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 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. 1
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified May 25 17:48 EDT 2022. Contains 354071 sequences. (Running on oeis4.)