A258825 a(n) = Number of times the k-th term is equal to k in the modified Collatz trajectory of n, when counting the initial term n as the term zero: n, A014682(n), A014682(A014682(n)), ...
0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 2, 0, 0
Offset: 1
Keywords
Examples
For n = 5, the Collatz function does the following: 5 -> 8 -> 4 -> 2 -> 1. Here, for k = 1, 2, 3, 4, applying k iterations to 5 does not yield k. So a(5) = 0. For n = 6, the Collatz function does the following: 6 -> 3 -> 5 -> 8 -> 4 -> 2 -> 1. After the 4th iteration, you arrive at 4. Since this is the only time this occurs, a(6) = 1.
Links
Programs
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Mathematica
A258825[n_]:=Count[MapIndexed[{#1}==#2-1&,NestWhileList[If[OddQ[#],(3#+1)/2,#/2]&,n,#>1&]],True];Array[A258825,100] (* Paolo Xausa, Nov 06 2023 *)
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PARI
Tvect(n)=v=[n]; while(n!=1, if(n%2, k=(3*n+1)/2; 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-1, c++)); print1(c, ", "))
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Scheme
(define (A258825 n) (let loop ((n n) (i 0) (s 0)) (if (= 1 n) (+ s (if (= i 1) 1 0)) (loop (A014682 n) (+ 1 i) (+ s (if (= i n) 1 0)))))) (define (A014682 n) (if (even? n) (/ n 2) (/ (+ n n n 1) 2))) ;; Antti Karttunen, Aug 18 2017
Extensions
Name changed by Antti Karttunen, Aug 18 2017
Comments