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A triangle sequence from a sum: t0(n,m)=(2 + PartitionsQ[n] - PartitionsQ[m] - PartitionsQ[n - m]); t1(n,k)=Sum[(-1)^j *t0[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; t(n,m)=If[n == 0, 1, t1(n, k) + t1(n, n - k)]
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%I #4 Oct 13 2012 14:55:14

%S 1,2,2,4,4,4,53,28,28,53,246,138,102,138,246,4207,1775,814,814,1775,

%T 4207,60922,25758,10035,5654,10035,25758,60922,1114857,457387,168720,

%U 63929,63929,168720,457387,1114857,10706482,4934925,2063635,791789

%N A triangle sequence from a sum: t0(n,m)=(2 + PartitionsQ[n] - PartitionsQ[m] - PartitionsQ[n - m]); t1(n,k)=Sum[(-1)^j *t0[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; t(n,m)=If[n == 0, 1, t1(n, k) + t1(n, n - k)]

%C Row sums are:

%C {1, 4, 12, 162, 870, 13592, 199084, 3609786, 37432146, 1169080936,

%C 28322293344,...}.

%C This sequence is equivalent to the Eulerian numbers sum for the symmetrical q partitions.

%F t0(n,m)=(2 + PartitionsQ[n] - PartitionsQ[m] - PartitionsQ[n - m]); t1(n,k)=Sum[(-1)^j *t0[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; t(n,m)=If[n == 0, 1, t1(n, k) + t1(n, n - k)]

%e {1},

%e {2, 2},

%e {4, 4, 4},

%e {53, 28, 28, 53},

%e {246, 138, 102, 138, 246},

%e {4207, 1775, 814, 814, 1775, 4207},

%e {60922, 25758, 10035, 5654, 10035, 25758, 60922},

%e {1114857, 457387, 168720, 63929, 63929, 168720, 457387, 1114857},

%e {10706482, 4934925, 2063635, 791789, 438484, 791789, 2063635, 4934925, 10706482},

%e {357283571, 146408798, 55255747, 18896106, 6696246, 6696246, 18896106, 55255747, 146408798, 357283571},

%e {8544327694, 3575342129, 1376619624, 477863600, 150928577, 72130096, 150928577, 477863600, 1376619624, 3575342129, 8544327694}

%t Clear[t, t0, n, k];

%t t0[n_, m_] = (2 + PartitionsQ[n] - PartitionsQ[m] - PartitionsQ[n - m]);

%t t[n_, k_] = Sum[(-1)^j *t0[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}];

%t Table[Table[If[n == 0, 1, t[n, k] + t[n, n - k]], {k, 0, n}], {n, 0, 10}];

%t Flatten[%]

%K nonn,tabl,uned

%O 0,2

%A _Roger L. Bagula_, Feb 04 2009