A055086 n appears 1+[n/2] times.

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

Offset: 0

Author

Michael Somos, Jun 13 2000

Keywords

• nonn
• easy

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

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Randell Heyman, Cardinality of a floor function set, arXiv:1905.00533 [math.NT], 2019.
• Michael Somos, Sequences used for indexing triangular or square arrays, 2003.

Crossrefs

Cf. A000267, A055087, A063524, A075993.

Cf. A067514.

Programs

Mathematica

Flatten[Table[Table[n,{Floor[n/2]+1}],{n,0,20}]] (* Harvey P. Dale, Mar 07 2014 *)

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

Python

from math import isqrt
def A055086(n): return isqrt((n<<2)|1)-1 # Chai Wah Wu, Nov 23 2024

Formula

a(n) = [sqrt(4*n + 1)] - 1 = A000267(n) - 1.

a(n) = Sum_{k=1..n} A063524(A075993(n, k)), for n>0. - Reinhard Zumkeller, Apr 06 2006

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