A292278 a(n) = (Fibonacci(3*n-1) + 1)/2 for n >= 1.

1, 3, 11, 45, 189, 799, 3383, 14329, 60697, 257115, 1089155, 4613733, 19544085, 82790071, 350704367, 1485607537, 6293134513, 26658145587, 112925716859, 478361013021, 2026369768941, 8583840088783, 36361730124071, 154030760585065, 652484772464329

Offset: 1

Author

Vincenzo Librandi, Sep 13 2017

Keywords

• nonn
• easy

Comments

Problem B-1211 proposed by Hideyuki Ohtsuka (see Links section): For n >= 1, prove that Fibonacci(n-1)^3 + Sum_{k=1..n} Fibonacci(k)^3 = (Fibonacci(3*n-1) + 1)/2.

Proof. Let F(n-1)^3 = (F(3*n-3) + 3*(-1)^n*F(n-1))/5 (see Ralf Stephan's formula in A056570) and Sum_{k=1..n} F(k)^3 = (F(3*n+2) - 6*(-1)^(n)*F(n-1) + 5)/10 (see Benjamin & Timothy's formula in A005968), where F=A000045, n>0. Therefore, (F(3*n-3) + 3*(-1)^n*F(n-1))/5 + (F(3*n+2) - 6*(-1)^(n)*F(n-1) + 5)/10 = (2*F(3*n-3) + F(3*n+2) + 5)/10 = (2*(F(3*n-1) - F(3*n-2)) + (3*F(3*n-1) + 2*F(3*n-2)) + 5)/10 = (5*F(3*n-1) + 5)/10 = a(n). - Bruno Berselli, Sep 14 2017

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Hideyuki Ohtsuka, Problem B-1211, The Fibonacci Quarterly, Volume 55, Number 3 (August 2017), p. 276 (see Comments section).
• Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Crossrefs

Cf. A000045; A005968, A056570.

Programs

Magma

[(Fibonacci(3*n-1)+1)/2: n in [1..30]];

Mathematica

Table[(Fibonacci[3 n - 1] + 1) / 2, {n, 40}]
LinearRecurrence[{5,-3,-1},{1,3,11},30] (* Harvey P. Dale, Mar 06 2024 *)

PARI

a(n) = (fibonacci(3*n-1)+1)/2; \\ Altug Alkan, Sep 13 2017

Formula

G.f.: x*(1 - 2*x - x^2)/((1 - x)*(1 - 4*x -x^2)).

a(n) = 5*a(n-1) - 3*a(n-2) - a(n-3).

Extensions

Edited by Bruno Berselli, Sep 14 2017