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%I #18 Sep 22 2017 11:37:40
%S 1,6,7,13,22,26,44,52,88,104,176,208,352,416,704,832,1408,1664,2816,
%T 3328,5632,6656,11264,13312,22528,26624,45056,53248,90112,106496,
%U 180224,212992,360448,425984,720896,851968,1441792,1703936,2883584,3407872,5767168
%N 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.
%C For n >= 2, a(n) - a(n-1) = A160824(n).
%C 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
%H Colin Barker, <a href="/A160825/b160825.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,2).
%F 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
%F From _Colin Barker_, Dec 29 2016: (Start)
%F a(n) = 2*a(n-2) for n>5.
%F G.f.: x*(1 + 6*x + 5*x^2 + x^3 + 8*x^4) / (1 - 2*x^2).
%F (End)
%t Join[{1, 6, 7}, LinearRecurrence[{0, 2}, {13, 22}, 38]] (* _Jean-François Alcover_, Sep 22 2017 *)
%o (PARI) zeros(n) = if(n<2,0,zeros(floor(n/2))+1-(n%2))
%o 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
%o (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
%Y Cf. A160824.
%K nonn,base,easy
%O 1,2
%A _Leroy Quet_, May 27 2009
%E More terms from _Hagen von Eitzen_, Jun 01 2009
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