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Search: id:A091023
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| A091023 |
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a(1)=1; for n >= 2, set a(n)=m, where n is the smallest unassigned index with exactly m-1 unassigned indices still remaining between m and m-1. |
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6
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| 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
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Suggested by Leroy Quet in SeqFan memo 3602 on Feb. 16, 2004, where he gave the terms with values 1-16, 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. - njas, Feb 23, 2004
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LINKS
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R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 28 2007, Table of n, a(n) for n = 1..78
Hans Havermann, Illustration of first 1600 terms (with some gaps)
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EXAMPLE
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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.
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MAPLE
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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 := prevn-1 ; tstdown := false ; while us > 0 and atst>0 do if op(atst, a) =0 then us := us-1 ; 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 := us-1 ; 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 (mathar(AT)strw.leidenuniv.nl), Apr 28 2007
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CROSSREFS
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Cf. A091052, A091053 (records), A091263 (inverse).
Sequence in context: A090954 A089778 A088253 this_sequence A120863 A093079 A095417
Adjacent sequences: A091020 A091021 A091022 this_sequence A091024 A091025 A091026
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KEYWORD
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nonn,nice
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AUTHOR
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John W. Layman (layman(AT)math.vt.edu), Feb 23 2004
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EXTENSIONS
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More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 28 2007
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