A281376 Total number of counts where floor(N/k) < floor((N+k)/n) for k = {1, 2, ..., n-1} and N >= n.

0, 0, 0, 1, 3, 6, 11, 17, 25, 35, 47, 60, 77, 95, 115, 138, 164, 191, 222, 254, 290, 329, 370, 412, 460, 510, 562, 617, 676, 736, 802, 869, 940, 1014, 1090, 1169, 1255, 1342, 1431, 1523, 1621, 1720, 1825, 1931, 2041, 2156, 2273, 2391, 2517, 2645, 2777

Offset: 1

Author

Lorenz H. Menke, Jr., Jan 20 2017

Keywords

• nonn

Examples

			For n = 5, we have counted the cases where floor(N/k) < floor((N+k)/5), k = {1,2,3,4} then a(5) = 3, since this is true only for k = 4 and N = 6, k = 4 and N = 7, and k = 4 and N = 11.
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..10000 (terms 1..200 from Lorenz H. Menke, Jr.)

Programs

Maple

A281376 := proc(n)
    local a,k,N;
    a := 0 ;
    for k from 1 to n-1 do
        for N from n do
            if floor(N/k) < floor((N+k)/n) then
                a := a +1 ;
            elif N >= (k+2*n)*k/(n-k) then
                break;
            end if;
        end do:
    end do:
    a ;
end proc:
seq(A281376(n),n=1..10) ; # R. J. Mathar, Feb 03 2017

Mathematica

a[n_] :=
  Block[{lhs, rhs, count},
   count = 0;
   Do[lhs = Floor[H1/k];
    rhs = Floor[(H1 + k)/n];
    If[lhs < rhs, count++], {k, 1, n - 1}, {H1,
     n, (n^2 - 3 n + 1) + 10}]; (* the 10 is simply guard counts *)
   Return[count]];
a281376[n_] :=
Sum[Floor[j/d], {d, Ceiling[(n - 3)/3]}, {j, n - (2 d + 1)}]
(* Hartmut F. W. Hoft, Jan 25 2017; based on the first formula above *)

PARI

a(n) = sum(d = 1, ceil((n-3)/3), sum(j = 1, n-(2*d+1), j\d)); \\ Michel Marcus, Jan 29 2017

Formula

a(n) = Sum_{d=1..ceiling((n-3)/3)} Sum_{j=1..n-(2*d+1)} floor(j/d). - Jon E. Schoenfield, Jan 23 2017

a(n) = Sum_{d=1..ceiling(n/3)-1} ((j+1)*(j*d/2 + n mod d)), where j = floor(n/d) - 3. - Jon E. Schoenfield, Jan 24 2017