A106281 -id:A106281 - cp's OEIS frontend

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

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

A106278 Number of distinct zeros of x^5-x^4-x^3-x^2-x-1 mod prime(n).

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 2, 3, 0, 2, 3, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 3, 1, 2, 3, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 3, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 3, 3, 1, 0, 1, 0, 0, 0, 1, 1, 1, 2, 1, 2, 0, 2, 0, 1, 1, 0, 1, 2, 0, 0, 2, 2, 1, 1, 2, 0, 0, 2, 1, 2, 2, 2, 1, 0, 0, 0, 0, 0, 0, 1, 0

Offset: 1

T. D. Noe, May 02 2005

nonn

This polynomial is the characteristic polynomial of the Fibonacci and Lucas 5-step sequences, A001591 and A074048. Similar polynomials are treated in Serre's paper. The discriminant of the polynomial is 9584=16*599 and 599 is the only prime for which the polynomial has 4 distinct zeros. The primes p yielding 5 distinct zeros, A106281, correspond to the periods of the sequences A001591(k) mod p and A074048(k) mod p having length less than p. The Lucas 5-step sequence mod p has one additional prime p for which the period is less than p: the 599 factor of the discriminant. For this prime, the Fibonacci 5-step sequence mod p has a period of p(p-1).

J.-P. Serre, On a theorem of Jordan, Bull. Amer. Math. Soc., 40 (No. 4, 2003), 429-440, see p. 433.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number

Cf. A106298 (period of the Lucas 5-step sequences mod prime(n)), A106284 (prime moduli for which the polynomial is irreducible).

Cf. A001591, A074048, A106281.

Table[p=Prime[n]; cnt=0; Do[If[Mod[x^5-x^4-x^3-x^2-x-1, p]==0, cnt++ ], {x, 0, p-1}]; cnt, {n, 150}]

from sympy import Poly, prime
from sympy.abc import x
def A106278(n): return len(Poly(x*(x*(x*(x*(x-1)-1)-1)-1)-1, x, modulus=prime(n)).ground_roots()) # Chai Wah Wu, Mar 14 2024

A106304 Period of the Fibonacci 5-step sequence A001591 mod prime(n).

Original entry on oeis.org

6, 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, Nov 19 2006

nonn

This sequence is the same as the period of the Lucas 5-step sequence (A106298) 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 in A106281.

Chai Wah Wu, Table of n, a(n) for n = 1..86
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).

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 A106304(n):
    a = b = (0,)*4+(1 % (p:= prime(n)),)
    s = 1 % 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) = A106303(prime(n)).

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

A211671 Least prime p such that the polynomial x^n - x^(n-1) - ... - 1 (mod p) has n distinct zeros.

Original entry on oeis.org

2, 11, 47, 137, 691, 25621, 59233, 2424511, 2607383, 78043403, 1032758989, 80051779

Offset: 1

T. D. Noe, Apr 18 2012

This is the characteristic polynomial of the n-step Fibonacci and Lucas sequences. For composite p, the polynomial can have more than n zeros! See A211672.

			For p = 11, x^2-x-1 = (x+3)(x+7) (mod p).
For p = 47, x^3-x^2-x-1 = (x+21)(x+30)(x+42) (mod p).
For p = 137, x^4-x^3-x^2-x-1 = (x+12)(x+79)(x+85)(x+97) (mod p).

Cf. A045468 (n=2), A106279 (n=3), A106280 (n=4), A106281 (n=5).

Cf. A211672 (for composite p).

Table[poly = x^n - Sum[x^k, {k, 0, n - 1}]; k = 1; While[p = Prime[k]; cnt = 0; Do[If[Mod[poly, p] == 0, cnt++], {x, 0, p - 1}]; cnt < n, k++]; p, {n, 5}]

a(n)={my(P=x^n-sum(k=0, n-1, x^k) ); forprime(p=2, oo, if(#polrootsmod(P,p)==n, return(p) ) );} \\ Joerg Arndt, Apr 15 2013

a(8)-a(10) from Joerg Arndt, Apr 15 2013

a(11)-a(12) from Jinyuan Wang, Apr 25 2025

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

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A106278 Number of distinct zeros of x^5-x^4-x^3-x^2-x-1 mod prime(n).

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A106304 Period of the Fibonacci 5-step sequence A001591 mod prime(n).

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A211671 Least prime p such that the polynomial x^n - x^(n-1) - ... - 1 (mod p) has n distinct zeros.

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