A119435 - OEIS

%I #31 Apr 20 2022 00:02:21

%S 1,2,5,3,9,7,13,4,17,12,23,10,22,18,29,6,33,24,43,16,40,31,51,14,41,

%T 30,53,25,49,38,61,8,65,45,83,32,76,58,95,21,74,55,94,42,87,68,107,19,

%U 78,56,100,39,91,70,113,34,89,66,112,52,104,81,125,11,129,86,163,60,148

%N a(n) = (binary reversal of n)-th integer among those positive integers not occurring earlier in the sequence.

%C This sequence is a permutation of the positive integers.

%C [Proof from _N. J. A. Sloane_, Apr 20 2022: a(n) always exists, so the sequence is infinite. Every time n is a power of 2, n-reversed is 1, and a(n) is the smallest missing number. Since there are infinitely many powers of 2, every number will eventually appear.]

%H Michael De Vlieger, <a href="/A119435/b119435.txt">Table of n, a(n) for n = 1..10000</a>, first 1000 terms from Diana L. Mecum.

%H Michael De Vlieger, <a href="/A119435/a119435.txt">Extended Table of n, a(n)</a> for n = 1..2^16.

%H Michael De Vlieger, <a href="/A119435/a119435.png">Scatterplot of a(n)</a>, n = 1..2^16.

%H Michael De Vlieger, <a href="/A119435/a119435_1.png">Log-log scatterplot of a(n)</a>, n = 1..2^16.

%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>

%e 12 in binary is 1100; so its binary reversal is 0011, which is 3 in decimal. Those positive integers not among the first 11 terms of the sequence are 6,8,10,11,14,..., and the third of these is 10, so a(12) = 10.

%t Block[{a = {1}, nn = 69}, Do[AppendTo[a, #] &@ Complement[Range[i + 2 nn], #][[FromDigits[#, 2] &@ Reverse@ IntegerDigits[i, 2]]] &@ a, {i, 2, nn}]; a] (* _Michael De Vlieger_, Sep 03 2017 *)

%Y Cf. A030101, A119436 (inverse).

%K easy,nonn,base,look

%O 1,2

%A _Leroy Quet_, May 19 2006

%E More terms from _Diana L. Mecum_, Jul 21 2008