A348637 -id:A348637 - cp's OEIS frontend

A347040 Times on a 12-hour digital clock with 6 digits at which the three continuously moving hands of an analog clock, in the best approximation, enclose the same angles with one another, i.e., have the smallest sum of squares of the deviations from 120 degrees. When interpreting the terms as times of the day in the form hh:mm:ss, padding to the left with zeros is assumed.

2142, 4324, 12647, 14929, 23253, 25435, 33758, 35940, 44404, 50546, 54909, 61051, 65414, 71556, 80020, 82202, 90525, 92707, 101031, 103313, 111636, 113818, 122142, 124324

Offset: 1

Hugo Pfoertner, Aug 13 2021

An exact hit in which all angles are exactly 120 degrees is impossible. The smallest possible deviation occurs at two points in each display cycle, namely at 02:54:34.5617..., and at 09:05:25.4383... . With rounding to the nearest integer second, this corresponds to the terms a(6)=25435 and a(17)=90525.

The least squares clock solution actually occurs at an exact rational time, namely 5333364000/509173 seconds after 00:00:00, or 285998/509173 seconds after 02:54:34; and the exact least squares sum (in units of squared rotations) is 1/3055038 or (in units of squared clock-second-ticks) 3600/3055038 = 600/509173. - Robert B Fowler, Oct 29 2021

Henry Ernest Dudeney, Amusements in Mathematics, London, New York, Nelson, 1917. (Problem #63: The Stop-Watch)

Cf. A120500, A347039, A348637, A350141.

a(8) corrected by Robert B Fowler, Oct 29 2021

A350141 Times when clock hands are most nearly equidistant - expressed exactly as a(n)*43200/1427 seconds after 00:00:00.

Original entry on oeis.org

43, 86, 172, 217, 303, 346, 432, 475, 563, 606, 692, 735, 821, 864, 952, 995, 1081, 1124, 1210, 1255, 1341, 1384

Offset: 1

Robert B Fowler, Dec 15 2021

The three hands of an analog 12-hour clock are never exactly equidistant (mutually 120 degrees apart), but they are very nearly so at 22 times during the 12-hour clock cycle. These 22 cases occur at exactly a(n)*43200/1427 seconds after 00:00, according to the simple "midpoint" definition given below, although three other (more complicated) definitions of near-equidistance are also described here. 11 cases occur in the first 6 hours, and each has a mirror image case in the second 6 hours. For all n = 1..22, a(n) mirrors a(23-n).
There are at least four ways to define the precise times when each case attains its nearest equidistance. For each case, the four definitions generate four distinct solutions, which all lie within a fraction of a second from each other. The best pair of cases occur near 02:54:34.5 and its mirror image 09:05:25.5, which correspond to a(6) = 346 and a(17) = 1081. The second-best pair is a(11) and a(12). The third to 11th best are a(n) for n = 1, 5, 7, 2, 10, 8, 4, 9, 3 (and their mirrors 23 - n).
MIDPOINT (definition #1). The simplest definition first locates each of the 22 times when the hour and minute hands are exactly 120 degrees apart, then determines the nearest time (earlier or later) when the second hand is exactly midway between them and around 120 (not 60) degrees from each of them. The midpoint solution times occur at exactly a(n)*43200/1427 seconds after 00:00. The a(n) are found using the FORMULA below. Midpoint times are used as first approximations when calculating solution times for the other three definitions. A347040 gives standard clock times for a(n), rounded to the nearest second, in integer format hh:mm:ss.
LARGEST TRIANGLE AREA (definition #2). The best solution occurs when the triangle whose vertices are the endpoints of the three clock hands has maximum area. See A348637.
LEAST SQUARES (definition #3). The best solution occurs when the sum of the squares of the deviations of the angles of each pair of clock hands from 120 degrees is minimum. To find this time, express time T in units of 12-hours (T = 0 at 00:00 and T = 1 at 12:00), and express all clock hand angles A in units of 120 degrees (3 units = 360 degrees). In these units, the midpoint times are T(n) = a(n)/1427, and the three clock hand angles at any time T are
    A(1) = hour-hand to second-hand angle   = 3*(720-1)*T  = 2157*T,
    A(2) = minute-hand to second-hand angle = 3*(720-12)*T = 2124*T,
    A(3) = hour-hand to minute-hand angle   = 3*(12-1)*T   = 33*T.
At any time near any T(n), all three A(j) will be near-integer, with their variances from 120 degrees represented by
    V(j) = abs(A(j) - round(A(j))).
At exactly time T(n), these variations are all integer multiples of 1/1427, given by
    V(3) = abs(33*a(n) - 1427*k)/1427,
    V(1) = V(2) = V(3)/2,
where k is defined in the FORMULA below, and is equal to floor((3n-1)/2).
The sum of the squared variances to be minimized near time T(n) is
    F(T) = V(1)^2 + V(2)^2 + V(3)^2.
The least-squares time t occurs when F'(t) = 0. The values of t are rational numbers.
LEAST VARIANCE (definition #4). The best solution occurs when the sum of the absolute values of the deviations of the angles of each pair of clock hands from 120 degrees is minimum. To find this time, use the same A(j) and V(j) as defined in the least-squares solution above, but instead minimize the sum of the variances
    G(T) = V(1) + V(2) + V(3).
The least variance time u occurs when G'(u) = 0. The values of u are rational numbers.

			For n = 6, we have k = 8, m = floor(11416/33) = 345, a(6) = 346.
This corresponds to a midpoint time of 346*43200/1427 seconds after 00:00:00, which is a clock time 02:54:34.562018220042046...
Comparison of solutions according to the four definitions:
  ---------------------  ---------------------------  ----------------
       DEFINITION        Conventional Clock Times     12-hour Cycles
  ---------------------  ---------------------------  ----------------
  (#1) Midpoint          02:54:34.562018220042046...  346/1427
  (#2) Largest triangle  02:54:34.561690797181984...  (no closed form)
  (#3) Least squares     02:54:34.561691213006188...  246915/1018346
  (#4) Least variances   02:54:34.547983310152990...  523/2157
  ---------------------  ---------------------------  ----------------

Henry Ernest Dudeney, Amusements in Mathematics, London, New York, Nelson, 1917. (Problem #63: The Stop-Watch)

Cf. A347040, A348637.

a := proc(n) n + iquo(n - 1, 2): iquo(1427 * %, 33): % + irem(% - %%, 2) end: # Peter Luschny, Dec 17 2021

a[n_] := Module[{k = n + Floor[(n - 1)/2], m}, m = Floor[1427*k/33]; If[EvenQ[m - k], m, m + 1]]; Array[a, 22] (* Amiram Eldar, Dec 16 2021 *)

a(n) = m + mod(m - k, 2) where k = n + floor((n-1)/2) and m = floor(k*1427/33).

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A350141 Times when clock hands are most nearly equidistant - expressed exactly as a(n)*43200/1427 seconds after 00:00:00.

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