A101122 XOR BINOMIAL transform of A101119.

7, 17, 0, 34, 0, 0, 0, 68, 0, 0, 0, 0, 0, 0, 0, 159, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 257, 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, 514, 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, 0, 0, 0, 0, 0, 0

Offset: 1

Author

Simon Plouffe and Paul D. Hanna, Dec 02 2004

Keywords

• nonn

Comments

Nonzero terms form A101121 and occur at positions 2^k for k >= 0. A101119 equals the nonzero differences of A006519 and A003484. See A099884 for the definition of the XOR BINOMIAL transform.

Crossrefs

Cf. A003484, A006519, A101119, A101120, A101121.

Programs

PARI

{a(n)=local(B);B=0;for(i=0,n-1,B=bitxor(B,binomial(n-1,i)%2* (16*2^valuation(n-i,2)-2^(valuation(n-i,2)%4)-8*(valuation(n-i,2)\4)-8)));B}

Python

from operator import xor
from functools import reduce
def A101122(n): return reduce(xor,(((1<<(m:=(~(k+1)&k).bit_length()+4))-((m&-4)<<1)-(1<<(m&3)))&-int(not k&~(n-1)) for k in range(n))) # Chai Wah Wu, Jul 10 2022

Formula

a(n) = SumXOR_{k=0..n} (C(n, k) mod 2)*A101119(k), where SumXOR is summation under XOR. A101119(n) = SumXOR_{k=0..n} (C(n, k) mod 2)*a(k). a(2^(n-1)) = A101121(n) for n >= 1 and a(k)=0 when k is not a power of 2.