A109451 a(1)=1; a(n) = smallest positive integer not already present such that a(n-1) and a(n) have a different number of 1's in their binary expansions.
1, 3, 2, 5, 4, 6, 7, 8, 9, 11, 10, 13, 12, 14, 15, 16, 17, 19, 18, 21, 20, 22, 23, 24, 25, 27, 26, 29, 28, 30, 31, 32, 33, 35, 34, 37, 36, 38, 39, 40, 41, 43, 42, 45, 44, 46, 47, 48, 49, 51, 50, 53, 52, 54, 55, 56, 57, 59, 58, 61, 60, 62, 63, 64, 65, 67, 66, 69, 68, 70, 71, 72
Offset: 1
Keywords
Examples
Among the positive integers (10, 11,12, 13,...) not among the first 9 terms of the sequence, 10 (decimal) has 2 1's in its binary form (1010), the same number of 1's as 9 in binary (1001). 11 (decimal), however, has 3 ones in its binary form (1011), so a(10) = 11.
Links
- Peter Kagey, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A000120.
Programs
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Maple
Cands:= [$2..100]: Ones:= map(t -> convert(convert(t,base,2),`+`), Cands): A[1]:= 1: dc:= 1: for n from 2 do found:= false; for k from 1 to nops(Cands) while not found do if Ones[k] <> dc then found:= true; A[n]:= Cands[k]; dc:= Ones[k]; Cands:= subsop(k=NULL,Cands); Ones:= subsop(k=NULL,Ones); fi od; if not found then break fi; od: seq(A[i],i=1..n-1); # Robert Israel, May 03 2016
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Mathematica
a = {1}; Nest[AppendTo[a, SelectFirst[Range@ 120, And[! MemberQ[a, #], First@ DigitCount[#, 2] != First@ DigitCount[Last@ a, 2]] &]] &, a, 71] (* Michael De Vlieger, May 03 2016, Version 10 *)
Formula
a(n) = n+1 if n == 2 or 4 (mod 8), n-1 if n == 3 or 5 (mod 8), n otherwise. - Peter Kagey, May 03 2016
G.f.: x*(1 + 2*x - 2*x^2 + x^3 + x^4 + x^5)/((1 - x)^2*(1 + x + x^4 + x^5)). - Ilya Gutkovskiy, May 04 2016
Extensions
Extended by Ray Chandler, Aug 27 2005
Comments