A211264 Number of integer pairs (x,y) such that 0 < x < y <= n and x*y <= n.

0, 1, 2, 3, 4, 6, 7, 9, 10, 12, 13, 16, 17, 19, 21, 23, 24, 27, 28, 31, 33, 35, 36, 40, 41, 43, 45, 48, 49, 53, 54, 57, 59, 61, 63, 67, 68, 70, 72, 76, 77, 81, 82, 85, 88, 90, 91, 96, 97, 100, 102, 105, 106, 110, 112, 116, 118, 120, 121, 127, 128, 130, 133, 136

Offset: 1

Author

Clark Kimberling, Apr 06 2012

Keywords

• nonn

Comments

Partial sums of A056924.

For a guide to related sequences, see A211266.

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A211266, A001222, A006218, A000196.

Cf. A066829, A094820, A181972.

Programs

Magma

[0] cat [&+[(&+[p[2]: p in Factorization(i)] mod 2) *Floor(n div i):i in [2..n] ]:n in [2..65]]; // Marius A. Burtea, Oct 17 2019

Maple

with(numtheory): seq(add((bigomega(i) mod 2)*floor(n/i), i=1..n), n=1..60); # Ridouane Oudra, Oct 17 2019
# Alternative:
ListTools:-PartialSums(map(t-> floor(numtheory:-tau(t)/2), [$1..100])); # Robert Israel, Oct 18 2019

Mathematica

a = 1; b = n; z1 = 120;
t[n_] := t[n] = Flatten[Table[x*y, {x, a, b - 1},
{y, x + 1, b}]]
c[n_, k_] := c[n, k] = Count[t[n], k]
Table[c[n, n], {n, 1, z1}]           (* A056924 *)
Table[c[n, n + 1], {n, 1, z1}]       (* A211159 *)
Table[c[n, 2*n], {n, 1, z1}]         (* A211261 *)
Table[c[n, 3*n], {n, 1, z1}]         (* A211262 *)
Table[c[n, Floor[n/2]], {n, 1, z1}]  (* A211263 *)
Print
c1[n_, m_] := c1[n, m] = Sum[c[n, k], {k, a, m}]
Table[c1[n, n], {n, 1, z1}]          (* A211264 *)
Table[c1[n, n + 1], {n, 1, z1}]      (* A211265 *)
Table[c1[n, 2*n], {n, 1, z1}]        (* A211266 *)
Table[c1[n, 3*n], {n, 1, z1}]        (* A211267 *)
Table[c1[n, Floor[n/2]], {n, 1, z1}] (* A181972 *)

Python

from math import isqrt
def A211264(n): return (lambda m: sum(n//k for k in range(1, m+1))-m*(m+1)//2)(isqrt(n)) # Chai Wah Wu, Oct 08 2021

Formula

a(n) = (1/2)*Sum_{i=1..n} (1 - A008836(i))*floor(n/i). - Enrique Pérez Herrero, Jul 10 2012 [Corrected by Ridouane Oudra, Oct 17 2019]

From Ridouane Oudra, Oct 17 2019: (Start)

a(n) = Sum_{i=1..n} A066829(i)*floor(n/i)

a(n) = (1/2)*(A006218(n) - A000196(n)). (End)

From Ridouane Oudra, Sep 28 2024: (Start)

a(n) = Sum_{k=1..n} floor((sqrt(k^2 + 4*n) - k)/2) ;

a(n) = A094820(n) - A000196(n) ;

a(n) = A181972(2*n). (End)