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%I #11 Dec 14 2023 15:39:16
%S 1,3,2,6,4,12,8,13,5,15,10,26,16,23,7,14,9,24,17,49,32,43,11,25,18,51,
%T 33,50,19,48,35,55,20,54,34,52,22,62,40,61,21,60,41,53,28,56,36,63,27,
%U 91,64,93,29,59,38,57,31,58,37,100,65,95,30,92,66,101,39,99,68,110,42
%N a(1)=1; a(odd n) = a(n-1) XOR a(n-2), for a(even n) we find the first i > 1 such that neither i nor (i XOR A116626(n-1)) is present in A116626(1..n-1), in which case a(n) = (i XOR A116626(n-1)).
%C This is a permutation of the natural numbers provided that A116625 is the complement of A116624. XOR is A003987.
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%o (MIT/GNU Scheme)
%o (define (A116626 n) (cond ((= 1 n) 1) ((odd? n) (A003987bi (A116626 (-1+ n)) (A116626 (- n 2)))) (else (let outloop ((i (A116648 (-1+ n)))) (let ((k (A003987bi i (A116626 (-1+ n))))) (let inloop ((j (- n 1))) (cond ((zero? j) k) ((= i (A116626 j)) (outloop (+ i 1))) ((= k (A116626 j)) (outloop (+ i 1))) (else (inloop (- j 1))))))))))
%Y Cf. a(2n) = a(2n-1) XOR a(2n+1), a(2n+1) = A116624(n+1). Inverse: A116627. Bisections: A116624, A116625. Cf. A116648.
%K nonn
%O 1,2
%A _Paul D. Hanna_ and _Antti Karttunen_, Feb 21 2006
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