

A332045


Numbers k such that ceiling(Pi/arctan(1/k)) = ceiling(k*Pi)+1.


1



6, 7, 14, 21, 28, 113, 226, 339, 452, 565, 678, 791, 904, 1017, 1130, 1243, 1356, 1469, 1582, 1695, 1808, 1921, 33215, 99532, 364913, 729826, 1725033, 3450066, 5175099, 27235615, 52746197, 131002976, 471265707, 811528438, 2774848045, 4738167652, 567663097408
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OFFSET

1,1


COMMENTS

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.
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).


LINKS



EXAMPLE

Pi/arctan(1/6) = 19.0228..., 6*Pi = 18.8495..., so 6 is a term.
113*t is here for t <= 17, because ceiling(Pi/arctan(1/(113*t))) = 355*t+1 and ceiling((113*t)*Pi) = 355*t.


PROG

(PARI) default(realprecision, 10000); isA332045(n) = ceil(Pi/atan(1/n))!=ceil(n*Pi)
(Magma) See Schoenfield link


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



