A360706 a(n) is the least positive number not yet used such that its binary representation has either all or none of its 1-bits in common with the XOR of a(1) to a(n-1).
1, 2, 3, 4, 8, 12, 5, 10, 6, 9, 7, 16, 17, 24, 14, 11, 18, 20, 13, 15, 19, 32, 36, 21, 25, 26, 22, 23, 27, 33, 37, 28, 34, 30, 29, 40, 42, 31, 64, 96, 35, 68, 38, 44, 41, 43, 39, 48, 56, 45, 65, 66, 46, 47, 67, 80, 52, 49, 57, 50, 82, 69, 97, 51, 53, 60, 54, 55, 58, 72, 73, 59, 61, 76, 70, 71, 74
Offset: 1
Examples
n a(n) a(n) in binary a(1) XOR ... XOR a(n-1) in binary ------------------------------------------------------------------ 1 1 1b 0b 2 2 10b 1b 3 3 11b 11b 4 4 100b 0b 5 8 1000b 100b 6 12 1100b 1100b 7 5 101b 0b ... Signed version of this sequence such that the arithmetic sum over the first k values equals the nim-sum over the first k values of the original sequence: 1, 2, -3, 4, 8, -12, 5, 10, -6, -9, 7, 16, -17, 24, -14, 11, -18, 20, -13, ...
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
- Thomas Scheuerle, Scatterplot red: a(n) blue: a(1) XOR...XOR a(n-1) for n = 1...50000
- Thomas Scheuerle, Proof that all positive integers will appear in this sequence
- Index entries for sequences that are permutations of the natural numbers
- Index entries for sequences related to binary expansion of n
Programs
-
MATLAB
function a = A360706( max_n ) s = 0; a = []; t = [1:max_n]; for n = 1:max_n k = 1; while (t(k) ~= bitand(s,t(k)))&&(0 ~= bitand(s,t(k))) k = k+1; end s = bitxor(s,t(k)); a(n) = t(k); t(k) = max(t)+1; t = sort(t); end end
-
PARI
{ m = s = 0; for (n = 1, 77, for (v = 1, oo, if (!bittest(s, v), x = bitand(m, v); if (x==0 || x==v, s += 2^v; m = bitxor(m, v); print1 (v", "); break;);););); } \\ Rémy Sigrist, Aug 31 2024
Formula
If a(m1) = 2^k and a(m2) = 2^k-1 then m1 - 2^k < 0 and m2 - (2^k-1) > 0 for k > 2.
Comments