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

1,1

Comments

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

Links

Table of n, a(n) for n=1..53.

Formula

Conjecture: a(n) = floor(1/(Pi^2/6 - Product{k=1..n} 1/(1-1/prime(k)^2) )). - Andrew Howroyd, Nov 24 2018

Example

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.

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]]

Prog

(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

Crossrefs

Cf. A007406, A007407, A013661 (zeta(2)), A261208.

Sequence in context: A226220 A218076 A308722 * A160732 A320608 A246147

Adjacent sequences: A320189 A320190 A320191 * A320193 A320194 A320195

Keyword

nonn

Author

Horst H. Manninger, Oct 07 2018

Status

approved