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%I A320192 #35 Feb 11 2019 21:30:39
%S A320192 3,6,12,20,27,37,46,59,72,84,98,111,125,140,157,172,188,205,221,239,
%T A320192 258,277,296,316,334,353,374,395,418,441,462,484,505,528,549,572,595,
%U A320192 618,641,664,688,712,736,761,786,813,838,862,886,912,937,963,991
%N A320192 Number of summation terms of the reciprocal integer squares series that gives the best approximation to n terms of the Euler product for zeta(2).
%C A320192 Conjecture: It appears that for integer n the number of summation terms of the reciprocal integer squares series that gives the best approximation to Pi^2/6 - 1/n is n. - _Andrew Howroyd_, Nov 24 2018
%F A320192 Conjecture: a(n) = floor(1/(Pi^2/6 - Product{k=1..n} 1/(1-1/prime(k)^2) )). - _Andrew Howroyd_, Nov 24 2018
%e A320192 First term of Euler product: 4/3 needs 3 terms for best approximation: 1 + 1/4 + 1/9.
%e A320192 Case n=2: The second term of the Euler product is 1/((1-1/2^2)*(1-1/3^2)) = 3/2 = 1.5 which lies between Sum_{k=1..6} 1/k^2 and Sum_{k=1..7} 1/k^2. The first of these is the better approximation so a(2) = 6.
%t A320192 x = 70;
%t A320192 y = Round[x^1.8];
%t A320192 eulerp = Table[Product[1./(1 - 1/Prime[n]^2), {n, 1, k}], {k, 1, y}];
%t A320192 sums = Table[{k, Sum[1./n^2, {n, 1, k}]}, {k, 1, y}];
%t A320192 diff = Table[Abs[eulerp[[r]] - sums[[All, 2]]], {r, x}];
%t A320192 count = Flatten[Array[Range[y] &, x]];
%t A320192 fldiff = Flatten[diff];
%t A320192 t5 = Transpose[{count, fldiff}];
%t A320192 t6 = Partition[t5, y];
%t A320192 t7 = Table[MinimalBy[t6[[i]], Last], {i, x}];
%t A320192 sol = Flatten[t7, 1][[All, 1]]
%o A320192 (PARI) \\ to make this faster use floating point, but beware precision.
%o A320192 c(r)={my(s=r-r, k=0); while(s < r, k++; s+=1/k^2); k - (2*(s-r) >= 1/k^2)}
%o A320192 a(n)={c(prod(k=1, n, 1/(1-1/prime(k)^2)))} \\ _Andrew Howroyd_, Nov 23 2018
%Y A320192 Cf. A007406, A007407, A013661 (zeta(2)), A261208.
%K A320192 nonn
%O A320192 1,1
%A A320192 _Horst H. Manninger_, Oct 07 2018

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