A316296 a(n) = Sum_{k=1..n} f(k, n), where f(i, j) is the number of multiples of i greater than j and less than 2*j.

0, 1, 3, 5, 9, 10, 15, 18, 21, 24, 31, 30, 38, 41, 44, 48, 55, 56, 64, 65, 70, 75, 84, 81, 90, 95, 98, 103, 112, 109, 120, 123, 129, 134, 139, 139, 150, 155, 160, 161, 173, 170, 183, 184, 187, 198, 205, 202, 212, 217, 223, 226, 239, 236, 245, 248, 255, 262, 271, 266, 282, 285, 288

Offset: 1

Author

Andrea La Rosa, Jun 29 2018

Keywords

• nonn
• easy

Comments

f(n, m) is the number of multiples of n that are > m and < 2*m. n and m must be both >= 0.

By definition, this means that f(n, m) =

0 if n >= 2m;

1 if m < n < 2m;

If n <= m, then m = kn + q, where 0 <= q < n.

It can be proven that in this case f(n, m) =

k - 1 if q = 0;

k if q > 0 and (n - q) >= q;

k + 1 if q > 0 and (n - q) < q.

Let sd(n) = A006218; then a(n) = sd(2n-1) - sd(n) - (n - 1).

Also, a(n) = Sum_{k=n+1..2n-1} (d(k) - 1), where d(k) is number of divisors (A000005).

Number of ways the numbers from 1..n divide the numbers from n+1..2n-1, n>=2. - Wesley Ivan Hurt, Feb 08 2022

Examples

			For n = 7, a(7) = f(1,7) + f(2,7) + f(3,7) + f(4,7) + f(5,7) + f(6,7) + f(7,7) = 6 + 3 + 2 + 2 + 1 + 1 = 15.
		

Programs

JavaScript

function f(n,m){
    var count = 0;
    for(var i=m+1; i<2*m; i++){
        if(i%n === 0) count++;
    }
    return count;
}
function sf(n){
    var sum = 0;
    for(var i=1; i<=n; i++){
        sum += f(i, n);
    }
    return sum;
}

PARI

a(n) = n + sum(m = 1, n, (floor((n<<1 - 1) / m) - ceil((n + 1) / m))) \\ David A. Corneth, Jun 29 2018

Python

from math import isqrt
def A316296(n): return ((s:=isqrt(n))+(t:=isqrt(m:=(n<<1)-1)))*(s-t)+(sum(m//k for k in range(1,t+1))-sum(n//k for k in range(1,s+1))<<1)-n+1 # Chai Wah Wu, Oct 24 2023

Formula

a(n) = Sum_{k=1..n} Sum_{i=n+1..2n-1} (1-ceiling(i/k)+floor(i/k)). - Wesley Ivan Hurt, Feb 08 2022