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%I #2 Mar 30 2012 17:34:28
%S 1,1,1,1,4,1,1,9,9,1,1,8,12,8,1,1,15,20,20,15,1,1,12,45,40,45,12,1,1,
%T 21,63,105,105,63,21,1,1,16,56,112,140,112,56,16,1,1,27,72,168,252,
%U 252,168,72,27,1,1,20,135,240,420,504,420,240,135,20,1
%N A modulo two parity function as a triangle sequence:k=1; t(n,m)=Binomial[n,m]+p(n,m); Always even parity function: p(n,m)=If[Mod[Binomial[n, m], 2] == 0, 2^(k - 1)*Binomial[n, m], If[Mod[Binomial[n, m], 2] == 1 && Binomial[n, m] > 1, 2^k* Binomial[n, m], 0]].
%C Row sums are: {1, 2, 6, 16, 30, 64, 128, 260, 510, 1024, 2048,...}. The k is added to give a quantum level to the resulting symmetrical functions.
%F t(n,m)=Binomial[n,m]+p(n,m);
%F k=1;
%F p(n,m)=If[Mod[Binomial[n, m], 2] == 0, 2^(k - 1)*Binomial[n, m], If[Mod[Binomial[n, m], 2] == 1 && Binomial[n, m] > 1, 2^k* Binomial[n, m], 0]].
%e {1},
%e {1, 1},
%e {1, 4, 1},
%e {1, 9, 9, 1},
%e {1, 8, 12, 8, 1},
%e {1, 15, 20, 20, 15, 1},
%e {1, 12, 45, 40, 45, 12, 1},
%e {1, 21, 63, 105, 105, 63, 21, 1},
%e {1, 16, 56, 112, 140, 112, 56, 16, 1},
%e {1, 27, 72, 168, 252, 252, 168, 72, 27, 1},
%e {1, 20, 135, 240, 420, 504, 420, 240, 135, 20, 1}
%t Clear[p];
%t k=1;
%t p[n_, m_] = If[Mod[Binomial[n, m], 2] == 0, 2^(k - 1)*Binomial[n, m], If[Mod[Binomial[n, m], 2] == 1 && Binomial[n, m] > 1, 2^k*Binomial[n, m], 0]];
%t Table[Table[Binomial[n, m] + p[n, m], {m, 0, n}], {n, 0, 10}];
%t Flatten[%]
%K nonn
%O 0,5
%A _Roger L. Bagula_, Nov 30 2008
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