Search: id:A154353
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%I A154353
%S A154353 1,1,5,15,5,16,101,101,16,42,483,1008,483,42,99,1926,7197,7197,1926,99,
%T A154353 219,6912,42549,75645,42549,6912,219,466,23272,224068,647239,647239,
%U A154353 224068,23272,466,968,75306,1094544,4847007,7830372,4847007,1094544
%N A154353 A difference triangular sequence: t(n,m)=(Eulerian[n,m]-Binomial[n,m]^2)/
               2: starting at n=4 and m={2,n-1}.
%C A154353 Row sums are:
%C A154353 {2, 25, 234, 2058, 18444, 175005, 1790090, 19866022, 239148084, 3112158322,
%C A154353 43583945300,...}.
%C A154353 Noticing the Eulerian numbers and the binomial squared were the same 
               for the first four rows,
%C A154353 I subtracted them and extracted the zeros to get this sequence.
%C A154353 The resulting fractal can be seen as:
%C A154353 a = Table[Table[(CoefficientList[FullSimplify[ExpandAll[p[x, n]]],x][[m]] 
               - Binomial[n - 1,m - 1]^2)/2, {m, 2, n - 1}], {n, 4, 34}];
%C A154353 b = Table[If[m <= n + 1, Mod[a[[n]][[m]], 2], 0], {m, 1, Length[a]}, 
               {n, 1, Length[a]}];
%C A154353 ListDensityPlot[b, Mesh -> False]
%F A154353 t(n,m)=(Eulerian[n,m]-Binomial[n,m]^2)/2:
%F A154353 starting at n=4 and m={2,n-1}.
%e A154353 {1, 1},
%e A154353 {5, 15, 5},
%e A154353 {16, 101, 101, 16},
%e A154353 {42, 483, 1008, 483, 42},
%e A154353 {99, 1926, 7197, 7197, 1926, 99},
%e A154353 {219, 6912, 42549, 75645, 42549, 6912, 219},
%e A154353 {466, 23272, 224068, 647239, 647239, 224068, 23272, 466},
%e A154353 {968, 75306, 1094544, 4847007, 7830372, 4847007, 1094544, 75306, 968},
%e A154353 {1981, 237623, 5080230, 33104787, 81149421, 81149421, 33104787, 5080230, 
               237623, 1981},
%e A154353 {4017, 737685, 22742525, 211518255, 752497122, 1137159114, 752497122, 
               211518255, 22742525, 737685, 4017},
%e A154353 {8100, 2265615, 99164495, 1285615475, 6420803247, 13984115718, 13984115718, 
               6420803247, 1285615475, 99164495, 2265615, 8100}
%t A154353 Clear[p, x, n]; p[x_, n_] = (x - 1)^(n + 1)*Sum[((-1)^(n + 1)*k^n)*x^k, 
               {k, 0, Infinity}]/x;
%t A154353 Table[Table[(CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x][[m]] 
               - Binomial[n - 1, m - 1]^2)/2, {m, 2, n - 1}], {n, 4, 14}];
%t A154353 Flatten[%]
%Y A154353 Sequence in context: A082269 A107776 A161202 this_sequence A114332 A077348 
               A113259
%Y A154353 Adjacent sequences: A154350 A154351 A154352 this_sequence A154354 A154355 
               A154356
%K A154353 nonn,uned,tabf
%O A154353 4,3
%A A154353 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 07 2009

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