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Numbers x such that there exists n in N : (x+1)^3-x^3=79*n^2.
2

%I #21 Apr 07 2024 17:43:05

%S 97,44489542,20300667221737,9263235054568775182,

%T 4226832681849540586103377,1928712206384045833465574584822,

%U 880075237193226049220159073617573017

%N Numbers x such that there exists n in N : (x+1)^3-x^3=79*n^2.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (456303,-456303,1).

%F a(n) = 456302*a(n-1) - a(n-2) + 228150.

%F G.f.: x*(98*x^2-228151*x-97)/((x-1)*(x^2-456302*x+1)). - _Colin Barker_, Aug 24 2012

%e a(1)=97 because the first relation is : 98^3-97^3=79*19^2.

%K easy,nonn

%O 1,1

%A _Richard Choulet_, Oct 06 2008