A114213 A generalized Pascal triangle modulo 2.
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1
Offset: 0
Examples
Triangle begins 1; 1, 1; 1, 1, 1; 1, 1, 1, 1; 1, 1, 0, 1, 1; 1, 1, 0, 0, 1, 1; 1, 1, 1, 0, 1, 1, 1; 1, 1, 1, 1, 1, 1, 1, 1; 1, 1, 0, 1, 0, 1, 0, 1, 1; 1, 1, 0, 0, 0, 0, 0, 0, 1, 1; 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1;
Links
- Jeffrey Shallit and Lukas Spiegelhofer, Continuants, run lengths, and Barry's modified Pascal triangle, arXiv:1710.06203 [math.CO], 2017.
Programs
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PARI
T(n,k) = sum(j=0, n-k, binomial(k, j)*binomial(n-k, j)*(1+(-1)^j)/2) % 2; \\ Michel Marcus, Jun 06 2021
Formula
T(n, k) = (Sum_{j=0..n-k} C(k, j)*C(n-k, j)*(1+(-1)^j)/2) mod 2.
Comments