

A091023


a(1)=1; for n >= 2, set a(n)=m, where n is the smallest unassigned index with exactly m1 unassigned indices still remaining between m and m1.


6



1, 2, 13, 3, 6, 26, 4, 11, 205, 9, 5, 24, 7, 51, 22, 102, 20, 49, 18, 8, 410, 10, 16, 12, 47, 14, 100, 45, 203, 43, 98, 41, 3277, 39, 96, 37, 201, 35, 94, 15, 33, 17, 408, 19, 31, 21, 92, 23, 29, 25, 199, 27, 90, 819, 88, 197, 86, 406, 84, 195, 82, 1638, 80, 193, 78, 404
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OFFSET

1,2


COMMENTS

Suggested by Leroy Quet in SeqFan memo 3602 on Feb 16 2004, where he gave the terms with values 116, with a(6) the first unassigned term.
Considering the number of unassigned indices to the left of the current position gives an equivalent sequence, A091068, which is easier to analyze.  N. J. A. Sloane, Feb 23 2004


LINKS

R. J. Mathar, Apr 28 2007, Table of n, a(n) for n = 1..78
Hans Havermann, Illustration of first 1600 terms (with some gaps)


EXAMPLE

After 1 has been assigned to a(1), the first unassigned term that is one term away from 1 is a(2), so a(2)=2;
the first unassigned term that is two terms away from 2 is a(4), so a(4)=3;
the first unassigned term that is 3 terms away from 3 is a(7), so a(7)=4;
the first unassigned term that is 4 terms away from 4 is a(11), so a(11)=5;
at this point we have 1,2,*,3,*,*,4,*,*,*,5,..., where * indicates a term to which a value has not yet been assigned.
The next value to assign is 6 which must be assigned to the first term of the sequence that is 5 terms away from a(11)=5; since a(5) has not yet been assigned a value and since at this point 5 terms with unassigned values lie between a(5) and a(11), we must assign 6 to a(5), i.e., a(5)=6.


MAPLE

nmax := 20000 : a := [seq(0, i=1..nmax)] : a := subsop(1=1, a) : a := subsop(2=2, a) : prevn := 2 : n := 3: while true do us := n ; atst := prevn1 ; tstdown := false ; while us > 0 and atst>0 do if op(atst, a) =0 then us := us1 ; if us = 1 then tstdown := true ; a := subsop(atst=n, a) ; prevn := atst ; break ; fi ; fi ; atst := atst 1 ; od ; if tstdown = false then us := n ; atst := prevn+1 ; while us > 0 do if op(atst, a) =0 then us := us1 ; if us = 1 then a := subsop(atst=n, a) ; prevn := atst ; break ; fi ; fi ; atst := atst +1 ; od ; fi ; for i from 1 to 150 do printf("%d, ", op(i, a)) ; od ; print() ; n := n+1 ; od : # R. J. Mathar, Apr 28 2007


CROSSREFS

Cf. A091052, A091053 (records), A091263 (inverse).
Sequence in context: A088253 A249224 A217010 * A120863 A286459 A335817
Adjacent sequences: A091020 A091021 A091022 * A091024 A091025 A091026


KEYWORD

nonn,nice


AUTHOR

John W. Layman, Feb 23 2004


EXTENSIONS

More terms from R. J. Mathar, Apr 28 2007


STATUS

approved



