Discussion
Sat Jul 23
14:52
Felix Fröhlich: Thanks for catching and correcting this.
CROSSREFS
Cf. A007814, A007949, A112762, A022337, A122840, A027868, A054899, A122841 , A160093, A160094, A196563, A196564.
Discussion
Sat Jul 23
14:49
Jianing Song: Added a missing comma
FORMULA
a(n) = (2*i*exp(-i*Pi/5)/sqrt(10 - 2*sqrt(5)))*Sum_{j=1..floor(log_5(n))}(1 - exp(2*Pi*i*binomial(n-1, 5^j-1)/5)). - Dario T. de Castro, Jul 19 2022
Discussion
Sat Jul 23
05:06
Kevin Ryde: Removed for collective opinion too obscure. At a no-change now I think.
Discussion
Wed Jul 20
09:44
Dario T. de Castro: Hi, Mr. Arndt. In fact this formula is the expression (29) in the paper published in Integers, that I included in the section Links of this page. I particularly think the expression with the 2*sin(Pi / p) in the denominator (with p = 5 here) would look better as it gives na idea of the general formula.
09:47
Dario T. de Castro: the link to the paper is
http://math.colgate.edu/~integers/w61/w61.pdf
Thu Jul 21
01:54
Kevin Ryde: Is this another 0 or 1 mod 5 and using on it exp, i, Pi, and a biquadrate root to turn that into 0 or 1 ? This is far too complicated.
07:23
Dario T. de Castro: this biquadrate root has to do with sin(Pi /5), since it is equal to (sqrt(10 - 2*sqrt(5)))/4
07:27
Dario T. de Castro: Regarding the fact that it is too complicated I agree, if you are looking from the perspective of determining v_p(n). But if one sees it as a way to determine the sum of certain binomial coefficients, using v_p(n) to know the answer. I think we can call it a simplification.
07:37
Kevin Ryde: No. You're bringing in complex roots of unity to do a mod 5 ? What's wrong with mod 5?
07:40
Kevin Ryde: In general, a lot of wild expressions simplify to simple sequences. It's impractical to list them all.
09:08
Dario T. de Castro: This is an interesting chapter of this discussion. But I think it will be dificult to make sure that all expressions shown in this site for instance are the simplest of all possible candidates. Equation (29) in the paper: http://math.colgate.edu/~integers/w61/w61.pdf are not what I would call a terribly complex formula, although I would call it impractical and inefficient computationally speaking. The presence of complex exponentials in it gives me an idea of "approachable by analytic number theory", for instance.
Fri Jul 22
07:57
Kevin Ryde: When it's been through analytic number theory and come out the other side with some wonderful result, please re-submit. For now it's going to be no.
13:04
Dario T. de Castro: It is ok.
FORMULA
a(n) = (2*i*exp(-i*Pi/5)/sqrt(10 - 2*sqrt(5)))*(floor(log_5(n)) - Sum_{j=1..floor(log_5(n))} (1 - exp(2*Pi*i*binomial(n-1, 5^j-1)/5)). - Dario T. de Castro, Jul 19 2022
