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%I #26 Aug 07 2013 16:53:11
%S 1,2,4,7,10,14,19,25,32,39,47,56,66,77,89,102,115,129,144,160,177,195,
%T 214,234,255,276,298,321,345,370,396,423,451,480,510,541,572,604,637,
%U 671
%N 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}.
%C 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.
%C 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, ....
%C 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).
%F 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).
%e 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),...}.
%e a(1) = 1 that is the number of elements in {(1,1)},
%e a(2) = 2 that is the number of elements in {(1,1),(2,2)} and
%e a(3) = 4 that is the number of elements in {(1,1),(2,2),(3,2),(3,3)}, ...
%t (* Calculates a(n) using the definition of the sequence. *)
%t data = Flatten[Table[Table[{k, n}, {k, n, n^2}], {n, 1, 40}], 1];
%t Table[Length[Select[data, #[[1]] <= m &]], {m, 1, 40}]
%t (* Calculates a(n) using a formula. *)
%t 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}]
%o (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
%K nonn,easy
%O 1,2
%A Taishi Inoue, Hiroshi Matsui, and _Ryohei Miyadera_, Sep 27 2011
%E Entry rewritten by _R. J. Mathar_, Jan 28 2012
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