A341671 - OEIS

%I #34 Feb 20 2021 16:06:44

%S 3,39,543,7563,105339,1467183,20435223,284625939,3964327923,

%T 55215964983,769059181839,10711612580763,149193516948843,

%U 2077997624703039,28942773228893703,403120827579808803,5614748812888429539,78203362552858204743,1089232326927126436863,15171049214426911911339

%N Solutions y of the Diophantine equation 3*(x^2+x+1) = y^2.

%C Corresponding x are in A028231.

%C This equation belongs to the family of equations studied by Kustaa A. Inkeri, y^m = a * (x^q-1)/(x-1) with here: m=2, a=3, q=3. This equation is exhibed in A307745 by _Giovanni Resta_ to prove that this sequence has infinitely many terms.

%C This Diophantine equation 3*(x^2+x+1) = y^2 has infinitely many solutions because the Pell-Fermat equation u^2 - 3*v^2 = -2 also has infinitely many solutions. The corresponding (u,v) are in (A001834, A001835) and for each pair (u,v), the corresponding solutions of 3*(x^2+x+1) = y^2 are x = (3*u*v-1)/2 and y = 3*(u^2+1)/2.

%C Note that if y = 3*z, this equation becomes 3*z^2 = x^2+x+1 with solutions (x, z) = (A028231, A001570).

%H Kustaa A. Inkeri, <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa21/aa21117.pdf">On the Diophantine equation a(x^n-1)/(x-1) = y^m</a>, Acta Arithmetica, Vol. 21, No. 1 (1972), pp. 299-311.

%F a(n) = 3*A001570(n). - _Hugo Pfoertner_, Feb 17 2021

%F a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).

%e The first few values for (x,y) are (1,3), (22,39), (313,543), (4366,7563), (60817,105339), ...

%t f[x_] := Sqrt[3*(x^2 + x + 1)]; f /@ LinearRecurrence[{15, -15, 1}, {1, 22, 313}, 20] (* _Amiram Eldar_, Feb 17 2021 *)

%Y Cf. A001834, A001835, A001570, A028231, A307745.

%Y Subsequence of A158235, for a(n)>3.

%K nonn

%O 1,1

%A _Bernard Schott_, Feb 17 2021

%E More terms from _Amiram Eldar_, Feb 17 2021