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 A112088 Number of leaf nodes in a binary tree. 77

%I

%S 2,3,5,7,11,16,24,36,54,81,122,183,274,411,617,925,1388,2082,3123,

%T 4684,7026,10539,15809,23713,35570,53355,80032,120048,180072,270108,

%U 405162,607743,911615,1367422,2051133,3076700,4615050,6922575,10383862

%N Number of leaf nodes in a binary tree.

%H Seiichi Manyama, <a href="/A112088/b112088.txt">Table of n, a(n) for n = 1..5678</a> (terms 1..1000 from David W. Wilson)

%F a(1)=2; a(n)=floor((5+sum(a(1) to a(n-1)))/2). - _Graeme McRae_, Jun 09 2006

%t f[n_] := Block[{a = 2, b = 0, c = 4}, Do[x = b + a; c -= x; If[c < 0, a *= 2; c += 2a]; b = Floor[(2a - c + 1)/2], {i, n}]; x]; Array[f, 40] (* _Robert G. Wilson v_, Jan 11 2006 *))

%t f[s_] := Append[s, Ceiling[2 + Plus @@ s/2]]; Nest[f, {2}, 38] (* _Robert G. Wilson v_, Jul 08 2006 *)

%o (Ruby) a,c=2,4;p Array.new(99){c-=x=(a*4-c+1)/2;c+=2*a*=2 if c<0;x}

%o a:=2;b:=0;c:=4;p := proc() local x; global a,b,c; x := b + a; c := c - x; if(c<0) then a := a*2; c := c + a*2; end if; b := floor((a*2-c+1) / 2); x end proc: seq(p(), i=0..40);

%o (PARI) first(n)=my(v=vector(n),s); v[1]=s=2; for(n=2,n, s+=v[n]=(s+5)\2); v \\ _Charles R Greathouse IV_, Nov 07 2016

%K easy,nonn

%O 1,1

%A Simon Strandgaard (neoneye(AT)gmail.com), Nov 29 2005

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