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A258772
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Number of fixed points in the Collatz (3x+1) trajectory of n.
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4
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1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0
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OFFSET
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1,12
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COMMENTS
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This sequence uses the definition in A006370: if n is odd, n -> 3n+1 and if n is even, n -> n/2.
The number 3 appears first at a(187561). Do all nonnegative numbers appear? See A258821.
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LINKS
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EXAMPLE
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For n = 5, the trajectory is T(5) = [5, 16, 8, 4, 2, 1]. Since the fourth term in this sequence is 4, this is a fixed point. Since there is only one fixed point, a(5) = 1.
For n = 6, the trajectory is T(6) = [6, 3, 10, 5, 16, 8, 4, 2, 1]. Here, the k-th term in this trajectory does not equal k for any possible k. So a(6) = 0.
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MATHEMATICA
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A258772[n_]:=Count[MapIndexed[{#1}==#2&, NestWhileList[If[OddQ[#], 3#+1, #/2]&, n, #>1&]], True]; Array[A258772, 100] (* Paolo Xausa, Nov 06 2023 *)
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PROG
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(PARI) Tvect(n)=v=[n]; while(n!=1, if(n%2, k=(3*n+1); v=concat(v, k); n=k); if(!(n%2), k=n/2; v=concat(v, k); n=k)); v
for(n=1, 200, d=Tvect(n); c=0; for(i=1, #d, if(d[i]==i, c++)); print1(c, ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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