A147649 Binary prejudiced single Sierpinski modulo two Pascal shift: Prejudice function: p(n,m)=If[Mod[Binomial[n - 2, m - 1], 2] == 0, Round[Log[2]]/2, 1]; t(n,m)=Binomial[n, m] + If[n > 2, 2*Binomial[n - 2, m - 1]*p(n, m), 0]; Mod[If[n > 2, 2*Binomial[n - 2, m - 1]*p(n,m), 0],2]=0.
1, 1, 1, 1, 2, 1, 1, 5, 5, 1, 1, 6, 8, 6, 1, 1, 7, 16, 16, 7, 1, 1, 8, 19, 26, 19, 8, 1, 1, 9, 31, 45, 45, 31, 9, 1, 1, 10, 34, 86, 90, 86, 34, 10, 1, 1, 11, 50, 126, 196, 196, 126, 50, 11, 1, 1, 12, 53, 148, 266, 322, 266, 148, 53, 12, 1
Offset: 0
Examples
{1}, {1, 1}, {1, 2, 1}, {1, 5, 5, 1}, {1, 6, 8, 6, 1}, {1, 7, 16, 16, 7, 1}, {1, 8, 19, 26, 19, 8, 1}, {1, 9, 31, 45, 45, 31, 9, 1}, {1, 10, 34, 86, 90, 86, 34, 10, 1}, {1, 11, 50, 126, 196, 196, 126, 50, 11, 1}, {1, 12, 53, 148, 266, 322, 266, 148, 53, 12, 1}
Programs
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Mathematica
p[n_, m_] := If[Mod[Binomial[n - 2, m - 1], 2] == 0, Round[Log[2]]/2, 1]; Table[Table[Binomial[n, m] + If[n > 2, 2*Binomial[n - 2, m - 1], 0], {m, 0, n}], {n, 0, 10}]; Flatten[%]
Formula
Prejudice function: p(n,m)=If[Mod[Binomial[n - 2, m - 1], 2] == 0, Round[Log[2]]/2, 1]; t(n,m)=Binomial[n, m] + If[n > 2, 2*Binomial[n - 2, m - 1]*p(n, m), 0].
Comments