|
| |
|
|
A217926
|
|
A sequence relating to the rational values zeta(2n)^2/zeta(4n), which are expressible in terms of Bernoulli's numbers (see comments for definition of the sequence).
|
|
2
|
|
|
|
7, 17, 29, 29, 41, 41, 59, 59, 71, 97, 97, 97, 101, 101, 101, 137, 137, 149, 149, 149, 179, 179, 179, 191, 191, 191, 191, 191, 227, 227, 227, 227, 227, 269, 269, 269, 269, 307, 307, 311, 311, 311, 311, 347, 347, 347, 347, 347, 347, 419, 419, 419, 419, 419
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
The rational value zeta(2n)^2 / zeta(4n) = A114363(n) / A114362(n) = Product(p^4n/(((p^2n) - 1)^2)) / Product(p^4n/((p^4n) - 1)) = Product(((p^2n) + 1)/((p^2n) - 1)), where the product is over all primes.
Denote Product(((p^2n) + 1)/((p^2n) - 1)), where the product is over xxx, by F(n, xxx). For example, zeta(2n)^2 / zeta(4n) = F(n, all primes).
By definition, F(n, all primes) = F(n,primes < P) x [(P^2n) + 1]/[(P^2n) - 1]}] x F(n, primes p > P).
For small enough P, F(n, primes p > P) will be much closer to 1 than ((P^2n) + 1)/((P^2n) - 1))), so if Q is the value for which ((Q^2n) + 1)/((Q^2n) - 1) = F(n, all primes) / F(n,primes < P), Q will only be slightly less than P, so rounding Q up to the next integer (for Q < 2) and rounding up to the next odd integer (for Q >= 2) should give P. The sequence identifies the lowest P for a particular n for which such rounding fails to give P, as follows:
The sequence entry a(n) for n > 0 = the lowest prime P for which Q < P - 2, where Q is the value for which ((Q^2n) + 1)/((Q^2n) - 1) = F(n, all primes) / F(n,primes < P).
For sufficiently large n, a(n)/(2n) is bounded below, and appears to be bounded above (see A217552).
The primes up to at least A217552(n) * (2n) can therefore be reliably generated from A114363(n) / A114362(n) as follows:
Find the value of Q such that ((Q^2n) + 1)/((Q^2n) - 1) = A114363(n) / A114362(n) and round up to the nearest integer, giving 2. Then find the value of Q such that ((Q^2n) + 1)/((Q^2n) - 1) = (A114363(n) / A114362(n))/(F(n, primes <= last value found, i.e., 2)) and round up to the nearest odd integer, giving 3. Then find the value of Q such that ((Q^2n) + 1)/((Q^2n) - 1) = (A114363(n) / A114362(n))/(F(n, primes <= last value found, i.e., 3)) and round up to the nearest odd integer, and so on.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
For n = 4, A217552(n) * (2n) = 26.4417..., so primes up to at least this value can be generated. Successive Q and rounded up Q values:
1.99029 2
2.99331 3
4.95780 5
6.96977 7
10.63524 11
12.73590 13
16.12527 17
18.42182 19
22.27250 23
26.81206 27 (Q < 29 - 2)
so a(4) = 29.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
