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A087256
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Number of different initial values for 3x+1 trajectories in which the largest term appearing in the iteration is 2^n.
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7
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1, 1, 1, 6, 1, 3, 1, 3, 1, 12, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 13, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 9, 1, 3, 1, 3, 1, 11, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 21, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 78, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 9, 1, 3, 1
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OFFSET
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1,4
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COMMENTS
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It would be interesting to know whether the ...1,3,1,3,1,x,1,3,1,3,1,... pattern persists. - John W. Layman, Jun 09 2004
The observed pattern should persist. Proof: [1] a(odd)=1 because -1+2^odd is not divisible by 3, so in Collatz-algorithm 2^odd is preceded by increasing inverse step. Thus 2^odd is the only suitable initial value; [2] a[2k]>=3 for k>1 because 2^(2k)-1=-1+4^k=3A so {b=2^2k, (b-1)/3 and (2a-2)/3} are three relevant initial values. No more case arises unless condition-[3] (see below) was satisfied; [3] a[6k+4]>=5 for k>=1, ..iv=c=2^(6k+4); here {c, (c-1)/3, 2(c-1)/3, (2c-5)/9, (4c-10)/9} is 5 suitable initial values, iff (2c-5)/9 is integer; e.g. at 6k+4=10, {1024<-341<-682<-227<-454} back-tracking the iteration. - Labos Elemer, Jun 17 2004
Except for a(2)=1 the sequence has the 6-element quasiperiod 1, 3, 1, x, 1, 3 where x>=6, but unequal to 7 and 10 (see links below and in A033496). Observe that for n=2^(6k+4)=16*2^(6k), n mod 9 = 7 so that (2n-5)/9 is an integer and a(n)>=6.
Conjecture: All numbers m > 10 occur as values in A087256 (see A233293).
The conjecture has been verified for all 10 < k < 133 for Collatz trajectories with maximum value through 2^(36000*6 + 4). The largest fan of initial values in this range, F(6*1993+4), has maximum 2^11962 and size 3958.
(End)
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LINKS
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FORMULA
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EXAMPLE
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n = 10: 2^10 = 1024 = peak for trajectories started with initial value taken from the list: {151, 201, 227, 302, 341, 402, 454, 604, 682, 804, 908, 1024};
a trajectory with peak=1024: {201, 604, 302, 151, 454, 227, 682, 341, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1}
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MATHEMATICA
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c[x_]:=c[x]=(1-Mod[x, 2])*(x/2)+Mod[x, 2]*(3*x+1); c[1]=1; fpl[x_]:=FixedPointList[c, x]; {$RecursionLimit=1000; m=0}; Table[Print[{xm-1, m}]; m=0; Do[If[Equal[Max[fpl[n]], 2^xm], m=m+1], {n, 1, 2^xm}], {xm, 1, 30}]
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PROG
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(PARI) f(n, m) = 1 + if(2*n <= m, f(2*n, m), 0) + if (n%6 == 4, f(n\3, m), 0);
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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