|
| |
|
|
A084242
|
|
Least k, 1<= k <=n, such that the number of elements of the continued fraction for n/k is maximum.
|
|
6
| |
|
|
1, 1, 2, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 11, 9, 10, 11, 11, 11, 13, 13, 14, 13, 14, 15, 17, 17, 18, 19, 18, 23, 19, 21, 22, 22, 23, 22, 25, 29, 23, 26, 25, 27, 26, 27, 29, 31, 30, 29, 28, 33, 33, 31, 34, 41, 32, 36, 33, 37, 33, 35, 37, 39, 47, 37, 41, 42, 40, 41, 41, 53, 45, 43
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,3
|
|
|
COMMENTS
| Also, for n > 1, the smallest number k such that the Euclidean algorithm for (n,k) requires the maximum number of steps, A034883(n). - T. D. Noe, Mar 24 2011
|
|
|
FORMULA
| k>1, a(F(k))=F(k-1) where F(k) denotes the k-th Fibonacci number; probably, limit n ->oo 1/n*sum(k=1, n, a(k)) = 1/phi where phi is the Golden ratio (1+sqrt(5))/2
|
|
|
PROG
| (PARI) a(n)=if(n<0, 0, s=1; while(abs(vecmax(vector(n, i, length(contfrac(n/i))))-length(contfrac(n/s)))>0, s++); s)
|
|
|
CROSSREFS
| Cf. A071677.
Sequence in context: A120835 A091374 A065603 * A194806 A156261 A071823
Adjacent sequences: A084239 A084240 A084241 * A084243 A084244 A084245
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 21 2003
|
| |
|
|
