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).
3, 6, 12, 20, 27, 37, 46, 59, 72, 84, 98, 111, 125, 140, 157, 172, 188, 205, 221, 239, 258, 277, 296, 316, 334, 353, 374, 395, 418, 441, 462, 484, 505, 528, 549, 572, 595, 618, 641, 664, 688, 712, 736, 761, 786, 813, 838, 862, 886, 912, 937, 963, 991
Offset: 1
Keywords
Examples
First term of Euler product: 4/3 needs 3 terms for best approximation: 1 + 1/4 + 1/9.
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.
Programs
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Mathematica
x = 70; y = Round[x^1.8]; eulerp = Table[Product[1./(1 - 1/Prime[n]^2), {n, 1, k}], {k, 1, y}]; sums = Table[{k, Sum[1./n^2, {n, 1, k}]}, {k, 1, y}]; diff = Table[Abs[eulerp[[r]] - sums[[All, 2]]], {r, x}]; count = Flatten[Array[Range[y] &, x]]; fldiff = Flatten[diff]; t5 = Transpose[{count, fldiff}]; t6 = Partition[t5, y]; t7 = Table[MinimalBy[t6[[i]], Last], {i, x}]; sol = Flatten[t7, 1][[All, 1]] -
PARI
\\ to make this faster use floating point, but beware precision. c(r)={my(s=r-r, k=0); while(s < r, k++; s+=1/k^2); k - (2*(s-r) >= 1/k^2)} a(n)={c(prod(k=1, n, 1/(1-1/prime(k)^2)))} \\ Andrew Howroyd, Nov 23 2018
Formula
Conjecture: a(n) = floor(1/(Pi^2/6 - Product{k=1..n} 1/(1-1/prime(k)^2) )). - Andrew Howroyd, Nov 24 2018
Comments