A329722 a(n) = Sum_{k=0..n} ((binomial(n+2k,2n-k)*binomial(n,k)) mod 2).

1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 3, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 2, 4, 1, 1, 3, 4, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 3, 2, 2, 2, 4, 2, 2, 4, 2, 1, 1, 1, 2, 3, 3, 4, 7, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 3, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 2, 4, 1, 1

Offset: 0

Author

Chai Wah Wu, Nov 19 2019

Keywords

• nonn

Comments

Run length transform of the coefficients of (1-2x^3)/(1-x-x^2), i.e., 1, 1, 2, 1, 3, 4, 7, 11, ... (1, 1 followed by the Lucas sequence A000032).

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..10000
• Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166 [math.CO], 2016.

Crossrefs

Cf. A000032, A329723.

Programs

PARI

a(n) = sum(k=0, n, lift(Mod((binomial(n+2*k,2*n-k)*binomial(n,k)), 2))) \\ Felix Fröhlich, Nov 25 2019

Python

def A329722(n): return sum(int(not (~(n+2*k) & 2*n-k) | (~n & k)) for k in range(n+1)) # Chai Wah Wu, Sep 28 2021