A106303 -id:A106303 - cp's OEIS frontend

A001591 Pentanacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5), a(0)=a(1)=a(2)=a(3)=0, a(4)=1.

0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464, 912, 1793, 3525, 6930, 13624, 26784, 52656, 103519, 203513, 400096, 786568, 1546352, 3040048, 5976577, 11749641, 23099186, 45411804, 89277256, 175514464, 345052351, 678355061, 1333610936, 2621810068

Offset: 0

N. J. A. Sloane

Number of permutations satisfying -k <= p(i) - i <= r, i=1..n-4, with k=1, r=4. - Vladimir Baltic, Jan 17 2005
a(n) is the number of compositions of n-4 with no part greater than 5. - Vladimir Baltic, Jan 17 2005
The pentanomial (A035343(n)) transform of a(n) is a(5n+4), n >= 0. - Bob Selcoe, Jun 10 2014
a(n) is the number of ways to tile a strip of length n-4 with squares, dominoes, trominoes (of length 3), and rectangles with length 4 (tetraminoes) and length 5 (pentaminoes). - Wajdi Maaloul, Jun 21 2022

			n=2: a(14) = (1*1 + 2*1 + 3*2 + 4*4 + 5*8 + 4*16 + 3*31 + 2*61 + 1*120) = 464. - _Bob Selcoe_, Jun 10 2014
G.f. = x^4 + x^5 + 2*x^6 + 4*x^7 + 8*x^8 + 16*x^9 + 31*x^10 + 120*x^11 + ...

Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
Abdullah Açikel, Amrouche Said, Hacene Belbachir, and Nurettin Irmak, On k-generalized Lucas sequence with its triangle, Turkish J. Math. (2023) Vol. 47, No. 4, Art. 6, 1129-1143. See p. 1130.
Tomás Aguilar-Fraga, Jennifer Elder, Rebecca E. Garcia, Kimberly P. Hadaway, Pamela E. Harris, Kimberly J. Harry, Imhotep B. Hogan, Jakeyl Johnson, Jan Kretschmann, Kobe Lawson-Chavanu, J. Carlos Martínez Mori, Casandra D. Monroe, Daniel Quiñonez, Dirk Tolson III, and Dwight Anderson Williams II, Interval and L-interval Rational Parking Functions, arXiv:2311.14055 [math.CO], 2023. See p. 14.
Joerg Arndt, Matters Computational (The Fxtbook), pp. 307-309.
Vladimir Baltic, On the number of certain types of strongly restricted permutations, Applicable Analysis and Discrete Mathematics Vol. 4, No 1 (April, 2010), 119-135.
Paul Barry, The Gamma-Vectors of Pascal-like Triangles Defined by Riordan Arrays, arXiv:1804.05027 [math.CO], 2018.
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.
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Ömür Deveci, Yesim Akuzum, Erdal Karaduman, and Ozgur Erdag, The Cyclic Groups via Bezout Matrices, Journal of Mathematics Research, Vol. 7, No. 2, 2015, pp. 34-41.
Ömür Deveci, Zafer Adıgüzel, and Taha Doğan, On the Generalized Fibonacci-circulant-Hurwitz numbers, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 1, 179-190.
G. P. B. Dresden and Z. Du, A Simplified Binet Formula for k-Generalized Fibonacci Numbers, J. Int. Seq. 17 (2014) # 14.4.7.
I. Flores, k-Generalized Fibonacci numbers, Fib. Quart., 5 (1967), 258-266.
Taras Goy and Mark Shattuck, Some Toeplitz-Hessenberg Determinant Identities for the Tetranacci Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.6.8.
Tian-Xiao He, Impulse Response Sequences and Construction of Number Sequence Identities, J. Int. Seq. 16 (2013) #13.8.2.
V. E. Hoggatt, Jr. and M. Bicknell, Diagonal sums of generalized Pascal triangles, Fib. Quart., 7 (1969), 341-358, 393.
F. T. Howard and Curtis Cooper, Some identities for r-Fibonacci numbers, Fibonacci Quart. 49 (2011), no. 3, 231-243.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 12
Omar Khadir, László Németh, and László Szalay, Tiling of dominoes with ranked colors, Results in Math. (2024) Vol. 79, Art. No. 253. See p. 2.
Sergey Kirgizov, Q-bonacci words and numbers, arXiv:2201.00782 [math.CO], 2022.
Vladimir Victorovich Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.
Michel Mollard, p-th order generalized Fibonacci cubes and maximal cubes in Fibonacci p-cubes, arXiv:2507.16387 [math.CO], 2025. See p. 3.
Elisa Heinrich Mora and Noah A. Rosenberg, An nth-cousin mating model and the n-anacci numbers, arXiv:2506.16577 [q-bio.PE], 2025.
László Németh and László Szalay, Explicit solution of system of two higher-order recurrences, arXiv:2408.12196 [math.NT], 2024. See p. 10.
Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Helmut Prodinger, Counting Palindromes According to r-Runs of Ones Using Generating Functions, J. Int. Seq. 17 (2014) # 14.6.2, odd length middle 0, r=4.
B. Sivakumar and V. James, A Notes on Matrix Sequence of Pentanacci Numbers and Pentanacci Cubes, Communications in Mathematics and Applications (2022) Vol. 13, Iss. 2, 603-611.
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.
Kai Wang, Identities for generalized enneanacci numbers, Generalized Fibonacci Sequences (2020).
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number
Eric Weisstein's World of Mathematics, Pentanacci Number
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1).

