A293983 a(n) = least k > 0 such that prime(k) XOR prime(k+1) XOR ... XOR prime(A293927(n)) = 0 (where XOR denotes the binary XOR operator, and prime(n) = A000040(n)).

8, 19, 15, 26, 33, 30, 26, 38, 22, 49, 47, 45, 58, 63, 69, 63, 65, 65, 71, 69, 69, 92, 92, 88, 123, 86, 123, 80, 132, 140, 80, 70, 153, 161, 56, 155, 176, 182, 145, 195, 143, 185, 133, 202, 125, 123, 216, 225, 235, 121, 237, 246, 235, 219, 227, 105, 260, 254

Offset: 1

Author

Rémy Sigrist, Oct 21 2017

Keywords

• nonn
• base

Comments

If a(n) = 1, then A126084(A293927(n)) = 0.

For any n > 0, A293927(n) - a(n) + 1 >= 4 (we need to XOR at least 4 consecutive prime numbers in order to obtain 0).

For any n > 0, if a(n) > 1 then A293927(n) - a(n) + 1 is even (we need to XOR an even number of odd prime numbers in order to obtain 0).

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A126084, A293927.

Programs

PARI

prev = vector(1774); s = 0; pi = 0; n = 0; forprime (p=1, 1697, pi++; s = bitxor(s, p); if (s==0 || prev[s], n++; print1 (prev[s]+1 ", "), prev[s] = pi));