

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}.


0



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
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OFFSET

1,2


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).


LINKS



FORMULA

a(n) = u*(u+1)*(2*u+1)/6  u*(u1)/2 + (nu)*(nu+1)/2, where u = floor(sqrt(n)) = A000196(n).


EXAMPLE

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)}, ...


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}]


PROG



CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



