Discussion
Sat Dec 05
04:18
Joerg Arndt: A comment in A218799 should suffice!
04:26
Bernard Schott: So, thanks to recycle this Olympiad sequence that is not enough interesting for you to be in OEIS. Merci.
05:39
Michel Marcus: "that is not enough interesting for you to be in OEIS" : this is not really the problem; but it is twice an existing sequence
08:35
Bernard Schott: Understood; if it is not really the problem, after examples of 02:27, some other sequences that are "twice an existing sequence": A233968(n) = 2*A138137(n), A071118(n) = 2*A002618(n), A305183(n) = 2*A187941(n).
08:39
Bernard Schott: So, if this Olympiad sequence with link, reference, comments is not enough interesting for OEIS because it is twice (for n>=1) an existing sequence, thanks to recycle.
Sun Dec 06
01:42
N. J. A. Sloane: I think add a comment to A218799, and recycle this.
COMMENTS
This sequence is the generalization of inspired by the 4th problem proposed during the second day of the final round of the 18th Austrian Mathematical Olympiad in 1987. The problem asked to find all triples solutions (x, y, z) only for n = 1987 (see Link, Reference and Examplelast example).
Discussion
Sat Dec 05
04:14
Bernard Schott: Thanks, replaced "generalization" by "inspired by".
Discussion
Sat Dec 05
02:27
Bernard Schott: Yes, there are many many sequences with this relation: A175395(n) = 2*A007598(n); A297842(n)=2*A001158(n); A162630(n)=2*A130517(n); A325042(n)=2*A301987(n)…
02:29
Bernard Schott: Please, if this Olympiad sequence is not enough interesting for OEIS, no problem to recycle this sequence. Thanks.
04:01
Kevin Ryde: I think you're got enough to say here to be a separate sequence. Not sure if "generalization" is the word though. The problem asked for the case n=1987 something something ...
NAME
Number of triples (x, y, z) of natural numbers satisfying x+y = n and 2*x*y = z^2.
Discussion
Sat Dec 05
01:51
Michel Marcus: a(n)=2*A218799(n) : so ---rather--- a comment in A218799 seems to make sense
FORMULA
a(0)=A218799(0); then for n>=1, a(n)=2*A218799(n) (observation remark from _Bernard Schott_, Dec 05 2020 Hugo Pfoertner_, Dec 02 2020).
NAME
Number of triples (x, y, z) of natural numbers satisfying x+y=n and 2*x*y = z^2.
FORMULA
a(0)=A218799(0); then for n>=1, a(n)=2*A218799(n). (observation from _Bernard Schott_, Dec 05 2020 Hugo Pfoertner
NAME
Number of triples (x, y, z) of natural numbers satisfying x + y = n and 2*x*y = z^2.