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%I #29 Jan 29 2024 09:01:38
%S 4,11,11,29,29,76,76,199,199,521,521,1364,1364,3571,3571,9349,9349,
%T 24476,24476,64079,64079,167761,167761,439204,439204,1149851,1149851,
%U 3010349,3010349,7881196,7881196,20633239,20633239,54018521,54018521,141422324
%N Pairs (x, y) of relatively prime positive integers such that (x^2 - 5)/y and (y^2 - 5)/x are both positive integers.
%C For x, y > 2, the solutions start with (4,11) -> (11, 29) -> (29, 76) -> ...
%C The sequence is infinite (see the proof in the second reference).
%C Consider the pairs of the form (a(2n-1), a(2n)). Limit_{n->oo} a(2n)/a(2n-1) = phi^2 = 2.618033988749894... (A104457).
%C Property: a(2n-1)^2 + a(2n)^2 = 3*a(2n-1)*a(2n) + 5.
%C Apparently a(2*n) = a(2*n+1) = A002878(n) for n >= 1. - _Georg Fischer_, Dec 05 2022
%H Art of Problem Solving, <a href="http://www.artofproblemsolving.com/community/c146">Problem A112</a>
%H Peter Vandendriessche and Hojoo Lee, <a href="https://web.archive.org/web/20150412114922/http://www.problem-solving.be/pen/published/pen-20070711.pdf">Problems in Elementary Number Theory</a> (see problem A112, p. 15). [Via Wayback Machine]
%F Conjectures from _Chai Wah Wu_, Jan 28 2024: (Start)
%F a(n) = 3*a(n-2) - a(n-4) for n > 4.
%F G.f.: x*(-4*x^3 - x^2 + 11*x + 4)/(x^4 - 3*x^2 + 1). (End)
%p nn:=10^6:a:=4:
%p for b from a+1 to nn do:
%p x:=(a^2-5)/b:y:=(b^2-5)/a:
%p if x>0 and y>0 and gcd(a,b)=1 and x=floor(x) and y=floor(y)
%p then
%p printf(`%d, `,a): printf(`%d, `,b):a:=b:
%p else fi:
%p od:
%t nn = 10^6; a = 4; Reap[For[b = a+1, b <= nn, b++, x = (a^2-5)/b; y = (b^2-5)/a; If[x>0 && y>0 && GCD[a, b] == 1 && x == Floor[x] && y == Floor[y], Print[a, " ", b]; Sow[a]; Sow[b]; a = b]]][[2, 1]] (* adapted from Maple *)
%t (* Second program: *)
%t Clear[a]; a[n_] := 2^(-n-2)*((7-3*Sqrt[5])*(1-Sqrt[5])^n-(-Sqrt[5]-1)^(n+1) - (Sqrt[5]-1)^(n+1) + (3*Sqrt[5]+7)*(Sqrt[5]+1)^n); Table[a[n] // Simplify, {n, 1, 36}] (* _Jean-François Alcover_, Jan 25 2017 *)
%Y Cf. A001622, A002878, A104457.
%K nonn
%O 1,1
%A _Michel Lagneau_, Jan 21 2017
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