

A086377


a(1)=1; a(n)=a(n1)+2 if n is in the sequence; a(n)=a(n1)+2 if n and (n1) are not in the sequence; a(n)=a(n1)+3 if n is not in the sequence but (n1) is in the sequence.


8



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
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OFFSET

1,2


COMMENTS

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) = (n1)^2/b(n1  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>oo? (3) How does the limit depend on the initial term b(1)? (End)
From the recurrence relation, it follows that the limit L = lim_{b>oo} b(n)/n satisfies the following quadratic equation: L^2  2*L  1 = 0 implying that L = 1+sqrt(2) or 1sqrt(2).  Max Alekseyev, 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 1sqrt(2).  Don Reble, 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?  Paul D. Hanna, May 02 2006
The three conjectures by respectively Biberstein, Hanna, and Kimberling have all been proved, see the paper by Bosma et al. in the Links.  Michel Dekking, Oct 05 2017


LINKS



FORMULA

a(n) = floor((1+sqrt(2))*n  1/sqrt(2)).


PROG

(Magma) [Floor((1+Sqrt(2))*n1/Sqrt(2)): n in [1..70]]; // Vincenzo Librandi, Oct 05 2017


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



