A106297 -id:A106297 - cp's OEIS frontend

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

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

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

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).

Cf. A000078, A001630, A001644, A000032, A073817, A106297 (Pisano Periods).
Essentially the same as A023424.
Cf. A123126, A123127, A074062.
Cf. A106273.

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 *)

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

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)

A106303 Period of the Fibonacci 5-step sequence A001591 mod n.

Original entry on oeis.org

1, 6, 104, 12, 781, 312, 2801, 24, 312, 4686, 16105, 312, 30941, 16806, 81224, 48, 88741, 312, 13032, 9372, 291304, 96630, 12166, 312, 3905, 185646, 936, 33612, 70728, 243672, 190861, 96, 1674920, 532446, 2187581, 312, 1926221, 13032

Offset: 1

T. D. Noe, May 02 2005

This sequence can differ from the corresponding Lucas sequence (A106297) only when n is a multiple of 2 or 599 because 9584 is the discriminant of the characteristic polynomial x^5-x^4-x^3-x^2-x-1 and the prime factors of 9584 are 2 and 599. [Corrected by Avery Diep, Aug 25 2025]

Chai Wah Wu, Table of n, a(n) for n = 1..388
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Cf. A001591, A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1), A106297 (period of Lucas 5-step sequence mod n).

n=5; Table[p=i; a=Join[{1}, Table[0, {n-1}]] a=Mod[a, p]; a0=a; k=0; While[k++; s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[n]]=s; a!=a0]; k, {i, 50}]

from itertools import count
def A106303(n):
    a = b = (0,)*4+(1 % n,)
    s = 1 % n
    for m in count(1):
        b, s = b[1:] + (s,), (s+s-b[0]) % n
        if a == b:
            return m # Chai Wah Wu, Feb 21-27 2022

Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).

Conjectures: a(5^k) = 781*5^(k-1) for k > 0. If a(p) != a(p^2) for p prime, then a(p^k) = p^(k-1)*a(p) for k > 0. - Chai Wah Wu, Feb 25 2022

A106298 Period of the Lucas 5-step sequence A074048 mod prime(n).

Original entry on oeis.org

1, 104, 781, 2801, 16105, 30941, 88741, 13032, 12166, 70728, 190861, 1926221, 2896405, 79506, 736, 8042221, 102689, 3720, 20151120, 2863280, 546120, 39449441, 48030024, 3690720, 29509760, 104060400, 37516960, 132316201, 28231632, 6384, 86714880, 2248090, 3128

Offset: 1

T. D. Noe, May 02 2005

nonn

This sequence is the same as the period of Fibonacci 5-step sequence (A106304) mod prime(n) except for n=1 and 109, which correspond to the primes 2 and 599 because 9584 is the discriminant of the characteristic polynomial x^5-x^4-x^3-x^2-x-1 and the prime factors of 9584 are 2 and 599. We have a(n) < prime(n) for the primes 2, 599 and A106281.

Chai Wah Wu, Table of n, a(n) for n = 1..82
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Cf. A106281 (primes p such that x^5-x^4-x^3-x^2-x-1 mod p has 5 distinct zeros), A106297.

n=5; Table[p=Prime[i]; a=Join[Table[ -1, {n-1}], {n}]; a=Mod[a, p]; a0=a; k=0; While[k++; s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[n]]=s; a!=a0]; k, {i, 40}]

from itertools import count
from sympy import prime
def A106298(n):
    a = b = (5%(p:=prime(n)),1%p,7%p,3%p,15%p)
    s = sum(b) % p
    for m in count(1):
        b, s = b[1:] + (s,), (s+s-b[0]) % p
        if a == b:
            return m # Chai Wah Wu, Feb 22-27 2022

a(n) = A106297(prime(n)).

a(31)-a(33) from Chai Wah Wu, Feb 27 2022

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

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A106303 Period of the Fibonacci 5-step sequence A001591 mod n.

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A106298 Period of the Lucas 5-step sequence A074048 mod prime(n).

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