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1
Least natural number which does not occur in A116626(1..n).
+20
5
2, 2, 4, 4, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 9, 9, 11, 11, 11, 11, 11, 11, 18, 18, 19, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 21, 21, 21, 21, 27, 27, 27, 27, 27, 27, 27, 27, 29, 29, 29, 29, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 39, 39, 39, 39, 42, 42, 42, 42, 44, 44, 44, 44
PROG
(MIT Scheme:) (define ( A116648 n) (if (< n 2) (+ n 1) (let ((prev ( A116648 (- n 1)))) (cond ((not (= ( A116626 n) prev)) prev) (else (let loop ((i (+ 1 prev))) (cond ((> ( A116627 i) n) i) (else (loop (+ 1 i))))))))))
a(n) = First integer i such that A116626(i) = n, 0 if n never occurs in A116626.
+20
2
1, 3, 2, 5, 9, 4, 15, 7, 17, 11, 23, 6, 8, 16, 10, 13, 19, 25, 29, 33, 41, 37, 14, 18, 24, 12, 49, 45, 53, 63, 57, 21, 27, 35, 31, 47, 59, 55, 67, 39, 43, 71, 22, 75, 79, 85, 89, 30, 20, 28, 26, 36, 44, 34, 32, 46, 56, 58, 54, 42, 40, 38, 48, 51, 61, 65, 73, 69, 81, 77, 95, 87
COMMENTS
If A116626 is really a permutation of natural numbers, then this is as well and no zeros are needed.
a(1)=1; for n>1, a(n) = least positive integer not appearing earlier such that {a(k) | 1 <= k <= n} and {a(k) XOR a(k-1) | 1 <= k <= n} are disjoint sets of distinct numbers.
+10
5
1, 2, 4, 8, 5, 10, 16, 7, 9, 17, 32, 11, 18, 33, 19, 35, 20, 34, 22, 40, 21, 41, 28, 36, 27, 64, 29, 38, 31, 37, 65, 30, 66, 39, 68, 42, 67, 44, 70, 45, 69, 128, 46, 72, 47, 77, 129, 71, 131, 73, 130, 74, 132, 75, 134, 79, 136, 80, 133, 81, 135, 84, 137, 82, 139, 85
MATHEMATICA
a = {1}; used = {}; Do[k = 1; While[MemberQ[Join[a, used], k] || MemberQ[Join[a, used], r = BitXor[a[[-1]], k]], k++]; AppendTo[a, k]; AppendTo[used, r], {n, 2, 66}]; a (* Ivan Neretin, Mar 13 2017 *)
PROG
(MIT/GNU Scheme)
(define ( A116624 n) (cond ((= 1 n) 1) (else (let outloop ((i 1)) (let ((k (A003987bi i ( A116624 (- n 1))))) (let inloop ((j (- n 1))) (cond ((zero? j) i) ((= i ( A116624 j)) (outloop (+ i 1))) ((= i ( A116625 (- j 1))) (outloop (+ i 1))) ((= k ( A116625 (- j 1))) (outloop (+ i 1))) ((= k ( A116624 j)) (outloop (+ i 1))) (else (inloop (- j 1))))))))))
3, 6, 12, 13, 15, 26, 23, 14, 24, 49, 43, 25, 51, 50, 48, 55, 54, 52, 62, 61, 60, 53, 56, 63, 91, 93, 59, 57, 58, 100, 95, 92, 101, 99, 110, 105, 111, 106, 107, 104, 197, 174, 102, 103, 98, 204, 198, 196, 202, 203, 200, 206, 207, 205, 201, 199, 216, 213, 212
1, 3, 5, 9, 15, 17, 23, 25, 29, 33, 41, 49, 53, 63, 67, 71, 75, 79, 85, 89, 95, 99, 103, 107, 139, 143, 163, 167, 175, 179, 199, 219, 223, 227, 235, 243, 267, 271, 275, 279, 283, 291, 299, 307, 323, 333, 337, 351, 357, 361, 371, 381, 395, 405, 425, 449, 457
COMMENTS
Equivalently, from i>1 onward those i where A116626(i) = A116648(i-1). Conjecture: all the terms are odd.
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