A099893 XOR BINOMIAL transform of A006068 (inverse Gray code).
0, 1, 3, 0, 7, 0, 0, 0, 15, 0, 0, 0, 0, 0, 0, 0, 31, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 63, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 127
Offset: 0
Programs
-
PARI
{a(n)=local(B);B=0;for(i=0,n,B=bitxor(B,binomial(n,i)%2*A006068(n-i) ));B}
Formula
a(2^n) = 2^(n+1)-1 for n>0, with a(0)=0 and a(k)=0 otherwise. a(n) = SumXOR_{i=0..n} (C(n, i)mod 2)*A006068(n-i) and SumXOR is summation under XOR.
Comments