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A101122 XOR BINOMIAL transform of A101119.

%I A101122 #16 Jul 10 2022 13:23:54
%S A101122 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,
%T A101122 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,
%U A101122 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
%N A101122 XOR BINOMIAL transform of A101119.
%C A101122 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.
%F A101122 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.
%o A101122 (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}
%o A101122 (Python)
%o A101122 from operator import xor
%o A101122 from functools import reduce
%o A101122 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
%Y A101122 Cf. A003484, A006519, A101119, A101120, A101121.
%K A101122 nonn
%O A101122 1,1
%A A101122 _Simon Plouffe_ and _Paul D. Hanna_, Dec 02 2004