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Lucas numbers of the form (x^3 + y^3) / 2 where x and y are distinct positive integers.
3

%I #41 Jul 23 2024 12:38:17

%S 76,1364,24476,439204,7881196,141422324,2537720636,45537549124,

%T 817138163596,14662949395604,263115950957276,4721424167835364,

%U 84722519070079276,1520283919093591604,27280388024614569596,489526700523968661124,8784200221406821330636

%N Lucas numbers of the form (x^3 + y^3) / 2 where x and y are distinct positive integers.

%C Lucas numbers that are the averages of 2 distinct positive cubes.

%C Inspired by relation between sequence A024851 and A188378.

%C Corresponding indices are 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, ...

%C 6*n + 3 is the corresponding form of indices.

%C Corresponding y values are listed by A188378, for n > 0. Note that corresponding x values are A188378(n) - 2, for n > 0.

%H Colin Barker, <a href="/A267797/b267797.txt">Table of n, a(n) for n = 1..750</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (18,-1).

%F a(n) = A000032(A016945(n)), for n > 0.

%F a(n) = A188378(n)^3 - 3*A188378(n)^2 + 6*A188378(n) - 4, for n > 0.

%F From _Colin Barker_, Jan 24 2016: (Start)

%F a(n) = (9+4*sqrt(5))^(-n)*(2-sqrt(5)+(2+sqrt(5))*(9+4*sqrt(5))^(2*n)).

%F a(n) = 18*a(n-1)-a(n-2) for n>2.

%F G.f.: 4*x*(19-x) / (1-18*x+x^2).

%F (End)

%e Lucas number 76 is a term because 76 = (3^3 + 5^3) / 2.

%e Lucas number 1364 is a term because 1364 = (10^3 + 12^3) / 2.

%e Lucas number 24476 is a term because 24476 = (28^3 + 30^3) / 2.

%e Lucas number 439204 is a term because 439204 = (75^3 + 77^3) / 2.

%e Lucas number 7881196 is a term because 7881196 = (198^3 + 200^3) / 2.

%e Lucas number 141422324 is a term because 141422324 = (520^3 + 522^3) / 2.

%t Table[Fibonacci[6 n + 4] + Fibonacci[6 n + 2], {n, 1, 20}] (* _Vincenzo Librandi_, Jan 24 2016 *)

%t LinearRecurrence[{18,-1},{76,1364},20] (* _Harvey P. Dale_, Jul 23 2024 *)

%o (PARI) l(n) = fibonacci(n+1) + fibonacci(n-1);

%o is(n) = for(i=ceil(sqrtn(n\2+1, 3)), sqrtn(n-.5, 3), ispower(n-i^3, 3) && return(1));

%o for(n=1, 120, if(is(2*l(n)), print1(l(n), ", ")));

%o (PARI) a(n) = ((5*fibonacci(n)*fibonacci(n+1) + 1 + (-1)^n)^3 + (5*fibonacci(n)*fibonacci(n+1) - 1 + (-1)^n)^3) / 2;

%o (PARI) a(n) = (fibonacci(6*n+4) + fibonacci(6*n+2));

%o (PARI) Vec(4*x*(19-x)/(1-18*x+x^2) + O(x^20)) \\ _Colin Barker_, Jan 24 2016

%o (Magma) [Fibonacci(6*n+4)+Fibonacci(6*n+2): n in [1..20]]; // _Vincenzo Librandi_, Jan 24 2016

%Y Cf. A000032, A016945, A024670, A024851, A049629, A188378.

%K nonn,easy

%O 1,1

%A _Altug Alkan_, Jan 24 2016