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%I #89 Mar 24 2024 04:23:13
%S 5,1,3,7,15,31,57,113,223,439,863,1695,3333,6553,12883,25327,49791,
%T 97887,192441,378329,743775,1462223,2874655,5651423,11110405,21842481,
%U 42941187,84420151,165965647,326279871,641449337,1261056193,2479171199,4873922247
%N Pentanacci numbers with initial conditions a(0)=5, a(1)=1, a(2)=3, a(3)=7, a(4)=15.
%C These pentanacci numbers follow the same pattern as Lucas, generalized tribonacci(A001644) and generalized tetranacci (A073817) numbers: Binet's formula is a(n)=r1^n+r^2^n+r3^n+r4^n+r5^n, with r1, r2, r3, r4, r5 roots of the characteristic polynomial. a(n) is also the trace of A^n, where A is the pentamatrix ((1,1,0,0,0),(1,0,1,0,0),(1,0,0,1,0),(1,0,0,0,1),(1,0,0,0,0)).
%C For n >= 5, a(n) is the number of cyclic sequences consisting of n zeros and ones that do not contain five consecutive ones provided the positions of the zeros and ones are fixed on a circle. This is proved in Charalambides (1991) and Zhang and Hadjicostas (2015). (For n=1,2,3,4 the statement is still true provided we allow the sequence to wrap around itself on a circle). - _Petros Hadjicostas_, Dec 18 2016
%C a(3407) has 1001 decimal digits. - _Michael De Vlieger_, Dec 28 2016
%H Michael De Vlieger, <a href="/A074048/b074048.txt">Table of n, a(n) for n = 0..3406</a>
%H Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, <a href="http://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 C. A. Charalambides, <a href="http://www.fq.math.ca/Scanned/29-4/charalambides.pdf">Lucas numbers and polynomials of order k and the length of the longest circular success run</a>, The Fibonacci Quarterly, 29 (1991), 290-297.
%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>, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.4.
%H S. Saito, T. Tanaka, and N. Wakabayashi, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Saito/saito22.html">Combinatorial Remarks on the Cyclic Sum Formula for Multiple Zeta Values</a>, Journal of Integer Sequences, 14 (2011), # 11.2.4, Table 3.
%H Yüksel Soykan, <a href="https://doi.org/10.9734/ARJOM/2019/v14i330129">On A Generalized Pentanacci Sequence</a>, Asian Research Journal of Mathematics (2019) Vol. 14, No. 3, 1-9.
%H Yüksel Soykan, <a href="https://doi.org/10.9734/JAMCS/2019/v34i530224">Sum Formulas for Generalized Fifth-Order Linear Recurrence Sequences</a>, Journal of Advances in Mathematics and Computer Science (2019) Vol. 34, No. 5, 1-14.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.
%H L. Zhang and P. Hadjicostas, <a href="http://www.appliedprobability.org/data/files/TMS%20articles/40_2_3.pdf">On sequences of independent Bernoulli trials avoiding the pattern '11..1'</a>, Math. Scientist, 40 (2015), 89-96.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,1,1,1).
%F a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5).
%F G.f.: (5-4*x-3*x^2-2*x^3-x^4) / (1-x-x^2-x^3-x^4-x^5).
%F a(n) = 2*a(n-1) -a(n-6), n>5. [_Vincenzo Librandi_, Dec 20 2010]
%F For k>0 and n>=0, a(n+5*k) = a(k)*a(n+4*k) - A123127(k-1)*a(n+3*k) + A123126(k-1)*a(n+2*k) - A074062(k)*a(n+k) + a(n). For example, if k=4, n=3, we have a(n+5*k) = a(23) = 5651423, a(4)*a(19) - A123127(3)*a(15) + A123126(3)*a(1695) - A074062(4)*a(7) + a(3) = (15)*(378329) - (1)*(25327) + (1)*(1695) - (-1)*(113) + (7) = 5651423. - _Kai Wang_, Sep 13 2020
%F From _Kai Wang_, Dec 16 2020: (Start)
%F For k >= 0,
%F | a(k+4) a(k+5) a(k+6) a(k+7) a(k+8) |
%F | a(k+3) a(k+4) a(k+5) a(k+6) a(k+7) |
%F det | a(k+2) a(k+3) a(k+4) a(k+5) a(k+6) | = 9584 = A106273(5).
%F | a(k+1) a(k+2) a(k+3) a(k+4) a(k+5) |
%F | a(k) a(k+1) a(k+2) a(k+3) a(k+4) |
%F (End)
%t CoefficientList[Series[(5-4*x-3*x^2-2*x^3-x^4)/(1-x-x^2-x^3-x^4-x^5), {x, 0, 30}], x]
%t LinearRecurrence[{1, 1, 1, 1, 1}, {5, 1, 3, 7, 15}, 60] (* _Vladimir Joseph Stephan Orlovsky_, Feb 08 2012 *)
%o (PARI) polsym(polrecip(1-x-x^2-x^3-x^4-x^5),33) \\ _Joerg Arndt_, Jan 28 2019
%Y Cf. A000078, A001630, A001644, A000032, A073817, A106297 (Pisano Periods).
%Y Essentially the same as A023424.
%Y Cf. A123126, A123127, A074062.
%Y Cf. A106273.
%K easy,nonn
%O 0,1
%A Mario Catalani (mario.catalani(AT)unito.it), Aug 14 2002
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