A292278 - OEIS

%I #62 Jan 05 2025 19:51:41

%S 1,3,11,45,189,799,3383,14329,60697,257115,1089155,4613733,19544085,

%T 82790071,350704367,1485607537,6293134513,26658145587,112925716859,

%U 478361013021,2026369768941,8583840088783,36361730124071,154030760585065,652484772464329

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

%C 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.

%C 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

%H Vincenzo Librandi, <a href="/A292278/b292278.txt">Table of n, a(n) for n = 1..1000</a>

%H Hideyuki Ohtsuka, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Problems/ElemProSolnAugust2017.pdf">Problem B-1211</a>, The Fibonacci Quarterly, Volume 55, Number 3 (August 2017), p. 276 (see Comments section).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-3,-1).

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

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

%t Table[(Fibonacci[3 n - 1] + 1) / 2, {n, 40}]

%t LinearRecurrence[{5,-3,-1},{1,3,11},30] (* _Harvey P. Dale_, Mar 06 2024 *)

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

%o (PARI) a(n) = (fibonacci(3*n-1)+1)/2; \\ _Altug Alkan_, Sep 13 2017

%Y Cf. A000045; A005968, A056570.

%K nonn,easy

%O 1,2

%A _Vincenzo Librandi_, Sep 13 2017

%E Edited by _Bruno Berselli_, Sep 14 2017