Row 5 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
Cf. A106303 (Pisano period lengths).
Cf. A035343 (pentanomial coefficients).
Cf. A074048, A123127, A123126, A074062.

a:=[0,0,0,0,1]; [n le 5 select a[n] else Self(n-1) + Self(n-2) + Self(n-3) + Self(n-4) + Self(n-5): n in [1..40]]; // Marius A. Burtea, Oct 03 2019

g:=1/(1-z-z^2-z^3-z^4-z^5): gser:=series(g, z=0, 49): seq((coeff(gser, z, n)), n=-4..32); # Zerinvary Lajos, Apr 17 2009
# second Maple program:
a:= n-> (<<0|1|0|0|0>, <0|0|1|0|0>, <0|0|0|1|0>, <0|0|0|0|1>, <1|1|1|1|1>>^n)[1, 5]:
seq(a(n), n=0..44);  # Alois P. Heinz, Apr 09 2021

CoefficientList[Series[x^4/(1 - x - x^2 - x^3 - x^4 - x^5), {x, 0, 50}], x]
a[0] = a[1] = a[2] = a[3] = 0; a[4] = a[5] = 1; a[n_] := a[n] = 2 a[n - 1] - a[n - 6]; Array[a, 37, 0]
LinearRecurrence[{1, 1, 1, 1, 1}, {0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)

a(n):=mod(floor(10^((n-4)*(n+1))*10^(5*(n+1))*(10^(n+1)-1)/(10^(6*(n+1))-2*10^(5*(n+1))+1)),10^n); /* Tani Akinari, Apr 10 2014 */

a=vector(100);a[4]=a[5]=1;for(n=6,#a,a[n]=a[n-1]+a[n-2]+a[n-3]+a[n-4]+a[n-5]);concat(0, a) \\ Charles R Greathouse IV, Jul 15 2011

A001591(n,m=5)=(matrix(m,m,i,j,i==j-1||i==m)^n)[1,m] \\ M. F. Hasler, Apr 20 2018

a(n)= {my(x='x, p=polrecip(1 - x - x^2 - x^3 - x^4 - x^5)); polcoef(lift(Mod(x, p)^n), 4); }
vector(41, n, a(n-1)) \\ Joerg Arndt, May 16 2021

def pentanacci():
    a, b, c, d, e = 0, 0, 0, 0, 1
    while True:
        yield a
        a, b, c, d, e = b, c, d, e, a + b + c + d + e
f = pentanacci()
print([next(f) for  in range(100)]) # _Reza K Ghazi Apr 09 2021

