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Search: id:A086377
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| A086377 |
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a(1)=1; a(n)=a(n-1)+2 if n is in the sequence; a(n)=a(n-1)+2 if n and (n-1) are not in the sequence; a(n)=a(n-1)+3 if n is not in the sequence but (n-1) is in the sequence. |
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5
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| 1, 4, 6, 8, 11, 13, 16, 18, 21, 23, 25, 28, 30, 33, 35, 37, 40, 42, 45, 47, 49, 52, 54, 57, 59, 62, 64, 66, 69, 71, 74, 76, 78, 81, 83, 86, 88, 91, 93, 95, 98, 100, 103, 105, 107, 110, 112, 115, 117, 120, 122, 124, 127, 129, 132, 134, 136, 139, 141, 144, 146, 148, 151
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Comment from Joseph Biberstine (jrbibers(AT)indiana.edu), May 02 2006 (Start). The continued fraction 4/Pi = 1 + 1/(3 + 4/(5 + 9/(7 + 16/(9 + 25/(11 + ...))))) (see A079037) suggests the recurrence b[n] = 2*n - 1 + n^2/b[n+1] with b[1] = 4/Pi. Solving the above recurrence in the other direction we would have b[n] = (n-1)^2/(b[n-1 - 2*n + 3) with b[1] = 4/Pi.
Now consider this last defined sequence b[n]. It appears to grow linearly, (1) does it? (2) What is the limit of b[n]/n as n->Infinity? (3) How does the limit depend on the initial term b[1]? (End)
Comment from Max Alekseyev, May 02 2006: From the recurrence relation, it follows that the L=lim b[n]/n satisfies the following quadratic equation: L^2 - 2*L - 1 = 0 implying that L = 1+sqrt(2) or 1-sqrt(2).
Comment from Don Reble, May 02 2006: Note that b[n]/n decreases, while b[n]/(n+1) increases. I speculate that 4/Pi is the only b[1] value such that b[n]/n converges to 1+sqrt(2) instead of 1-sqrt(2).
Comment from Paul D. Hanna (pauldhanna(AT)juno.com), May 02 2006: It appears that round( b(n) ) = floor((1+sqrt(2))*n-1/sqrt(2)) = A086377(n) = a(n). This is certainly true for the first 190 terms. Is there a formal proof?
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FORMULA
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a(n)=floor((1+sqrt(2))*n-1/sqrt(2))
n is in sequence if A004641(n)==1 or A001030(n)==2. a(n) = A080652(n)-1 = A064437(n+1)-2 = A081841(n+2)-3. - Ralf Stephan, Feb 23 2004
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CROSSREFS
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Sequence in context: A152019 A091181 A047290 this_sequence A003662 A132635 A134779
Adjacent sequences: A086374 A086375 A086376 this_sequence A086378 A086379 A086380
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KEYWORD
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nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 13 2003
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