|
| |
|
|
A074840
|
|
Numerators a(n) of fractions slowly converging to sqrt(2): let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < sqrt(2), then a(n+1) = a(n) + 1, else a(n+1)= a(n).
|
|
5
| |
|
|
0, 1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 15, 16, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 22, 23, 24, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40, 41, 41, 42, 42, 43, 43
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,4
|
|
|
COMMENTS
| a(n) + b(n) = n and as n -> +infinity, a(n) / b(n) converges to sqrt(2). For all n, a(n) / b(n) < sqrt(2).
Is a(n)=2n-ceiling(n*sqrt(2)) for n>=1? [From Clark Kimberling, Sep 9 2011]
|
|
|
FORMULA
| a(1) = 0. b(n) = n - a(n). If (a(n) + 1) / b(n) < sqrt(2), then a(n+1) = a(n) + 1, else a(n+1) = a(n).
a(n) = floor(n*(2-sqrt(2))). - Vladeta Jovovic (vladeta(AT)eunet.rs), Oct 04 2003
|
|
|
EXAMPLE
| a(6)= 3 so b(6) = 6 - 3 = 3. a(7) = 4 because (a(6) + 1) / b(6) = 4/3 which is < sqrt(2). So b(7) = 7 - 4 = 3. a(8) = 4 because (a(7) + 1) / b(7) = 5/3 which is not < sqrt(2).
|
|
|
CROSSREFS
| Cf. A001601.
Sequence in context: A139327 A076905 A098295 * A064542 A076935 A019446
Adjacent sequences: A074837 A074838 A074839 * A074841 A074842 A074843
|
|
|
KEYWORD
| easy,frac,nonn
|
|
|
AUTHOR
| Robert A. Stump (bee_ess107(AT)msn.com), Sep 09 2002
|
| |
|
|
