A051201 Sum of elements of the set { [ n/k ] : 1 <= k <= n }.

1, 3, 4, 7, 8, 12, 13, 15, 19, 21, 22, 28, 29, 31, 33, 39, 40, 43, 44, 51, 53, 55, 56, 60, 66, 68, 70, 73, 74, 83, 84, 87, 89, 91, 93, 103, 104, 106, 108, 112, 113, 123, 124, 127, 130, 132, 133, 138, 146, 149, 151, 154, 155, 159, 161, 172, 174, 176, 177, 183, 184, 186

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006218, A000267, A055086.

Programs

Maple

seq(add(j, j in {seq(floor(n/i), i=1..n)}), n=1..100); # Ridouane Oudra, Nov 02 2024

Mathematica

a[n_] := With[{m = Quotient[Floor@Sqrt[4n+1]-1, 2]}, m(m+1)/2 + Sum[ Quotient[n, k], {k, 1, Quotient[n, m+1]}]];
Array[a, 100] (* Jean-François Alcover, Nov 20 2020, after Max Alekseyev *)

PARI

{ a(n) = m=(sqrtint(4*n+1)-1)\2; m*(m+1)/2 + sum(k=1,n\(m+1),n\k) } \\ Max Alekseyev, Feb 12 2012

Python

from math import isqrt
def A051201(n): return ((m:=isqrt((n<<2)+1)+1>>1)*(m-1)>>1)+sum(n//k for k in range(1,n//m+1)) # Chai Wah Wu, Oct 31 2023

Formula

a(n) = m*(m+1)/2 + Sum_{k=1..floor(n/(m+1))} floor(n/k), where m is the largest number such that m*(m+1) <= n, i.e., m=floor((sqrt(4*n+1)-1)/2 ). - Max Alekseyev, Feb 12 2012

From Ridouane Oudra, Nov 02 2024: (Start)

a(n) = r*(r-1)/2 + Sum_{k=1..floor(sqrt(n))} floor(n/k), where r = floor(sqrt(n+1) + 1/2).

a(n) = (1/4)*t*(t-1) + (1/2)*Sum_{k=1..n} floor(n/k), where t = floor(sqrt(4*n + 1)).

a(n) = (1/4)*A000267(n)*A055086(n) + (1/2)*A006218(n). (End)