A222802 When A114183 decreases in value for the n-th time, dropping to k (say), a(n) is the number of steps earlier that floor(k/2) appeared in A114183.

5, 8, 12, 18, 19, 21, 25, 33, 36, 44, 53, 37, 53, 64, 14, 31, 32, 69, 71, 76, 77, 108, 120, 39, 93, 105, 123, 125, 157, 170, 52, 91, 93, 99, 190, 192, 89, 225, 238, 121, 72, 158, 251, 238, 251, 270, 205, 50, 209, 282, 284, 286, 287, 288, 289, 361, 385, 370, 281, 282, 340, 342, 344, 346, 309, 310, 312, 367, 460, 275

Offset

1,1

Comments

The fact that, when a number k occurs in A114183, floor(k/2) has already appeared, is a key step in the proof that A114183 is a permutation of the natural numbers. This fact is obvious if k is the result of a doubling step. The present sequence is an attempt to gain insight into why it is true when k occurs at a square root step.

Links

N. J. A. Sloane, Table of n, a(n) for n = 1..7322

Example

The first 50 terms of A114183 are:
1, 2, 4, 8, 16, 32, 5, 10, 3, 6, 12, 24, 48, 96, 9, 18, 36, 72, 144, 288, 576, 1152, 33, 66, 132, 11, 22, 44, 88, 176, 13, 26, 52, 7, 14, 28, 56, 112, 224, 448, 21, 42, 84, 168, 336, 672, 25, 50, 100, 200.
The sequence decreases from 32 to 5, from 10 to 3, from 96 to 9, and so on.
The values of k are therefore 5, 3, 9, 33, 11, 13, 7, 21, 25, ...
and the corresponding values of floor(k/2) are 2, 1, 4, 16, 5, 6, 3, 10, 12, ...
Since 2 appeared in A114183 5 steps before 5, a(1) = 5,
since 1 appeared 8 steps before 3, a(2) = 8,
since 4 appeared 12 steps before 9, a(3) = 12, and so on.

Crossrefs

Cf. A114183, A221715, A221716, A213220.

Sequence in context: A145285 A314409 A314410 * A215284 A325438 A314411

Adjacent sequences: A222799 A222800 A222801 * A222803 A222804 A222805

Keyword

nonn

Author

N. J. A. Sloane, Mar 08 2013

Status

approved