

A281376


Total number of counts where floor(N/k) < floor((N+k)/n) for k = {1, 2, ..., n1} and N >= n.


1



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
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OFFSET

1,5


LINKS

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


FORMULA

a(n) = Sum_{d=1..ceiling((n3)/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


EXAMPLE

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.


MAPLE

A281376 := proc(n)
local a, k, N;
a := 0 ;
for k from 1 to n1 do
for N from n do
if floor(N/k) < floor((N+k)/n) then
a := a +1 ;
elif N >= (k+2*n)*k/(nk) 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 *)


PROG

(PARI) a(n) = sum(d = 1, ceil((n3)/3), sum(j = 1, n(2*d+1), j\d)); \\ Michel Marcus, Jan 29 2017


CROSSREFS

Sequence in context: A273140 A320272 A119639 * A247586 A107957 A000603
Adjacent sequences: A281373 A281374 A281375 * A281377 A281378 A281379


KEYWORD

nonn


AUTHOR

Lorenz H. Menke, Jr., Jan 20 2017


STATUS

approved



