A074048 Pentanacci numbers with initial conditions a(0)=5, a(1)=1, a(2)=3, a(3)=7, a(4)=15.

5, 1, 3, 7, 15, 31, 57, 113, 223, 439, 863, 1695, 3333, 6553, 12883, 25327, 49791, 97887, 192441, 378329, 743775, 1462223, 2874655, 5651423, 11110405, 21842481, 42941187, 84420151, 165965647, 326279871, 641449337, 1261056193, 2479171199, 4873922247

Offset: 0

Author

Mario Catalani (mario.catalani(AT)unito.it), Aug 14 2002

Keywords

• easy
• nonn

Comments

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)).

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

a(3407) has 1001 decimal digits. - Michael De Vlieger, Dec 28 2016

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..3406
• Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
• C. A. Charalambides, Lucas numbers and polynomials of order k and the length of the longest circular success run, The Fibonacci Quarterly, 29 (1991), 290-297.
• Spiros D. Dafnis, Andreas N. Philippou, and Ioannis E. Livieris, An Alternating Sum of Fibonacci and Lucas Numbers of Order k, Mathematics (2020) Vol. 9, 1487.
• Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.4.
• S. Saito, T. Tanaka, and N. Wakabayashi, Combinatorial Remarks on the Cyclic Sum Formula for Multiple Zeta Values, Journal of Integer Sequences, 14 (2011), # 11.2.4, Table 3.
• Yüksel Soykan, On A Generalized Pentanacci Sequence, Asian Research Journal of Mathematics (2019) Vol. 14, No. 3, 1-9.
• Yüksel Soykan, Sum Formulas for Generalized Fifth-Order Linear Recurrence Sequences, Journal of Advances in Mathematics and Computer Science (2019) Vol. 34, No. 5, 1-14.
• Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
• L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.
• Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1).

Crossrefs

Cf. A000078, A001630, A001644, A000032, A073817, A106297 (Pisano Periods).

Essentially the same as A023424.

Cf. A123126, A123127, A074062.

Cf. A106273.

Programs

Mathematica

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]
LinearRecurrence[{1, 1, 1, 1, 1}, {5, 1, 3, 7, 15}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

PARI

polsym(polrecip(1-x-x^2-x^3-x^4-x^5),33) \\ Joerg Arndt, Jan 28 2019

Formula

a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5).

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

a(n) = 2*a(n-1) -a(n-6), n>5. [Vincenzo Librandi, Dec 20 2010]

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

From Kai Wang, Dec 16 2020: (Start)

For k >= 0,

| a(k+4) a(k+5) a(k+6) a(k+7) a(k+8) |

| a(k+3) a(k+4) a(k+5) a(k+6) a(k+7) |

det | a(k+2) a(k+3) a(k+4) a(k+5) a(k+6) | = 9584 = A106273(5).

| a(k+1) a(k+2) a(k+3) a(k+4) a(k+5) |

| a(k) a(k+1) a(k+2) a(k+3) a(k+4) |

(End)