G.f.: x^4/(1 - x - x^2 - x^3 - x^4 - x^5). - Simon Plouffe in his 1992 dissertation.
G.f.: Sum_{n >= 0} x^(n+4) * (Product_{k = 1..n} (k + k*x + k*x^2 + k*x^3 + x^4)/(1 + k*x + k*x^2 + k*x^3 + k*x^4)). - Peter Bala, Jan 04 2015
Another form of the g.f.: f(z) = (z^4-z^5)/(1-2*z+z^6); then a(n) = Sum_{i=0..floor((n-4)/6)} ((-1)^i*binomial(n-4-5*i,i)*2^(n-4-6*i)) - Sum_{i=0..floor((n-5)/6)} ((-1)^i*binomial(n-5-5*i,i)*2^(n-5-6*i)) with convention Sum_{i=m..n} alpha(i) = 0 for m > n. - Richard Choulet, Feb 22 2010
a(n) = Sum_{k=1..n} (Sum_{r=0..k} (binomial(k,r) * Sum_{m=0..r} (binomial(r,m) * Sum_{j=0..m} (binomial(m,j)*binomial(j,n-m-k-j-r))))), n > 0. - Vladimir Kruchinin, Aug 30 2010
Sum_{k=0..4*n} a(k+b)*A035343(n,k) = a(5*n+b), b >= 0.
a(n) = 2*a(n-1) - a(n-6). - Vincenzo Librandi, Dec 19 2010
a(n) = (Sum_{i=0..n-1} a(i)*A074048(n-i))/(n-4) for n > 4. - Greg Dresden and Advika Srivastava, Oct 01 2019
For k>0 and n>0, a(n+5*k) = A074048(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). - Kai Wang, Sep 06 2020
lim n->oo a(n)/a(n-1) = A103814. - R. J. Mathar, Mar 11 2024

A106297 Period of the Lucas 5-step sequence A074048 mod n.

Original entry on oeis.org

1, 1, 104, 6, 781, 104, 2801, 12, 312, 781, 16105, 312, 30941, 2801, 81224, 24, 88741, 312, 13032, 4686, 291304, 16105, 12166, 312, 3905, 30941, 936, 16806, 70728, 81224, 190861, 48, 1674920, 88741, 2187581, 312, 1926221, 13032, 3217864, 9372, 2896405

Offset: 1

T. D. Noe, May 02 2005, Nov 19 2006

This sequence can differ from the corresponding Fibonacci sequence (A106303) 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]

a(n) divides A106303(n). - Avery Diep, Aug 25 2025

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

Cf. A074048, A106303 (period of Fibonacci 5-step sequence mod n), A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).

n=5; Table[p=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, 50}]

from itertools import count
def A106297(n):
    a = b = (5%n,1%n,7%n,3%n,15%n)
    s = sum(b) % 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 22-27 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. a(2^k) = 3*2^(k-1) for k > 1. If a(p) != a(p^2) for prime p > 2, then a(p^k) = p^(k-1)*a(p) for k > 0. - Chai Wah Wu, Feb 25 2022

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

A351657 Period of the Fibonacci n-step sequence mod n.

Original entry on oeis.org

1, 3, 13, 10, 781, 728, 137257, 36, 273, 212784, 28531167061, 42640

Offset: 1

Ilya Gutkovskiy, Feb 16 2022

From Chai Wah Wu, Feb 23 2022: (Start)
a(14) = 92269645680
a(15) = 4976066589192413
a(16) = 136
a(18) = 306281976
(End)

			For n = 4, take the tetranacci sequence (A000078), 0, 0, 0, 1, 1, 2, 4, 8, 15, 29, 56, 108, 208, ... (mod 4), which gives 0, 0, 0, 1, 1, 2, 0, 0, 3, 1, 0, 0, 0, 1, 1, 2, ... This repeats a pattern of length 10, so a(4) = 10.

