|
|
A086377
|
|
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.
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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] = (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)
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). - 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 1-sqrt(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)).
|
|
MATHEMATICA
|
t = Nest[Flatten[# /. {0->{0, 1, 1}, 1->{0, 1}}] &, {0}, 5] (*A189687*)
f[n_] := t[[n]]
Flatten[Position[t, 0]] (* A086377 conjectured *)
Flatten[Position[t, 1]] (* A081477 conjectured *)
s[n_] := Sum[f[i], {i, 1, n}]; s[0] = 0;
Table[s[n], {n, 1, 120}] (*A189688*)
Table[Floor[(1 + Sqrt[2]) n - 1/Sqrt[2]], {n, 80}] (* Vincenzo Librandi, Oct 05 2017 *)
|
|
PROG
|
(Magma) [Floor((1+Sqrt(2))*n-1/Sqrt(2)): n in [1..70]]; // Vincenzo Librandi, Oct 05 2017
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|