%I
%S 12,44,95,166,256,367,497,647,816,1006,1215,1444,1692,1961,2249,2557,
%T 2884,3231,3598,3985,4392,4818,5264,5730,6215,6720,7245,7790
%N First point of the discrete Theodorus spiral in the fourth quadrant for the nth revolution, for n >= 1.
%C This sequence is used in a conjecture on points z_k of the discrete (outer) Theodorus spiral living on quadrant IV of the complex plane of sheet S_n, where S_n := {r*exp(i*phi), r > 0, 2*Pi*(n1) <= phi < 2*Pi*n}. This corresponds to the nth revolution, for n >= 1.
%C This conjecture is 2*Pi  varphi(A072895(n)) > arctan(a(n)), n >= 1, with varphi(k) = phi(k)  2*Pi*floor(phi(k)/(2*Pi)) where z_k = sqrt(k)*exp(i*phi(k)).
%C This conjecture implies a conjecture relating points of the discrete inner spiral to those of the outer ones, namely Khat(k2) := floor(phihat(k2)/(2*Pi)) = K(k) =: floor(phi(k)/(2*Pi)) for k >= 3, where zhat_k = sqrt(k)*exp(i*phihat(k)) is a point of the discrete inner Theodorus spiral, given in terms of z_k by zhat(k) = ((k1 + 2*sqrt(k)*i )/(k+1))*z_k. This implies phihat(k) = phi(k) + arctan((sqrt(k1)  sqrt(k2))/(1 + sqrt((k1)*(k2)))). The implied conjecture Khat(k2) = K(k), k >= 3, for the other three quadrants of each sheet S_n can be proved. For the inner spiral see the Waldvogel link.
%C If the implied conjecture is true then A295339(n) = A072895(n)  2, for n >= 1, hence A296179(n) = A295338(n), for n >= 2.
%C For the conjecture and the proof for the first three quadrants for each sheet S_n see the W. Lang link.  _Wolfdieter Lang_, Jan 24 2018
%H Wolfdieter Lang, <a href="/A296181/a296181.pdf">Notes on the Discrete Theodorus Spiral</a>
%H Joerg Waldvogel, <a href="http://www.sam.math.ethz.ch/~joergw/Papers/theopaper.pdf">Analytic Continuation of the Theodorus Spiral</a>.
%F a(n) is the smallest index k for which KIV(k) = n, with KIV(k):= floor((phi(k)  3*Pi/2)/(2*Pi)) + 1, for k >= 1, where phi(k) is the polar angle of the point z_k = sqrt(n)*exp(i*phi(k)) of the (outer) discrete Theodorus spiral.
%e a(1) = 12 because phi(11)  3*Pi/2 is about 0.1869017440 (Maple 10 digits), that is, KIV(11) = 1 + 1 = 0 (not n = 1) but phi(12)  3*Pi/2 is about +0.1059410277, that is, KIV(12) = 0 + 1 = 1 (on sheet S_1).
%e a(2) = 44 because phi(43)  3*Pi/2 is about 6.270091849, that is KIV(43) = 0 + 1 = 1 (not n = 2) but varphi(44)  3*Pi/2 is about 6.421424486, that is KIV(44) = 1 + 1 = 2 (on sheet S_2).
%Y Cf. A072895, A295338, A295339, A296179.
%K nonn,more
%O 1,1
%A _Wolfdieter Lang_, Jan 05 2018
