cp's OEIS Frontend

A248121 a(n) = floor(1 / (1/n - Pi^2/6 + Sum_{h=1..n} 1/h^2)).

%I A248121 #18 Nov 11 2024 01:57:01
%S A248121 2,9,20,34,53,76,102,133,168,206,249,296,346,401,460,522,589,660,734,
%T A248121 813,896,982,1073,1168,1266,1369,1476,1586,1701,1820,1942,2069,2200,
%U A248121 2334,2473,2616,2762,2913,3068,3226,3389,3556,3726,3901,4080,4262,4449,4640
%N A248121 a(n) = floor(1 / (1/n - Pi^2/6 + Sum_{h=1..n} 1/h^2)).
%C A248121 It is well known that Sum_{h>=1} 1/h^2 = Pi^2/6; this sequence provides insight into the manner of convergence.
%D A248121 Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 20.
%H A248121 Clark Kimberling, <a href="/A248121/b248121.txt">Table of n, a(n) for n = 1..1000</a>
%F A248121 a(n) ~ 2*n^2. - _Vaclav Kotesovec_, Oct 09 2014
%F A248121 Conjectures from _Chai Wah Wu_, Aug 03 2022: (Start)
%F A248121 a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5) for n > 5.
%F A248121 G.f.: -x*(x + 1)^2*(x + 2)/((x - 1)^3*(x^2 + x + 1)). (End)
%e A248121 Let d(n) = Pi^2/6 - Sum_{h=1..n} 1/h^2. Approximations are shown here:
%e A248121 n ... 1/n .... d(n) ....... 1/n - d(n) ... a(n)
%e A248121 1 ... 1 ...... 0.644934 ... 0.355066 ..... 2
%e A248121 2 ... 0.5 .... 0.394934 ... 0.105066 ..... 9
%e A248121 3 ... 0.33 ... 0.283823 ... 0.04951 ...... 20
%e A248121 4 ... 0.25 ... 0.221323 ... 0.028677 ..... 34
%t A248121 z = 200; p[k_] := p[k] = Sum[1/h^2, {h, 1, k}];
%t A248121 N[Table[Pi^2/6 - p[n], {n, 1, z/4}]]
%t A248121 f[n_] := f[n] = Select[Range[z], Pi^2/6 - p[#] < 1/n &, 1]
%t A248121 u = Flatten[Table[f[n], {n, 1, z}]]  (* A000027 *)
%t A248121 v = Floor[Table[1/(1/n - (Pi^2/6 - p[n])), {n, 1, z}]]  (* A248121 *)
%Y A248121 Cf. A000027, A264938 (second conjecture).
%K A248121 nonn,easy
%O A248121 1,1
%A A248121 _Clark Kimberling_, Oct 02 2014
%E A248121 Typo in name corrected by _Vaclav Kotesovec_, Oct 09 2014