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A126084
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a(n) = XOR of first n primes.
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5
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0, 2, 1, 4, 3, 8, 5, 20, 7, 16, 13, 18, 55, 30, 53, 26, 47, 20, 41, 106, 45, 100, 43, 120, 33, 64, 37, 66, 41, 68, 53, 74, 201, 64, 203, 94, 201, 84, 247, 80, 253, 78, 251, 68, 133, 64, 135, 84, 139, 104, 141, 100, 139, 122, 129, 384, 135, 394, 133, 400, 137, 402, 183, 388, 179
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history;
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OFFSET
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0,2
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COMMENTS
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The values at odd positive indices are even and the values at even positive indices are odd.
Does this sequence contain any zeros for n > 0? Probabilistically, one would expect so; but none in first 10000 terms. - Franklin T. Adams-Watters, Jul 17 2011
None below 1.5 * 10^11: any prime p such that a(pi(p)) = 0 is 43 bits or longer. Heuristic chances that a prime below 2^100 yields 0 are about 45%. Note that an n-bit prime can yield 0 only if a(pi(p)) is odd, where p is the smallest n-bit prime. That is, for n > 1, there are no zeros from pi(2^n) to pi(2^(n+1)) if A007053(n) is even. - Charles R Greathouse IV, Jul 17 2011
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LINKS
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FORMULA
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a(0) = 0; a(n) = a(n-1) XOR prime(n).
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EXAMPLE
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a(4) = 3 because ((2 XOR 3) XOR 5) XOR 7 = (1 XOR 5) XOR 7 = 4 XOR 7 = 3
[Or, in base 2]
((10 XOR 11) XOR 101) XOR 111 = (1 XOR 101) XOR 111 = 100 XOR 111 = 11
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MATHEMATICA
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Module[{nn=70, prs}, prs=Prime[Range[nn]]; Table[BitXor@@Take[prs, n], {n, 0, nn}]] (* Harvey P. Dale, Jun 23 2016 *)
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PROG
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(PARI) al(n)=local(m); vector(n, k, m=bitxor(m, prime(k))) /* Produces a vector without a(0) = 0; Franklin T. Adams-Watters, Jul 17 2011 */
(PARI) v=primes(300); for(i=2, #v, v[i]=bitxor(v[i], v[i-1])); concat(0, v) \\ Charles R Greathouse IV, Aug 26 2014
(PARI) q=0; forprime(p=2, 313, print1(q, ", "); q=bitxor(q, p)) /* Klaus Brockhaus, Mar 06 2007; adapted by Rémy Sigrist, Oct 23 2017 */
(Python)
from operator import xor
from functools import reduce
from sympy import primerange, prime
def A126084(n): return reduce(xor, primerange(2, prime(n)+1)) if n else 0 # Chai Wah Wu, Jul 09 2022
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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