Wikipedia:Reference desk/Archives/Mathematics/2022 August 18

= August 18 =

Lucas number questions
Naraht (talk) 14:02, 18 August 2022 (UTC)
 * 1) And this may belong on the WikiProject, should there be something in the article indicating that No Lucas Number is divisible by 5 (terms mod 5 go 2,1,3,4,2,1,...)
 * 2) What other Fibonacci like sequences have this property, in that no terms of the sequence are equal to 0 mod N for a given N.


 * Let the Fibonacci-like integer sequence $$S^{a,b}_0, S^{a,b}_1, ...$$ be defined by:
 * $$S^{a,b}_0=a,$$
 * $$S^{a,b}_1=b,$$
 * $$S^{a,b}_{n{+}2}=S^{a,b}_{n}+S^{a,b}_{n{+}1}.$$
 * So the original Fibonacci numbers are the elements of $$S^{0,1},$$ wile the Lucas numbers are the elements of $$S^{2,1}.$$ Then the following holds:
 * $$\forall~n\cdot S^{a,b}_n\not\equiv 0~(\hbox{mod}~5)\iff a\not\equiv 0~(\hbox{mod}~5)\,\land\,b\equiv 3a~(\hbox{mod}~5).$$
 * We can only add anything of the nature if it has been published in a reliable source we can cite. An obvious place to look is H. J. A. Duparc, Divisibility properties of recurring sequences (Ph.D. thesis, Amsterdam, 1953). I am not close to a library that might have this. I should add that the result is somewhat trivial: since $$S^{a,b}_{n{+}5}\equiv 3S^{a,b}_n~(\hbox{mod}~5),$$ you have to check only the first five terms $$a,b,a{+}b,a{+}2b,2a{+}3b.$$ --Lambiam 16:25, 18 August 2022 (UTC)


 * Also 13, 17, 37, ... . If x and y are consecutive entries, then x2+xy-y2 = ±5. If y=0 mod p then x2 = ±5 mod p, and you can use the theory of quadratic residues to determine the primes where this has no solution, basically all primes congruent to 13 or 17 mod 20. The case p = 5 is special, but then you get x=y=0 mod p which is also impossible. The upshot is that there are infinite number of primes p for which no Lucas number is divisible by p. --RDBury (talk) 16:54, 18 August 2022 (UTC)
 * It is not just primes. No Lucas number is divisible by 8, 12 or 21. --Lambiam 20:43, 18 August 2022 (UTC)
 * Good point; I was doing primes because they're easier. --RDBury (talk) 03:17, 19 August 2022 (UTC)
 * Also among the primes there are other non-dividers than 5 and those congruent to 13 or 17 modulo 20, for example 61, 89 and 109. --Lambiam 15:56, 20 August 2022 (UTC)
 * This is also true. I don't know if there is a way to prove this except by following out the sequence until there is a repeat mod p, then verifying that that there are no entries divisible by p. There are certainly values of x with x2 = 5 mod 61, namely 26 and 35. But even limited to pairs (x, y) with x2+xy-y2 = ±5, you can have multiple "orbits", and the question is whether (2, 1) is in the same orbit as (26, 0) and/or (35, 0). --RDBury (talk) 10:51, 22 August 2022 (UTC)
 * I found these indeed by following out the sequence until a repeat. There is an optimization possible, stopping as soon as the previous term is twice the current one mod p, but I didn't bother to use it since the repeat occurs within p2 steps. --Lambiam 11:43, 22 August 2022 (UTC)
 * As a side remark, everything may be restated in terms of the Fibonacci numbers, since by linearity $$S_n^{a,b}=aS_n^{1,0}+bS_n^{0,1}=aF_n+bF_{n-1}. $$ pm a 17:25, 23 August 2022 (UTC)

Sequence of no Lucas numbers divide bases?
Looks like we have 5,8,12,13,17, is there any way to generate a complete list?Naraht (talk) 19:13, 23 August 2022 (UTC)
 * Complete, no, since the list is infinite. For some more terms, see . --Lambiam 21:10, 23 August 2022 (UTC)
 * Well I think that leads to the potentially infinite versus the actual infinite question which is still a matter of debate for some! :-) NadVolum (talk) 22:53, 23 August 2022 (UTC)
 * I missed 10, that's why I missed the list in OEIS. :(.Naraht (talk) 23:52, 23 August 2022 (UTC)
 * I suspect that the asymptotic density of these non-dividers is in fact 1, just as for the composite numbers. --Lambiam 03:17, 24 August 2022 (UTC)
 * Let $$N_\operatorname{div}(n)$$ stand for the number of dividers (which divide some Lucas number: 1, 2, 3, 4, 6, 7, 9, 11, 14, 18, 19, ...) less than or equal to $$n.$$ It appears to be approximated by the formula
 * $$N_\operatorname{div}(n)\sim \frac{an}{\log (bn)},a\approx 2.08,b\approx 4.7,$$
 * which is reminiscent of the prime-counting function. --Lambiam 04:43, 25 August 2022 (UTC)

Is this plot of hours of daylight by latitude and day of year correct?
Long ago, I drew File:Hours_of_daylight_vs_latitude_vs_day_of_year_cmglee.svg ultimately based on File:Hours_of_daylight_vs_latitude_vs_day_of_year.png.

I've just realised that latitudes beyond the Arctic/Antarctic Circles experience continuous days and nights, so shouldn't the singularities be at ~67 instead of 90 degrees?

Thanks, cm&#610;&#671;ee&#9094;&#964;a&#671;&#954; 23:34, 18 August 2022 (UTC)


 * Apparently the calculations were originally based on Sunrise equation. I haven't checked the calculations, but the aspect you're talking about looks correct to me. Between the arctic circle and north pole there are times of the year with no sunlight (check), times with some sunlight and some dark (check), and times with all dark (check). The singularities are more apparent if you look at the animation on the Sunrise equation page for latitudes above and below the polar circles. (The original.) The graph of sunlight vs. time of years is has two flattened intervals representing periods of all day and all dark, and between them is a continuously varying curve. The corners are the singularities since the curve is not differentiable there, and they do start to appear at about 67 degrees. --RDBury (talk) 02:35, 19 August 2022 (UTC)
 * Thanks, RDBury. I mistakenly thought that beyond the circles, one experienced 6 months of daylight and 6 months of night, where it is only so at 90°. At other latitudes, some days have continuous daylight/night, but other days are "normal". cm&#610;&#671;ee&#9094;&#964;a&#671;&#954; 08:18, 20 August 2022 (UTC)