A152237 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]].

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

Offset

0,5

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.

Links

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

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]].

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}

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[%]

Crossrefs

Sequence in context: A060102 A308359 A199065 * A176282 A082043 A177944

Adjacent sequences: A152234 A152235 A152236 * A152238 A152239 A152240

Keyword

nonn

Author

Roger L. Bagula, Nov 30 2008

Status

approved