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XOR BINOMIAL transform of A101119.
4

%I #16 Jul 10 2022 13:23:54

%S 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 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 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 XOR BINOMIAL transform of A101119.

%C 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 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 (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 (Python)

%o from operator import xor

%o from functools import reduce

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

%K nonn

%O 1,1

%A _Simon Plouffe_ and _Paul D. Hanna_, Dec 02 2004