cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A295339 Least k for the inner Theodorus spiral to complete n revolutions.

Original entry on oeis.org

15, 52, 108, 184, 279, 394, 530, 684, 859, 1053, 1267, 1501, 1755, 2028, 2321, 2634, 2966, 3318, 3690, 4082, 4493, 4925, 5375, 5846, 6336, 6847, 7376, 7926, 8495, 9085, 9693, 10322, 10970, 11638, 12326, 13034, 13761, 14508, 15275
Offset: 1

Views

Author

Wolfdieter Lang, Dec 13 2017

Keywords

Comments

Here the points of the inner discrete Theodurus spiral in the complex plane are zhat(k) = rho(k)*exp(i*phihat(k)) with rho(k) = sqrt(k) and phihat(k) starts with phihat(1) = Pi/2 and is not restricted to be <= 2*Pi, it is phihat(k) = Sum_{j=0..k-1} (2*alpha(j+1) - alpha(j)) with alpha(j) = arctan(1/sqrt(j)), for k >= 1. The formula is phihat(k) = phi(k) + alpha(k), with the recurrence for the arguments of the outer spiral phi(k) = phi(k-1) + alpha(k-1), k >= 2, with phi(1) = 0.
If one considers punctured sheets S_n = rho*exp(i*phi_n), with rho > 0 and 2*Pi*(n-1) <= phi_n < 2*Pi*n, for n >= 1, then on sheet S_n there are a(n) - a(n-1) = A296179(n) points zhat, where a(0) = 0.
An analytic continuation of Davis's interpolation of the outer spiral is given in the Waldvogel link (see Figure 2 there). The point zhat(k) (called G_k on Figure 1 there) on the inner spiral is obtained from mirroring the point z(k) (called F_k there) of the outer spiral on the hypotenuse O,z(k+1), for k >= 1. In the present case the arguments phihat(k) of zhat(k) are taken positive.
Conjecture: a(n) = A072895(n) - 2, n >= 1. This follows from the conjecture that the sequences K := {floor(phi(k)/(2*Pi)}{k >= 1} with phi given above, and Khat:= {floor(phihat(k)/(2*Pi)}{k >= 1} with phihat given above satisfy Khat(k-2) = K(k), for k >= 3. Note that phihat(k-2) - phi(k) = alpha(k-2) - alpha(k-1) =: delta(k) = arctan((sqrt(k-1) - sqrt(k-2))/(1 + sqrt((k-1)*(k-2)))) > 0, for k >= 3. Therefore the conjecture is that delta(k) < 2*Pi*(1 - frac(phi(k)/(2*Pi))), for k >= 3, or, equivalently, phihat(k-2) < 2*Pi*(K(k) + 1), for k >= 3.

References

  • P. J. Davis, Spirals from Theodorus to Chaos, A K Peters, Wellesley, MA, 1993.

Crossrefs

Cf. A072895 (outer spiral), A296179.

Formula

a(n) = -1 + first position of n in the sequence
Khat:= {floor(phihat(k)/(2*Pi))}_{ k>= 1}, with phihat given in a comment above in terms of phi.
Conjecture: a(n) = A072895(n) - 2, n >= 1 (see the comment above).