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%I #2 Mar 30 2012 17:34:28
%S 1,1,1,1,6,1,1,15,15,1,1,12,18,12,1,1,25,30,30,25,1,1,18,75,60,75,18,
%T 1,1,35,105,175,175,105,35,1,1,24,84,168,210,168,84,24,1,1,45,108,252,
%U 378,378,252,108,45,1,1,30,225,360,630,756,630,360,225,30,1
%N A modulo two parity function as a triangle sequence:k=2; 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, 8, 32, 44, 112, 248, 632, 764, 1568, 3248,...}. 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=2;
%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, 6, 1},
%e {1, 15, 15, 1},
%e {1, 12, 18, 12, 1},
%e {1, 25, 30, 30, 25, 1},
%e {1, 18, 75, 60, 75, 18, 1},
%e {1, 35, 105, 175, 175, 105, 35, 1},
%e {1, 24, 84, 168, 210, 168, 84, 24, 1},
%e {1, 45, 108, 252, 378, 378, 252, 108, 45, 1},
%e {1, 30, 225, 360, 630, 756, 630, 360, 225, 30, 1}
%t Clear[p];
%t k=2;
%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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