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A258828
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Least number k such that A258825(k) = n.
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1
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OFFSET
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0,2
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COMMENTS
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If it exists, a(n) > 10^6 for n > 3.
For n = 2, after 17 and 20 iterations, you arrive at 17 and 20, respectively. It appears the total number of iterations of the possible k values is either 26 or 33.
For n = 3, after 14, 17, and 20 iterations, you arrive at 14, 17, and 20, respectively. It appears the total number of iterations of the possible k values is 26.
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LINKS
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EXAMPLE
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For n = 105, the Collatz function does the following: 105 -> 158 -> 79 -> 119 -> 179 -> 269 -> 404 -> 202 -> 101 -> 152 -> 76 -> 38 -> 19 -> 29 -> 44 -> 22 -> 11 -> 17 -> 26 -> 13 -> 20 -> 10 -> 5 -> 8 -> 4 -> 2 -> 1. After the 17th and 20th iteration, we can see we reach 17 and 20, respectively. Since 105 is the smallest number to have exactly two occurrences, a(2) = 105. Note that there are 26 iterations before you reach 1. It appears that all numbers with exactly two occurrences have either 26 or 33 total iterations to get to 1.
For n = 305, the Collatz function does the following: 305 -> 458 -> 229 -> 344 -> 172 -> 86 -> 43 -> 65 -> 98 -> 49 -> 74 -> 37 -> 56 -> 28 -> 14 -> 7 -> 11 -> 17 -> 26 -> 13 -> 20 -> 10 -> 5 -> 8 -> 4 -> 2 -> 1. After the 14th, 17th, and 20th iteration, we can see we reach 14, 17, and 20, respectively. Since 305 is the smallest number to have exactly 3 occurrences, a(3) = 305. Note that there are 26 iterations before you reach 1. It appears that all numbers with exactly three occurrences have 26 total iterations to get to 1.
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PROG
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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
n=0; m=1; while(m<10^3, d=Tvect(m); c=0; for(i=1, #d, if(d[i]==i-1, c++)); if(c==n, print1(m, ", "); m=0; n++); m++)
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CROSSREFS
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KEYWORD
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nonn,more,hard,bref
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AUTHOR
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STATUS
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approved
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