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Esanacci (hexanacci or "6-anacci") numbers.
18

%I #62 Aug 21 2024 11:39:15

%S 6,1,3,7,15,31,63,120,239,475,943,1871,3711,7359,14598,28957,57439,

%T 113935,225999,448287,889215,1763832,3498707,6939975,13766015,

%U 27306031,54163775,107438335,213112838,422726969,838513963,1663261911,3299217791,6544271807

%N Esanacci (hexanacci or "6-anacci") numbers.

%C These esanacci numbers follow the same pattern as Lucas, generalized tribonacci (A001644), generalized tetranacci (A073817), and generalized pentanacci (A074048) numbers.

%C The closed form is a(n) = r1^n + r^2^n + r3^n + r4^n + r5^n + r6^n, with r1, r2, r3, r4, r5, r6 roots of the characteristic polynomial.

%C a(n) is also the trace of A^n, where A is the matrix ((1, 1, 0, 0, 0, 0), (1, 0, 1, 0, 0, 0), (1, 0, 0, 1, 0, 0), (1, 0, 0, 0, 1, 0), (1, 0, 0, 0, 0, 1), (1, 0, 0, 0, 0, 0)).

%H T. D. Noe, <a href="/A074584/b074584.txt">Table of n, a(n) for n=0..200</a>

%H Martin Burtscher, Igor Szczyrba, and RafaƂ Szczyrba, <a href="https://www.emis.de/journals/JIS/VOL18/Szczyrba/sz3.html">Analytic Representations of the n-anacci Constants and Generalizations Thereof</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.

%H Mario Catalani, <a href="http://arxiv.org/abs/math/0210201">Polymatrix and Generalized Polynacci Numbers</a>, arXiv:math/0210201 [math.CO], 2002.

%H Spiros D. Dafnis, Andreas N. Philippou, and Ioannis E. Livieris, <a href="https://doi.org/10.3390/math8091487">An Alternating Sum of Fibonacci and Lucas Numbers of Order k</a>, Mathematics (2020) Vol. 9, 1487.

%H Tony D. Noe and Jonathan Vos Post, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Noe/noe5.html">Primes in Fibonacci n-step and Lucas n-step Sequences,</a> J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,1,1,1,1).

%F a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) + a(n-6), a(0)=6, a(1)=1, a(2)=3, a(3)=7, a(4)=15, a(5)=31.

%F G.f.: (6-5*x-4*x^2-3*x^3-2*x^4-x^5)/(1-x-x^2-x^3-x^4-x^5-x^6).

%F a(n) = 2*a(n-1) - a(n-7) for n >= 7. - _Vincenzo Librandi_, Dec 20 2010

%t CoefficientList[Series[(6-5*x-4*x^2-3*x^3-2*x^4-x^5)/(1-x-x^2-x^3-x^4-x^5-x^6), {x, 0, 40}], x]

%t LinearRecurrence[{1,1,1,1,1,1},{6,1,3,7,15,31},40] (* _Harvey P. Dale_, Nov 08 2011 *)

%o (PARI) polsym(polrecip(1-x-x^2-x^3-x^4-x^5-x^6), 40) \\ _G. C. Greubel_, Apr 22 2019

%o (Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (6-5*x-4*x^2-3*x^3-2*x^4-x^5)/(1-x-x^2-x^3-x^4-x^5-x^6) )); // _G. C. Greubel_, Apr 22 2019

%o (Sage) ((6-5*x-4*x^2-3*x^3-2*x^4-x^5)/(1-x-x^2-x^3-x^4-x^5-x^6)).series(x, 40).coefficients(x, sparse=False) # _G. C. Greubel_, Apr 22 2019

%o (Python)

%o def aupton(nn):

%o alst = [6, 1, 3, 7, 15, 31]

%o for n in range(6, nn+1): alst.append(sum(alst[n-6:n]))

%o return alst[:nn+1]

%o print(aupton(33)) # _Michael S. Branicky_, Jun 01 2021

%Y Cf. A000078, A001630, A001644, A000032, A073817, A074048.

%K easy,nonn

%O 0,1

%A Mario Catalani (mario.catalani(AT)unito.it), Aug 26 2002