cp's OEIS Frontend

Since n*Pi is an irrational number, all tie-breaking rounding methods yield the same integer sequence. - Artur Twarecki (numbers(AT)twarecki.ca), Dec 19 2006

Or around of length of oval of constant width n. - Vladimir Shevelev, Oct 07 2013

Note that ceiling(Pi/arctan(1/k)) - ceiling(k*Pi) is equal to either 0 or 1, that is, for all other k we have ceiling(Pi/arctan(1/k)) = ceiling(k*Pi).

Numbers k such that there exists some integer m such that Pi/arctan(1/k) > m > k*Pi.

Numbers k such that A331859(k^2) = A121854(k^2)+1 = A121855(k^2).

In A331859 there is a remark that A331859(100^n) = A011545(n). I'm in doubt of this, because if k = 10^n is here, then A331859(100^n) = ceiling(k*Pi), while A011545(n) = ceiling(k*Pi)-1, this equality would be violated.

Note that for k >= 3 we have 1/k < Pi/arctan(1/k)-k*Pi < (Pi/3)/k. As a result, a necessary condition for k being a term here is that there exists some m such that 0 < m/k - Pi < (Pi/3)/k^2, and a sufficient condition is that there exists some m such that 0 < m/k - Pi < 1/k^2.

Let P(n) = A002485(n), Q(n) = A002486(n), then it is known that 1/(Q(n)*(Q(n)*Q(n+1))) < |P(n)/Q(n) - Pi| < 1/(Q(n)*Q(n+1)) for n >= 2; furthermore, P(n)/Q(n) - Pi is positive for odd n and negative for even n. As a result, let n >= 3, then we have:

- If n is even, then Q(n) can never be a term.

- If n is odd, then k = Q(n)*t is a term if t <= sqrt(Q(n+1)/Q(n)), in which case ceiling(Pi/arctan(1/k)) = P(n)*t+1 and ceiling(k*Pi) = P(n)*t. The converse is not true (e.g., n = 3, t = 4745).