|
| |
|
|
A055086
|
|
n appears 1+[n/2] times.
|
|
21
|
|
|
|
0, 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
The PARI functions t1, t2 can be used to read a triangular array T(n,k) (n >= 0, 0 <= k <= floor(n/2)) by rows from left to right: n -> T(t1(n), t2(n)).
a(n) gives the number of distinct positive values taken by [n/k]. E.g., a(5)=3: [5/{1,2,3,4,5}]={5,2,1,1,1}. - Marc LeBrun, May 17 2001
This sequence gives the elements in increasing order of the set {i+2j} where i>=0, j>=0. - Benoit Cloitre, Sep 22 2012
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = [sqrt(4*n + 1)] - 1 = A000267(n) - 1.
a(n) = ceiling(2*sqrt(n+1)) - 2. - Mircea Merca, Feb 05 2012
a(0) = 0, then for n>=1 a(n) = 1 + a(n-1-floor(a(n-1)/2)). - Benoit Cloitre, May 08 2017
a(n) = floor(b) + floor(n/(floor(b)+1)) where b = (sqrt(4*n+1)-1)/2. - Randell G Heyman, May 08 2019
Sum_{k>=1} (-1)^(k+1)/a(k) = Pi/8 + 3*log(2)/4. - Amiram Eldar, Jan 26 2024
|
|
|
MATHEMATICA
|
Flatten[Table[Table[n, {Floor[n/2]+1}], {n, 0, 20}]] (* Harvey P. Dale, Mar 07 2014 *)
|
|
|
PROG
|
(PARI) {a(n) = floor(sqrt(4*n + 1)) - 1}
(PARI) t1(n)=floor(sqrt(1+4*n)-1) /* A055086 */
(PARI) t2(n)=(1+4*n-sqr(floor(sqrt(1+4*n))))\4 /* A055087 */
(PARI) a(n)=if(n<1, 0, a(n-1-a(n-1)\2)+1) \\ Benoit Cloitre, May 09 2017
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
