A098703 a(n) = (3^n + phi^(n-1) + (-phi)^(1-n)) / 5, where phi denotes the golden ratio A001622.

0, 1, 2, 6, 17, 50, 148, 441, 1318, 3946, 11825, 35454, 106328, 318929, 956698, 2869950, 8609617, 25828474, 77484812, 232453449, 697358750, 2092073666, 6276216817, 18828643686, 56485920112, 169457742625, 508373199218, 1525119551286

Offset: 0

Author

Ross La Haye, Oct 27 2004

Keywords

• nonn
• easy

Comments

Sums of antidiagonals of A090888.

Partial sums of A099159.

Form an array with m(0,n) = A000045(n), the Fibonacci numbers, and m(i,j) = Sum_{kJ. M. Bergot, May 27 2013

Examples

			a(2) = 2 because 3^2 = 9, Luc(1) = 1 and (9 + 1) / 5 = 2.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Eric Weisstein, Golden Ratio
• Eric Weisstein, Lucas Number
• Eric Weisstein, Fibonacci Number
• Index entries for linear recurrences with constant coefficients, signature (4,-2,-3).

Crossrefs

Cf. A000032, A000045, A000244, A000285, A001622, A001654, A022383, A052964, A084178, A090888, A094688, A099159.

Programs

Magma

I:=[0,1,2]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2)-3*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 18 2018

Mathematica

f[n_] := (3^n + Fibonacci[n] + Fibonacci[n - 2])/5; Table[ f[n], {n, 0, 27}] (* Robert G. Wilson v, Nov 04 2004 *)
LinearRecurrence[{4, -2, -3}, {0, 1, 2}, 30] (* Jean-François Alcover, Feb 17 2018 *)

SageMath

def A098703(n): return (3**n + lucas_number2(n-1,1,-1))//5
print([A098703(n) for n in range(21)]) # G. C. Greubel, Jun 02 2025

Formula

a(n) = (((1 + sqrt(5))^n - (1 - sqrt(5))^n) / (2^n*sqrt(5))) + ((3^n - (((1 + sqrt(5)) / 2)^(n+1) + ((1 - sqrt(5)) / 2)^(n+1))) / 5).

a(n) = (3^n + (((1 + sqrt(5)) / 2)^(n-1) + ((1 - sqrt(5)) / 2)^(n-1))) / 5.

a(n) = (3^n + A000032(n-1))/5 = A000045(n) + (3^n - A000032(n+1))/5.

a(n) = (3^n + A000045(n) + A000045(n-2))/5.

a(n) = (3^n + 4*A000045(n) - A000045(n+2))/5.

a(n) = Sum_{k=0...n-1} (A000045(k)*3^(n-k-1) - A000045(k-2)*2^(n-k-1)).

a(n) = 4*a(n-1) - 2*a(n-2) - 3*a(n-3).

a(n) = A000045(n) + A094688(n-1).

a(n) = 3^1 * a(n-1) - A000045(n-3), for n > 2.

a(n) = 3^2 * a(n-2) - A000285(n-4), for n > 3.

a(n) = 3^3 * a(n-3) - A022383(n-5), for n > 4.

Limit_{n -> oo} a(n) / a(n-1) = 3.

From Ross La Haye, Dec 21 2004: (Start)

a(n) = A101220(1,3,n).

Binomial transform of unsigned A084178.

Binomial transform of signed A084178 gives the Fibonacci oblongs (A001654). (End)

G.f.: x*(1-2*x)/((1-3*x)*(1-x-x^2)). - Ross La Haye, Aug 09 2005

a(0) = 0, a(1) = 1, a(n) = a(n-1) + a(n-2) + 3^(n-2) for n > 1. - Ross La Haye, Aug 20 2005

Binomial transform of A052964 beginning {0,1,0,3,1,10,...}. - Ross La Haye, May 31 2006

Extensions

More terms from Robert G. Wilson v, Nov 04 2004

More terms from Ross La Haye, Dec 21 2004