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Search: id:A088442
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| A088442 |
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A linear version of the Josephus problem. |
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+0
7
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| 1, 3, 1, 3, 9, 11, 9, 11, 1, 3, 1, 3, 9, 11, 9, 11, 33, 35, 33, 35, 41, 43, 41, 43, 33, 35, 33, 35, 41, 43, 41, 43, 1, 3, 1, 3, 9, 11, 9, 11, 1, 3, 1, 3, 9, 11, 9, 11, 33, 35, 33, 35, 41, 43, 41, 43, 33, 35, 33, 35, 41, 43, 41, 43, 129, 131, 129, 131, 137, 139, 137, 139, 129, 131
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Or a(n) is in A145812 such that (2n+3-a(n))/2 is in A145812 as well. Note also that a(n)+2A090569(n+1)=2n+3. [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Oct 20 2008]
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REFERENCES
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C. Groer, The mathematics of survival ..., Amer. Math. Monthly, 110 (No. 9, 2003), 812-825. (This is the sequence W(2n+1).)
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FORMULA
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To get a(n), write 2n+1 as Sum b_j 2^j, then a(n) = 1 + Sum_{j odd} b_j 2^j.
Also, a(n) = 2*ceiling(n/2)+1-2*a(ceiling(n/2)).
Equals A004514 + 1. - Chris Groer (cgroer(AT)math.uga.edu), Nov 10, 2003
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EXAMPLE
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If n=4, 2n+1 = 9 = 1 + 0*2 + 0*2^2 + 1*2^3, so a(4) = 1 + 0*2 + 1*2^3 = 9.
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MAPLE
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a:=proc(n) local b: b:=convert(2*n+1, base, 2): 1+sum(b[2*j]*2^(2*j-1), j=1..nops(b)/2) end: seq(a(n), n=0..100);
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MATHEMATICA
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a[n_] := a[n]= 2*Ceiling[n/2]+1-2a[Ceiling[n/2]]
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CROSSREFS
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Cf. A006257, A088443, A088452.
Sequence in context: A143453 A164308 A082511 this_sequence A037095 A146436 A058842
Adjacent sequences: A088439 A088440 A088441 this_sequence A088443 A088444 A088445
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KEYWORD
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nonn,easy
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), Nov 09 2003
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EXTENSIONS
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More terms from Emeric Deutsch (deutsch(AT)duke.poly.edu), May 27 2004
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