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A070883
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Bitwise XOR of n and n-th prime.
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9
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3, 1, 6, 3, 14, 11, 22, 27, 30, 23, 20, 41, 36, 37, 32, 37, 42, 47, 80, 83, 92, 89, 68, 65, 120, 127, 124, 119, 112, 111, 96, 163, 168, 169, 182, 179, 184, 133, 128, 133, 154, 159, 148, 237, 232, 233, 252, 239, 210, 215, 218, 219, 196, 205, 310, 319, 308, 309, 302
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,1
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COMMENTS
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For any integer k, XOR(n,k) = 2*OR(n,k) - (n+k). - Gary Detlefs, Oct 26 2013
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LINKS
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Eric Weisstein's World of Mathematics, aut.
Eric Weisstein's World of Mathematics, XOR.
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FORMULA
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a(n) = 2*OR(p,n) - (p+n), for n-th prime p. - Gary Detlefs, Oct 26 2013
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EXAMPLE
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A000040(25)=97, [25]2 = '00011001', [97]2 = '01100001' '00011001' XOR '01100001' = '01111000', therefore a(25)=120.
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MATHEMATICA
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Table[ BitXor[ n, Prime[n]], {n, 1, 55}]
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PROG
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(Haskell)
import Data.Bits (xor)
a070883 n = a070883_list !! (n-1)
a070883_list = zipWith xor [1..] a000040_list
(PARI) a(n) = bitxor(n, prime(n));
(Python)
from sympy import prime
def a(n): return n^prime(n)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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