1, m]
%e {1},
%e {1, 1},
%e {1, 5, 1},
%e {1, 18, 18, 1},
%e {1, 52, 144, 52, 1},
%e {1, 131, 766, 766, 131, 1},
%e {1, 303, 3273, 6743, 3273, 303, 1},
%e {1, 664, 12312, 45422, 45422, 12312, 664, 1},
%e {1, 1406, 42844, 261230, 463348, 261230, 42844, 1406, 1},
%e {1, 2913, 141936, 1358100, 3915312, 3915312, 1358100, 141936, 2913, 1},
%e {1, 5953, 455481, 6595734, 29172972, 47114784, 29172972, 6595734, 455481, 5953, 1}
%t << DiscreteMath`Combinatorica`
%t t[n_, m_, k_] = Binomial[n, m] - k*(Binomial[n, m]* Binomial[n + 1, m]/(m + 1)) + k*Eulerian[n + 1, m];
%t Table[Flatten[Table[Table[t[n, m, k], {m, 0, n}], {n, 0, 10}]], {k, 0, 10}]
%Y Cf. A007318, A008292, A001263, A060187
%K nonn,tabl,uned
%O 0,5
%A _Roger L. Bagula_, May 25 2010
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