Eric Weisstein's World of Mathematics, Fibonacci n-Step Number
Eric Weisstein's World of Mathematics, Pisano Period

Cf. A000040, A001039, A001175, A007582, A046738, A106295, A106303.

from math import lcm
from itertools import count
from sympy import factorint
def f(n,pe): # period of the Fibonacci n-step sequence mod pe
    a = b = (0,)*(n-1)+(1%pe,)
    s = 1 % pe
    for m in count(1):
        b, s = b[1:] + (s,), (s+s-b[0]) % pe
        if a == b:
            return m
def A351657(n): return 1 if n == 1 else lcm(*(f(n,p**e) for p, e in factorint(n).items())) # Chai Wah Wu, Feb 23-27 2022

From Chai Wah Wu, Feb 22 2022: (Start)
Conjecture 1: a(p) = (p^p-1)/(p-1) for p prime, i.e., a(A000040(n)) = A001039(n).
Conjecture 2: a(2^k) = 2^(k-1)*(1+2^k) = A007582(k).
Conjecture 3 (which implies Conjectures 1 and 2): a(p^k) = (p^(p*k)-1)*p^(k-1)/(p^k-1) for k > 0 and prime p.
(End)

a(11)-a(12) from Chai Wah Wu, Feb 22 2022

A106290 Number of different orbit lengths of the 5-step recursion mod n.

Original entry on oeis.org

1, 3, 4, 4, 2, 9, 2, 6, 7, 6, 2, 11, 2, 6, 8, 8, 2, 9, 3, 8, 8, 6, 4, 12, 3, 6, 10, 8, 3, 18, 2, 10, 8, 6, 4, 11, 2, 6, 8, 12, 2, 18, 4, 8, 14, 9, 4, 16, 3, 9, 8, 8, 2, 12, 4, 12, 10, 6, 3, 22

Offset: 1

T. D. Noe, May 02 2005

Consider the 5-step recursion x(k) = (x(k-1)+x(k-2)+x(k-3)+x(k-4)+x(k-5)) mod n. For any of the n^5 initial conditions x(1), x(2), x(3), x(4) and x(5) in Zn, the recursion has a finite period. Each of these n^5 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths for each n. For n=8, there are 6 different lengths: 1, 2, 3, 6, 12 and 24. The maximum possible length of an orbit is A106303(n), the period of the Fibonacci 5-step sequence mod n.

Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Cf. A106287 (orbits of 5-step sequences), A106309 (primes that yield a simple orbit structure in 5-step recursions).

from itertools import count,product
def A106290(n):
    bset, tset = set(), set()
    for t in product(range(n),repeat=5):
        t2 = t
        for c in count(1):
            t2 = t2[1:] + (sum(t2)%n,)
            if t == t2:
                bset.add(c)
                tset.add(t)
                break
            if t2 in tset:
                tset.add(t)
                break
    return len(bset) # Chai Wah Wu, Feb 22 2022

A181190 Maximal length of chain-addition sequence mod 10 with window of size n.

