|
%I #94 Feb 13 2024 06:55:37
%S 1,6,7,13,20,33,53,86,139,225,364,589,953,1542,2495,4037,6532,10569,
%T 17101,27670,44771,72441,117212,189653,306865,496518,803383,1299901,
%U 2103284,3403185,5506469,8909654,14416123,23325777,37741900,61067677,98809577,159877254
%N Fibonacci sequence beginning 1, 6.
%C a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(6;n-1-k,k), n>=1, with a(-1)=5. These are the sums of the SW-NE diagonals in P(6;n,k), the (6,1) Pascal triangle A093563. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also sums of SW-NE diagonals in (1,5)-Pascal triangle A096940.
%C Subsequence of primes: 7, 13, 53, 139, 953, 44771, 189653, 1494692464747, ... - _R. J. Mathar_, Aug 09 2012
%C a(n) is the sum of seven consecutive Fibonacci numbers. a(n) = F(n-4) + F(n-3) + F(n-2) + F(n-1) + F(n) + F(n+1) + F(n+2), where F(n)=A000045(n), extended so that F(-1)=1, F(-2)=-1, F(-3)=2, and F(-4)=-3. - _Graeme McRae_, Apr 24 2014
%H Vincenzo Librandi, <a href="/A022096/b022096.txt">Table of n, a(n) for n = 0..1000</a>
%H Jia Huang, <a href="https://arxiv.org/abs/1902.11139">Hecke algebras of simply-laced type with independent parameters</a>, arXiv:1902.11139 [math.RT], 2019.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, <a href="http://arxiv.org/abs/1212.1368">A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake</a>, arXiv preprint arXiv:1212.1368 [cs.DM], 2012-2014.
%H J. L. Ramírez and G. N. Rubiano, <a href="http://www.mathematica-journal.com/2014/02/properties-and-generalizations-of-the-fibonacci-word-fractal/">Properties and Generalizations of the Fibonacci Word Fractal</a>, The Mathematica Journal, Vol. 16 (2014).
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).
%F a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=6.
%F G.f.: (1+5*x)/(1-x-x^2).
%F a(n) = 5*Fibonacci(n+2) - 4*Fibonacci(n+1). - _Gary Detlefs_, Dec 21 2010
%F a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(-11 + sqrt(5)) + (1 + sqrt(5))^n*(11 + sqrt(5))))/sqrt(5). - _Herbert Kociemba_, Dec 18 2011
%F a(n) = Fibonacci(n+3) - Fibonacci(n-4). - _Greg Dresden_ and Sam Neale, Mar 08 2022
%t CoefficientList[Series[(1 + 5 x)/(1 - x - x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Apr 25 2014 *)
%t LinearRecurrence[{1,1},{1,6},40] (* _Harvey P. Dale_, Aug 07 2023 *)
%o (Magma) a0:=1; a1:=6; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // _Bruno Berselli_, Feb 12 2013
%o (PARI) a(n)=([0,1; 1,1]^n*[1;6])[1,1] \\ _Charles R Greathouse IV_, Jan 29 2016
%Y a(n) = A101220(5, 0, n+1).
%Y a(n) = A109754(5, n+1).
%Y Cf. A000045.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
%E Spelling correction by _Jason G. Wurtzel_, Aug 22 2010
|
