A024851 -id:A024851 - cp's OEIS frontend

A188378 Partial sums of A005248.

2, 5, 12, 30, 77, 200, 522, 1365, 3572, 9350, 24477, 64080, 167762, 439205, 1149852, 3010350, 7881197, 20633240, 54018522, 141422325, 370248452, 969323030, 2537720637, 6643838880, 17393796002, 45537549125, 119218851372, 312119004990, 817138163597, 2139295485800

Offset: 0

Gabriele Fici, Mar 29 2011

Different from A024851.

Luo proves that these integers cannot be uniquely decomposed as the sum of distinct and nonconsecutive terms of the Lucas number sequence. - Michel Marcus, Apr 20 2020

Robert Israel, Table of n, a(n) for n = 0..1000
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
David C. Luo, Nonuniqueness Properties of Zeckendorf Related Decompositions, arXiv:2004.08316 [math.NT], 2020.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000032, A002878, A027941, A001654.

Cf. A016945, A267797.

[5*Fibonacci(n)*Fibonacci(n+1)+1+(-1)^n: n in [0..40]]; // Vincenzo Librandi, Jan 24 2016

f:= gfun:-rectoproc({a(n+3)-4*a(n+2)+4*a(n+1)-a(n), a(0) = 2, a(1) = 5, a(2) = 12}, a(n), remember):
map(f, [$0..60]); # Robert Israel, Feb 02 2016

LinearRecurrence[{4,-4,1},{2,5,12},30] (* Harvey P. Dale, Oct 05 2015 *)
Accumulate@ LucasL@ Range[0, 58, 2] (* Michael De Vlieger, Jan 24 2016 *)

a(n) = 5*fibonacci(n)*fibonacci(n+1) + 1 + (-1)^n; \\ Michel Marcus, Aug 26 2013

Vec((-2+3*x)/((x-1)*(x^2-3*x+1)) + O(x^100)) \\ Altug Alkan, Jan 24 2016

a(n) = A000032(2n+1)+1 = A002878(n)+1 = 2*A027941(n+1)-3*A027941(n).
G.f.: ( -2+3*x ) / ( (x-1)*(x^2-3*x+1) ). - R. J. Mathar, Mar 30 2011
a(n) = 5*A001654(n) + 1 + (-1)^n, n>=0. [Wolfdieter Lang, Jul 23 2012]
(a(n)^3 + (a(n)-2)^3) / 2 = A000032(A016945(n)) = Lucas(6*n+3) = A267797(n), for n>0. - Altug Alkan, Jan 31 2016
a(n) = 2^(-1-n)*(2^(1+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n). - Colin Barker, Nov 02 2016

A267797 Lucas numbers of the form (x^3 + y^3) / 2 where x and y are distinct positive integers.

Original entry on oeis.org

76, 1364, 24476, 439204, 7881196, 141422324, 2537720636, 45537549124, 817138163596, 14662949395604, 263115950957276, 4721424167835364, 84722519070079276, 1520283919093591604, 27280388024614569596, 489526700523968661124, 8784200221406821330636

Offset: 1

Altug Alkan, Jan 24 2016

Lucas numbers that are the averages of 2 distinct positive cubes.
Inspired by relation between sequence A024851 and A188378.
Corresponding indices are 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, ...
6*n + 3 is the corresponding form of indices.
Corresponding y values are listed by A188378, for n > 0. Note that corresponding x values are A188378(n) - 2, for n > 0.

			Lucas number 76 is a term because 76 = (3^3 + 5^3) / 2.
Lucas number 1364 is a term because 1364 = (10^3 + 12^3) / 2.
Lucas number 24476 is a term because 24476 = (28^3 + 30^3) / 2.
Lucas number 439204 is a term because 439204 = (75^3 + 77^3) / 2.
Lucas number 7881196 is a term because 7881196 = (198^3 + 200^3) / 2.
Lucas number 141422324 is a term because 141422324 = (520^3 + 522^3) / 2.

Colin Barker, Table of n, a(n) for n = 1..750
Index entries for linear recurrences with constant coefficients, signature (18,-1).

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

[Fibonacci(6*n+4)+Fibonacci(6*n+2): n in [1..20]]; // Vincenzo Librandi, Jan 24 2016

Table[Fibonacci[6 n + 4] + Fibonacci[6 n + 2], {n, 1, 20}] (* Vincenzo Librandi, Jan 24 2016 *)
LinearRecurrence[{18,-1},{76,1364},20] (* Harvey P. Dale, Jul 23 2024 *)

l(n) = fibonacci(n+1) + fibonacci(n-1);
is(n) = for(i=ceil(sqrtn(n\2+1, 3)), sqrtn(n-.5, 3), ispower(n-i^3, 3) && return(1));
for(n=1, 120, if(is(2*l(n)), print1(l(n), ", ")));

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

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

Vec(4*x*(19-x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016

a(n) = A000032(A016945(n)), for n > 0.
a(n) = A188378(n)^3 - 3*A188378(n)^2 + 6*A188378(n) - 4, for n > 0.
From Colin Barker, Jan 24 2016: (Start)
a(n) = (9+4*sqrt(5))^(-n)*(2-sqrt(5)+(2+sqrt(5))*(9+4*sqrt(5))^(2*n)).
a(n) = 18*a(n-1)-a(n-2) for n>2.
G.f.: 4*x*(19-x) / (1-18*x+x^2).
(End)

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A188378 Partial sums of A005248.

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A267797 Lucas numbers of the form (x^3 + y^3) / 2 where x and y are distinct positive integers.

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