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

Author

N. J. A. Sloane, Mar 08 2013

Keywords

• nonn

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A114183, A221715, A221716, A213220.