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A141689 Average of Eulerian numbers (A008292) and Pascal's triangle (A007318): t(n,m) = (A008292(n,m) + A007318(n,m))/2.

%I A141689 #7 Jun 05 2018 22:33:57
%S A141689 1,1,1,1,3,1,1,7,7,1,1,15,36,15,1,1,31,156,156,31,1,1,63,603,1218,603,
%T A141689 63,1,1,127,2157,7827,7827,2157,127,1,1,255,7318,44145,78130,44145,
%U A141689 7318,255,1,1,511,23938,227638,655240,655240,227638,23938,511,1
%N A141689 Average of Eulerian numbers (A008292) and Pascal's triangle (A007318): t(n,m) = (A008292(n,m) + A007318(n,m))/2.
%C A141689 Row sums are: {1, 2, 5, 16, 68, 376, 2552, 20224, 181568, 1814656, ...}.
%C A141689 If Pascal's triangle and the Eulerian numbers are both fundamental arrays, then there should be a combinatorial set "between" them.
%H A141689 G. C. Greubel, <a href="/A141689/b141689.txt">Rows n=1..100 of triangle, flattened</a>
%e A141689 {1},
%e A141689 {1, 1},
%e A141689 {1, 3, 1},
%e A141689 {1, 7, 7, 1},
%e A141689 {1, 15, 36, 15, 1},
%e A141689 {1, 31, 156, 156, 31, 1},
%e A141689 {1, 63, 603, 1218, 603, 63, 1},
%e A141689 {1, 127, 2157, 7827, 7827, 2157, 127, 1},
%e A141689 {1, 255, 7318, 44145, 78130, 44145, 7318, 255, 1},
%e A141689 {1, 511, 23938, 227638, 655240, 655240, 227638, 23938, 511, 1}
%t A141689 Table[Table[(Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}] + Binomial[n - 1, k])/2, {k, 0, n - 1}], {n, 1, 10}]; Flatten[%]
%Y A141689 Cf. A008292, A007318.
%K A141689 nonn,tabl
%O A141689 1,5
%A A141689 _Roger L. Bagula_, Sep 09 2008
%E A141689 Edited by _N. J. A. Sloane_, Dec 13 2008