|
Search: id:A122458
|
|
|
| A122458 |
|
"Dropping time" of the reduced Collatz iteration starting with 2n+1. |
|
+0
10
|
|
| 0, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 37, 1, 35, 1, 2, 1, 5, 1, 3, 1, 34, 1, 2, 1, 3, 1, 4, 1, 34, 1, 2, 1, 32, 1, 3, 1, 5, 1, 2, 1, 3, 1, 28, 1, 5, 1, 2, 1, 26, 1, 3, 1, 19, 1, 2, 1, 3, 1, 5, 1, 9, 1, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 25, 1, 13, 1, 2, 1, 18, 1, 3, 1, 5, 1, 2, 1, 3, 1, 4, 1, 8, 1, 2, 1, 5
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
We count only the 3x+1 steps of the usual Collatz iteration. We stop counting when the iteration produces a number less than the initial 2n+1. For a fixed dropping time k, let N(k)=A100982(k) and P(k)=2^(A020914(k)-1). There are exactly N(k) odd numbers less than P(k) with dropping time k. Moreover, the sequence is periodic: if d is one of the N(k) odd numbers, then k=a(d)=a(d+i*P(k)) for all i>0. This periodicity makes it easy to compute the average dropping time of the reduced Collatz iteration: sum_{k>0} k*N(k)/P(k) = 3.492651852186...
|
|
LINKS
|
T. D. Noe, Table of n, a(n) for n=0..10000
|
|
EXAMPLE
|
a(3)=4 because, starting with 7, the iteration produces 11,17,13,5 and the last term is less than 7.
|
|
MATHEMATICA
|
nextOddK[n_]:=Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; dt[n_]:=Module[{m=n, cnt=0}, If[n>1, While[m=nextOddK[m]; cnt++; m>n]]; cnt]; Table[dt[n], {n, 1, 301, 2}]
|
|
CROSSREFS
|
Cf. A060445, A075677 (one step of the reduced Collatz iteration), A075680.
Sequence in context: A088964 A124331 A095248 this_sequence A127461 A101261 A067614
Adjacent sequences: A122455 A122456 A122457 this_sequence A122459 A122460 A122461
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
T. D. Noe (noe(AT)sspectra.com), Sep 08 2006
|
|
|
Search completed in 0.003 seconds
|
