A064664 Regard A064413 as giving a permutation of the positive integers; sequence gives inverse permutation.

1, 2, 5, 3, 10, 4, 14, 8, 6, 9, 20, 7, 28, 13, 11, 17, 33, 12, 37, 18, 15, 19, 43, 16, 24, 27, 22, 26, 57, 23, 61, 31, 21, 32, 25, 30, 67, 36, 29, 40, 74, 35, 81, 41, 39, 42, 89, 45, 50, 46, 34, 47, 100, 48, 53, 49, 38, 56, 107, 52, 115, 60, 51, 64, 54, 59, 128, 65, 44, 55

Offset: 1

Author

N. J. A. Sloane, Oct 11 2001

Keywords

• nonn

Comments

The Mathematica program computes the EKG sequence A064413 and then determines the inverse permutation. - T. D. Noe, Nov 13 2002

Examples

			Original permutation has cycles (1) (2) (3, 4, 6, 9, 10, 5) (..., 20, 18, 12, 7, 14, 13, 28, 26, ...) (8) ...
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
• J. C. Lagarias, E. M. Rains, and N. J. A. Sloane, The EKG sequence, Exper. Math. 11 (2002), 437-446; arXiv preprint, arXiv:math/0204011 [math.NT], 2002.
• Eric Weisstein's World of Mathematics, EKG Sequence.
• Index entries for sequences related to EKG sequence.
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A064413, A064665, A064666, A064667, A064668, A064669.

Equals A064421 + 1.

Programs

Haskell

import Data.List (elemIndex); import Data.Maybe (fromJust)
a064664 = (+ 1) . fromJust . (`elemIndex` a064413_list)
-- Reinhard Zumkeller, May 01 2014, Sep 17 2011

Mathematica

maxN=200; lst={1, 2}; unused=Range[3, maxN]; found=True; While[found, found=False; i=0; While[ !found&&i1, found=True; AppendTo[lst, unused[[i]]]; unused=Delete[unused, i]]]]; Take[Ordering[lst], unused[[1]]-1]