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

1,1

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.

Links

Table of n, a(n) for n=1..101.

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.

Prog

(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

Crossrefs

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

Sequence in context: A332573 A129422 A184062 * A090535 A107778 A122735

Adjacent sequences: A101119 A101120 A101121 * A101123 A101124 A101125

Keyword

nonn

Author

Simon Plouffe and Paul D. Hanna, Dec 02 2004

Status

approved