OFFSET
1,1
COMMENTS
Through a(n) natural numbers 1,2,3...a(n), A007677(n-1) of those terms are members of the upper level Beatty sequence A000572; while A007676(n) of those terms are in the lower level Beatty sequence A006594.
Check: a(5) = 26, which has 7 (= A007677(4)) terms in A000572: 3, 7, 11, 14, 18, 22 and 26; while the remaining 19 (= A007676(5)) are members of the lower level Beatty sequence A006594.
A085368(n)/A007677(n-1) converge upon (1 + e), as n approaches infinity. Check: A085368(6)/A007677(5) = 119/32 = 3.71875... where (1 + e) =3.718281828... A085368(n)/A007676(n) converge upon (1 + 1/e). Check: A085368(5)/A007676(5) = 119/87 = 1.3678.., where (1 + 1/e) = 1.367879441... A006594 and A000572 form Beatty pairs, with floor n*(1 + e) being the generator for A000572(n) and floor n*(1 + 1/e) the generator for A006594(n).
The cutting sequence for y = (1/e)x is generated from the line starting at (0,0), passing through an array of squares, giving "1" to an intersection with a vertical line and "0" to an intersection with a horizontal line. The cutting sequence for y = (1/e)x is 0, then (terms 1 through 26): 1 1 0 1 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0. In this sequence, n's for 0's are all members of the upper Beatty pair: A000572 (check: n's for the 0's are 3, 7, 11, 14, 18, 22 and 26 (the 7 being A007677(4)); while 19 terms (19 = A007676(5)) are members of the lower Beatty pair A006594, being denoted by "1" and thus intersecting vertical lines.
FORMULA
Convergents to 1/e are generated from the partial quotients of the continued fraction form of 1/e: [2, 1, 2, 1, 1, 4, 1, 1, 6...], where below each partial quotient, the first 9 convergents are 1/2, 1/3, 3/8...(i.e. 1/2 = [2], 1/3 = [2, 1], 3/8 = [2, 1, 2], etc;...then 4/11, 7/19, 32/87, 39/106, 71/193, 465/1264, where a(n) = sum of numerator and denominator of n-th convergent to 1/e with 1/2 = first convergent.
EXAMPLE
CROSSREFS
KEYWORD
nonn
AUTHOR
Gary W. Adamson, Jun 26 2003
EXTENSIONS
More terms from Colin Barker, Mar 11 2014
STATUS
approved