cp's OEIS Frontend

A339377 Number of triples (x, y, z) of natural numbers satisfying x+y = n and 2*x*y = z^2.

%I A339377 #40 Dec 06 2020 01:42:03
%S A339377 1,2,2,4,2,2,4,2,2,6,2,4,4,2,2,4,2,4,6,4,2,4,4,2,4,2,2,8,2,2,4,2,2,10,
%T A339377 4,2,6,2,4,4,2,4,4,4,4,6,2,2,4,2,2,10,2,2,8,4,2,10,2,4,4,2,2,6,2,2,10,
%U A339377 4,4,4,2,2,6,4,2,4,4,4,4,2,2,10,4,4,4,4,4,4
%N A339377 Number of triples (x, y, z) of natural numbers satisfying x+y = n and 2*x*y = z^2.
%C A339377 This sequence is 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 last example).
%C A339377 Some properties:
%C A339377 -> Inequalities, 0 <= x, y <= n; 0 <= z <= floor(n*sqrt(2)/2)
%C A339377 -> z is even and (x,y) are not together even.
%C A339377 -> a(n) = 1 iff n = 0, and the only solution is (0,0,0).
%C A339377 -> for n >= 1, a(n) >= 2 because (0,n,0) and (n,0,0) are always solutions.
%C A339377 -> a(n) is even for n >= 1.
%C A339377 -> If n = 3k, then (k,2k,2k) and (2k,k,2k) are solutions.
%C A339377 -> If 2*(n-1) = m^2, then (1,n-1,m) and (n-1,1,m) are solutions (with n in A058331).
%C A339377 -> The formula for n>0 comes from (x+y=n and 2*x*y=z^2) <==> n^2 = |x-y|^2 + 2*z^2.
%D A339377 Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 4 of Austrian Mathematical Olympiad 1987, page 29 [Warning: solution proposed in this book has a mistake with (x, y, z) = ([0, 1987], 1987-x, sqrt(2xy))].
%H A339377 The IMO compendium, <a href="https://imomath.com/othercomp/Aut/AutMO87.pdf"> Problem 4</a>, 18th Austrian Mathematical Olympiad, 1987.
%H A339377 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F A339377 a(0)=A218799(0); then for n>=1, a(n)=2*A218799(n) (remark from _Hugo Pfoertner_, Dec 02 2020).
%e A339377 a(9) = 6 and these 6 solutions are: (0, 9, 0), (1, 8, 4), (3, 6, 6), (6, 3, 6), (8, 1, 4), (9, 0, 0).
%e A339377 a(1987) = 4 and these 4 solutions are: (0, 1987, 0), (529, 1458, 1242), (1458, 529, 1242), (1987, 0, 0); this is the answer to the Olympiad problem in link.
%Y A339377 Cf. A058331, A218799, A339378 (variant with x+y = n and x*y = z^2).
%K A339377 nonn
%O A339377 0,2
%A A339377 _Bernard Schott_, Dec 02 2020