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A152237
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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]].
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0
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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, 21, 63, 105, 105, 63, 21, 1, 1, 16, 56, 112, 140, 112, 56, 16, 1, 1, 27, 72, 168, 252, 252, 168, 72, 27, 1, 1, 20, 135, 240, 420, 504, 420, 240, 135, 20, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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COMMENTS
| 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.
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FORMULA
| t(n,m)=Binomial[n,m]+p(n,m);
k=1;
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]].
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EXAMPLE
| {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, 21, 63, 105, 105, 63, 21, 1},
{1, 16, 56, 112, 140, 112, 56, 16, 1},
{1, 27, 72, 168, 252, 252, 168, 72, 27, 1},
{1, 20, 135, 240, 420, 504, 420, 240, 135, 20, 1}
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MATHEMATICA
| Clear[p];
k=1;
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]];
Table[Table[Binomial[n, m] + p[n, m], {m, 0, n}], {n, 0, 10}];
Flatten[%]
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CROSSREFS
| Sequence in context: A082950 A060102 A199065 * A176282 A082043 A177944
Adjacent sequences: A152234 A152235 A152236 * A152238 A152239 A152240
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KEYWORD
| nonn
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AUTHOR
| Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Nov 30 2008
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