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%I #25 Oct 23 2017 01:14:29
%S 17,28,30,33,36,43,45,47,51,52,56,58,65,66,72,74,76,80,84,90,94,107,
%T 111,119,126,129,130,133,137,143,145,155,156,166,169,174,179,185,192,
%U 200,202,204,208,213,214,216,219,228,238,246,248,249,250,254,258,262
%N Numbers n such that prime(k) XOR prime(k+1) XOR ... XOR prime(n) = 0 for some k < n (where XOR denotes the binary XOR operator, and prime(n) = A000040(n)).
%C Equivalently, numbers n such that A126084(n) = A126084(m) for some m < n.
%C See A293983(n) for the least k such that prime(k) XOR prime(k+1) XOR ... XOR prime(a(n)) = 0.
%H Robert Israel, <a href="/A293927/b293927.txt">Table of n, a(n) for n = 1..10000</a>
%e prime(33) XOR prime(34) XOR prime(35) XOR prime(36) = 137 XOR 139 XOR 149 XOR 151 = 0, hence 36 appears in the sequence.
%p N:= 1000: # to get all terms <= N
%p R[0]:= 0: T:= 2: p:= 2;
%p Res:= NULL:
%p for n from 2 to N do
%p p:= nextprime(p);
%p T:= Bits:-Xor(T,p);
%p if assigned(R[T]) then Res:= Res, n
%p else R[T]:= n
%p fi
%p od:
%p Res; # _Robert Israel_, Oct 22 2017
%o (PARI) s = 0; seen = 2^0; for (i = 1, 262, s = bitxor(s, prime(i)); if (bittest(seen, s), print1 (i ", "), seen += 2^s))
%Y Cf. A000040, A126084, A293983.
%K nonn,base
%O 1,1
%A _Rémy Sigrist_, Oct 21 2017