Original entry on oeis.org

1, 60, 124, 1560, 4686, 1456, 18744, 585936, 4882810, 212784

Offset: 1

Alexander Dashevsky (atanvarnoalda(AT)gmail.com), Oct 10 2010

Chain addition mod 10 with window n: take an n-digit 'seed'. Take the sum of its digits mod 10 and append to the seed. Repeat with the last n digits of the string, until the seed appears again.
This sequence shows the lengths of the longest sequences for different window sizes.
a(1)-a(10) all occur for seed 1 (among others). If this is always true, the sequence continues: 406224, 12695306, 4272460934, 380859180, 122070312496, 518798826, 3433227539058. - Lars Blomberg, Feb 12 2013
Comment from Michel Lagneau, Jan 20 2017, edited by N. J. A. Sloane, Jan 24 2017: (Start)
If seed 1 is always as good as or better than any other, as seems likely, then this sequence has the following alternative description.
Consider the n initial terms of an infinite sequence S(k, n) of decimal digits given by 0, 0,..., 0, 1. The succeeding terms are given by the final digits in the sum of the n immediately preceding terms. The sequence lists the period of each sequence corresponding to n = 2, 3, ...
a(2) = period of A000045 mod 10 (Fibonacci numbers mod 10) = A001175(10).
a(3) = period of A000073 mod 10 (tribonacci numbers mod 10) = A046738(10).
a(4) = period of A000078 mod 10 (tetranacci numbers mod 10) = A106295(10).
a(5) = period of A001591 mod 10 (pentanacci numbers mod 10) = A106303(10).
a(6) = period of A001592 mod 10 (hexanacci numbers mod 10).
a(7) = period of A122189 mod 10 (heptanacci numbers mod 10).
a(8) = period of A079262 mod 10 (octanacci numbers mod 10).
a(4) = 1560 because the four initial terms 0, 0, 0, 1 => S(k, 4) = 0, 0, 0, 1, 1, 2, 4, 8, 5, 9, 6, 8, 8, 1, 3, 0, 2, 6, 1, 9, 8, ... (tetranacci numbers mod 10). This sequence is periodic with period 1560:
S(1560 + 1, 4) = S(1, 4) = 0,
S(1560 + 2, 4) = S(2, 4) = 0,
S(1560 + 3, 4) = S(3, 4) = 0,
S(1560 + 4, 4) = S(4, 4) = 1.
(End)

			For n=2, the longest sequence begins with '01' (among others):
01123583145943707741561785381909987527965167303369549325729101.
It is 60 digits long (not counting the second '01' at the end).
For n=3, one of the longest sequences begins again with '001':
00112473441944756893025746770415061742394699425184352079627546556679289964992013
48570291225960516297849144970639807524172091001 (124 digits long without the second '001').

Cf. A000045, A000073, A000078, A001591, A001592, A003893, A079262, A122189

a(8)-a(10) from Lars Blomberg, Feb 12 2013

A193994 Number of zeros in the period of Fibonacci 5-step sequence A001591 mod n.

Original entry on oeis.org

1, 4, 35, 7, 156, 70, 400, 12, 45, 624, 1580, 61, 2380, 1600, 5460, 18, 5220, 33, 684, 1092, 14000, 6320, 523, 52, 185, 9520, 48, 2800, 2465, 10920, 6075, 22, 55300, 20880, 62400, 28, 52060, 464, 83300, 1872, 70180, 28000, 1903, 11060, 7020, 1046, 22, 79

Offset: 1

T. D. Noe, Aug 18 2011

nonn

Cf. A106303 (period of Fibonacci 5-step sequence).

n = 5; Table[a = Join[{1}, Table[0, {n - 1}]]; a = Mod[a, i]; a0 = a; k = 0; zeros = 0; While[k++; s = Mod[Plus @@ a, i]; a = RotateLeft[a]; If[s == 0, zeros++]; a[[n]] = s; a != a0]; zeros, {i, 100}]

from itertools import count
def A193994(n):
    a = b = (0,)*4+(1 % n,)
    c, s = 0, 1 % n
    for m in count(1):
        b, s = b[1:] + (s,), (s+s-b[0])% n
        c += int(s==0)
        if a == b:
            return c # Chai Wah Wu, Feb 22-27 2022

A351889 Table T(n,k) read by downward antidiagonals: period of n-step Fibonacci numbers mod k, n >= 1, k >= 1.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 8, 4, 1, 1, 6, 13, 5, 1, 1, 20, 8, 26, 6, 1, 1, 24, 31, 10, 104, 7, 1, 1, 16, 52, 312, 12, 728, 8, 1, 1, 12, 48, 130, 781, 14, 364, 9, 1, 1, 24, 16, 342, 312, 208, 16, 80, 10, 1, 1, 60, 39, 20, 2801, 728, 9372, 18, 91, 11, 1, 1, 10, 124, 78, 24, 342, 728, 195312

