A160825 a(1) = 1. For n >=2, a(n) = the smallest integer > a(n-1) such that both a(n) and a(n)-a(n-1) have the same number of (non-leading) 0's when they are represented in binary.

1, 6, 7, 13, 22, 26, 44, 52, 88, 104, 176, 208, 352, 416, 704, 832, 1408, 1664, 2816, 3328, 5632, 6656, 11264, 13312, 22528, 26624, 45056, 53248, 90112, 106496, 180224, 212992, 360448, 425984, 720896, 851968, 1441792, 1703936, 2883584, 3407872, 5767168

Offset

1,2

Comments

For n >= 2, a(n) - a(n-1) = A160824(n).

It is easy to see that: If a(n) = 2^m - 2^k with k>0, then a(n+1) = 2^m - 2^(k-1); in all other cases 2^k|a(n) implies 2^k|a(n+1). From this one concludes that a(n) = 11*2^k implies a(n+1) = 13*2^k and a(n) = 13*2^k implies a(n+1) = 11*2^(k+1). - Hagen von Eitzen, Jun 01 2009

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0,2).

Formula

a(2*n) = 13*2^(n-2) and a(2*n+1) = 11*2^(n-1) if n >= 2. - Hagen von Eitzen, Jun 01 2009

From Colin Barker, Dec 29 2016: (Start)

a(n) = 2*a(n-2) for n>5.

G.f.: x*(1 + 6*x + 5*x^2 + x^3 + 8*x^4) / (1 - 2*x^2).

(End)

Mathematica

Join[{1, 6, 7}, LinearRecurrence[{0, 2}, {13, 22}, 38]] (* Jean-François Alcover, Sep 22 2017 *)

Prog

(PARI) zeros(n) = if(n<2, 0, zeros(floor(n/2))+1-(n%2))
n=1; L=1; while(n<=100, print1(L, ", "); step = bitand(L, -L); if(step>1 && bitand(L+step, L+step-1)==0, step/=2); ofs=step; while(zeros(ofs)!=zeros(L+ofs), ofs += step); n+=1; L+=ofs; )  \\ Hagen von Eitzen, Jun 01 2009
(PARI) Vec(x*(1 + 6*x + 5*x^2 + x^3 + 8*x^4) / (1 - 2*x^2) + O(x^50)) \\ Colin Barker, Dec 29 2016

Crossrefs

Cf. A160824.

Sequence in context: A041074 A041749 A042723 * A020689 A128844 A128850

Adjacent sequences: A160822 A160823 A160824 * A160826 A160827 A160828

Keyword

nonn,base,easy

Author

Leroy Quet, May 27 2009

Extensions

More terms from Hagen von Eitzen, Jun 01 2009

Status

approved