

A152238


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]].


0



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, 1, 1, 35, 105, 175, 175, 105, 35, 1, 1, 24, 84, 168, 210, 168, 84, 24, 1, 1, 45, 108, 252, 378, 378, 252, 108, 45, 1, 1, 30, 225, 360, 630, 756, 630, 360, 225, 30, 1
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OFFSET

0,5


COMMENTS

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.


LINKS

Table of n, a(n) for n=0..65.


FORMULA

t(n,m)=Binomial[n,m]+p(n,m);
k=2;
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]].


EXAMPLE

{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, 1},
{1, 35, 105, 175, 175, 105, 35, 1},
{1, 24, 84, 168, 210, 168, 84, 24, 1},
{1, 45, 108, 252, 378, 378, 252, 108, 45, 1},
{1, 30, 225, 360, 630, 756, 630, 360, 225, 30, 1}


MATHEMATICA

Clear[p];
k=2;
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[%]


CROSSREFS

Sequence in context: A230073 A143210 A205133 * A086645 A168291 A154980
Adjacent sequences: A152235 A152236 A152237 * A152239 A152240 A152241


KEYWORD

nonn


AUTHOR

Roger L. Bagula, Nov 30 2008


STATUS

approved