Offset: 1

Chai Wah Wu, Feb 24 2022

			Table T(n,k) starts:
  1   1     1   1       1      1          1   1      1        1
  1   3     8   6      20     24         16  12     24       60
  1   4    13   8      31     52         48  16     39      124
  1   5    26  10     312    130        342  20     78     1560
  1   6   104  12     781    312       2801  24    312     4686
  1   7   728  14     208    728        342  28   2184     1456
  1   8   364  16    9372    728     137257  32   1092    18744
  1   9    80  18  195312    720      13680  36    240   585936
  1  10    91  20  488281    910    5764800  40    273  4882810
  1  11  8744  22   19344  96184      19152  44  26232   212784
  1  12  3851  24  406224  46212  109531200  48  11553   406224

Chai Wah Wu, Table of n, a(n) for n = 1..226
M. E. Waddill, Some properties of a generalized Fibonacci sequence modulo m, The Fibonacci Quarterly, vol. 16, no. 4, pp. 344-353 (1978).
Marcellus E. Waddill, Some Properties of the Tetranacci Sequence Modulo m, The Fibonacci Quarterly, vol. 30, no. 3, 232-238 (1992).
D. D. Wall, Fibonacci Series Modulo m, The American Mathematical Monthly, vol. 67, no. 6, pp. 525-532, (1960).
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number
Eric Weisstein's World of Mathematics, Pisano Period

Cf. A000045, A001175, A046738, A106295, A106303, A337212, A351657 (diagonal).

from functools import lru_cache
from math import lcm
from itertools import count
from sympy import factorint
@lru_cache(maxsize=None)
def A351889_T(n,k): # computes the period of the n-step Fibonacci sequence mod k
    if len(fs := factorint(k)) <= 1:
        a = b = (0,)*(n-1)+(1 % k,)
        s = 1 % k
        for m in count(1):
            b, s = b[1:] + (s,), (s + s - b[0]) % k
            if a == b:
                return m
    else:
        return lcm(*(A351889_T(n,p**e) for p, e in fs.items()))

T(1,k) = T(n,1) = 1.
T(2,k) = A001175(k).
T(3,k) = A046738(k).
T(4,k) = A106295(k) for k not a multiple of 563.
T(5,k) = A106303(k).
T(n,2) = n + 1 for n > 1.
T(n,3) = A337212(n).
T(n,n) = A351657(n).
T(n,p_1^e_1*...*p_m^e_m) = lcm(T(n,p_1^e_1),...,T(n,p_m^e_m)) for p_i distinct primes.
Conjecture 1: T(n,2^m) = (n+1)*2^(m-1) for n > 1.
Conjecture 2: For p prime, if T(n,p) != T(n,p^2) then T(n,p^k) = p^(k-1)T(n,p).
Conjecture 2 is true for n = 2, n = 3 and n = 4 (see [Wall, 1960], [Waddill, 1978] and [Waddill, 1992] resp.). It is easy to show that T(n,4) != n+1 for all n, and thus Conjecture 2 implies Conjecture 1.
Conjecture 3: T(p^m,p^k) = (p^(pm)-1)*p^(k-1)/(p^m-1) for p prime and k, m > 0.

cp's OEIS Frontend

A001591 Pentanacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5), a(0)=a(1)=a(2)=a(3)=0, a(4)=1.

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A106297 Period of the Lucas 5-step sequence A074048 mod n.

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

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A351657 Period of the Fibonacci n-step sequence mod n.

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A106290 Number of different orbit lengths of the 5-step recursion mod n.

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A181190 Maximal length of chain-addition sequence mod 10 with window of size n.

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A193994 Number of zeros in the period of Fibonacci 5-step sequence A001591 mod n.

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