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a(n) = floor(1 / (1/n - Pi^2/6 + Sum_{h=1..n} 1/h^2)).
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%I #14 Aug 03 2022 21:12:53

%S 2,9,20,34,53,76,102,133,168,206,249,296,346,401,460,522,589,660,734,

%T 813,896,982,1073,1168,1266,1369,1476,1586,1701,1820,1942,2069,2200,

%U 2334,2473,2616,2762,2913,3068,3226,3389,3556,3726,3901,4080,4262,4449,4640

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

%C It is well known that Sum_{h>=1} 1/h^2 = Pi^2/6; this sequence provides insight into the manner of convergence.

%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 20.

%H Clark Kimberling, <a href="/A248121/b248121.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) ~ 2*n^2. - _Vaclav Kotesovec_, Oct 09 2014

%F Conjectures from _Chai Wah Wu_, Aug 03 2022: (Start)

%F a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5) for n > 5.

%F G.f.: -x*(x + 1)^2*(x + 2)/((x - 1)^3*(x^2 + x + 1)). (End)

%e Let d(n) = Pi^2/6 - sum{1/(h^2}, h = 1..k}. Approximations are shown here:

%e n ... 1/n .... d(n) ....... 1/n - d(n) ... a(n)

%e 1 ... 1 ...... 0.644934 ... 0.355066 ..... 2

%e 2 ... 0.5 .... 0.394934 ... 0.105066 ..... 9

%e 3 ... 0.33 ... 0.283823 ... 0.04951 ...... 20

%e 4 ... 0.25 ... 0.221323 ... 0.028677 ..... 34

%t z = 200; p[k_] := p[k] = Sum[1/h^2, {h, 1, k}];

%t N[Table[Pi^2/6 - p[n], {n, 1, z/4}]]

%t f[n_] := f[n] = Select[Range[z], Pi^2/6 - p[#] < 1/n &, 1]

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A000027 *)

%t v = Floor[Table[1/(1/n - (Pi^2/6 - p[n])), {n, 1, z}]] (* A248121 *)

%Y Cf. A000027, A264938 (second conjecture).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Oct 02 2014

%E Typo in name corrected by _Vaclav Kotesovec_, Oct 09 2014