A196126 Let A = {(x,y): x, y positive natural numbers and y <= x <= y^2}. a(n) is the cardinality of the subset {(x,y) in A such that x <= n}.

1, 2, 4, 7, 10, 14, 19, 25, 32, 39, 47, 56, 66, 77, 89, 102, 115, 129, 144, 160, 177, 195, 214, 234, 255, 276, 298, 321, 345, 370, 396, 423, 451, 480, 510, 541, 572, 604, 637, 671

Offset: 1

Author

Taishi Inoue, Hiroshi Matsui, and Ryohei Miyadera, Sep 27 2011

Keywords

• nonn
• easy

Comments

The set A locates integer points in the first quadrant above the parabola y=sqrt(x) up to the diagonal y=x. a(n) counts them up to a sliding right margin.

The first differences of the sequence are 1, 2, 3, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 31, ....

In that way the sequence is constructed from first differences which are the natural numbers and repetitions for 3, 7, 13, 21, 31, 43, 57, 73, 91,...., (apparently the elements of A002061 starting at 3).

Examples

			The set is A = {(1,1),(2,2),(3,2),(4,2),(3,3),(4,3),(5,3),(6,3),(7,3),(8,3),(9,3),(4,4),(5,4),...}.
a(1) = 1 that is the number of elements in {(1,1)},
a(2) = 2 that is the number of elements in {(1,1),(2,2)} and
a(3) = 4 that is the number of elements in {(1,1),(2,2),(3,2),(3,3)},  ...
		

Programs

Mathematica

(* Calculates a(n) using the definition of the sequence. *)
data = Flatten[Table[Table[{k, n}, {k, n, n^2}], {n, 1, 40}], 1];
Table[Length[Select[data, #[[1]] <= m &]], {m, 1, 40}]
(* Calculates a(n) using a formula. *)
ff[t_] := Block[{u}, u = Floor[Sqrt[t]]; u (u + 1) (2 u + 1)/6 - u (u - 1)/2 + (t - u) (t - u + 1)/2]; Table[ff[t], {t, 1, 40}]

PARI

a(n)=my(u=sqrtint(n));u*(u^2+2)/3+(n-u)*(n-u+1)/2 \\ Charles R Greathouse IV, Oct 05 2011

Formula

a(n) = u*(u+1)*(2*u+1)/6 - u*(u-1)/2 + (n-u)*(n-u+1)/2, where u = floor(sqrt(n)) = A000196(n).

Extensions

Entry rewritten by R. J. Mathar, Jan 28 2012