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A332045 Numbers k such that ceiling(Pi/arctan(1/k)) = ceiling(k*Pi)+1.

%I A332045 #39 Mar 31 2024 12:05:01
%S A332045 6,7,14,21,28,113,226,339,452,565,678,791,904,1017,1130,1243,1356,
%T A332045 1469,1582,1695,1808,1921,33215,99532,364913,729826,1725033,3450066,
%U A332045 5175099,27235615,52746197,131002976,471265707,811528438,2774848045,4738167652,567663097408
%N A332045 Numbers k such that ceiling(Pi/arctan(1/k)) = ceiling(k*Pi)+1.
%C A332045 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).
%C A332045 Numbers k such that there exists some integer m such that Pi/arctan(1/k) > m > k*Pi.
%C A332045 Numbers k such that A331859(k^2) = A121854(k^2)+1 = A121855(k^2).
%C A332045 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.
%C A332045 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.
%C A332045 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:
%C A332045 - If n is even, then Q(n) can never be a term.
%C A332045 - 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).
%H A332045 Jon E. Schoenfield, <a href="/A332045/b332045.txt">Table of n, a(n) for n = 1..2259</a>
%H A332045 Jon E. Schoenfield, <a href="/A332045/a332045.txt">Magma program</a>
%H A332045 Wikipedia, <a href="https://en.wikipedia.org/wiki/Continued_fraction">Continued fraction</a>
%H A332045 Zhihu, <a href="https://zhuanlan.zhihu.com/p/55768956">A counting puzzle of colliding blocks which is related to Pi (in Chinese)</a>
%e A332045 Pi/arctan(1/6) = 19.0228..., 6*Pi = 18.8495..., so 6 is a term.
%e A332045 113*t is here for t <= 17, because ceiling(Pi/arctan(1/(113*t))) = 355*t+1 and ceiling((113*t)*Pi) = 355*t.
%o A332045 (PARI) default(realprecision, 10000); isA332045(n) = ceil(Pi/atan(1/n))!=ceil(n*Pi)
%o A332045 (Magma) // See Schoenfield link.
%Y A332045 Cf. A121381 (ceiling(n*Pi)), A121854 (floor(sqrt(n)*Pi)), A121855 (ceiling(sqrt(n)*Pi)), A011545 (floor(10^n*Pi)).
%Y A332045 Cf. A331859, A002486, A002485.
%K A332045 nonn
%O A332045 1,1
%A A332045 _Jianing Song_, Feb 05 2020
%E A332045 a(27)-a(32) from _Jon E. Schoenfield_, Feb 12 2020
%E A332045 a(33)-a(36) from _Giovanni Resta_, Feb 12 2020
%E A332045 a(37) from _Jon E. Schoenfield_, Feb 15 2020