A175184 -id:A175184 - cp's OEIS frontend

A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 5, 3, 0, 1, 4, 10, 12, 5, 0, 1, 5, 17, 33, 29, 8, 0, 1, 6, 26, 72, 109, 70, 13, 0, 1, 7, 37, 135, 305, 360, 169, 21, 0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 0, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 0, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89, 0, 1, 11, 101, 747, 4289, 18200, 53353, 98145, 98209, 42837, 5741, 144

Offset: 1

Roger L. Bagula, Jan 29 2010

Equals A073133 with an additional column A(.,0).
If the first column and top row are deleted, antidiagonal reading yields A118243.
Adding a top row of 1's and antidiagonal reading downwards yields A157103.
Antidiagonal sums are 0, 1, 2, 5, 12, 32, 93, 297, 1035, 3911, 15917, 69350, ....
From Jianing Song, Jul 14 2018: (Start)
All rows have strong divisibility, that is, gcd(A(n,k_1), A(n,k_2)) = A(n,gcd(k_1,k_2)) holds for all k_1, k_2 >= 0.
Let E(n,m) be the smallest number l such that m divides A(n,l), we have: for odd primes p that are not divisible by n^2 + 4, E(n,p) divides p - ((n^2+4)/p) if p == 3 (mod 4) and (p - ((n^2+4)/p))/2 if p == 1 (mod 4). E(n,p) = p for odd primes p that are divisible by n^2 + 4. E(n,2) = 2 for even n and 3 for odd n. Here ((n^2+4)/p) is the Legendre symbol. A prime p such that p^2 divides T(n,E(n,p)) is called an n-Wall-Sun-Sun prime.
E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.
Let pi(n,m) be the Pisano period of A(n, k) modulo m, i.e, the smallest number l such that A(n, k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p - 1 if ((n^2+4)/p) = 1 and 2(p+1) if ((n^2+4)/p) = -1. pi(n,p) = 4p for odd primes p that are divisible by n^2 + 4. pi(n,2) = 2 even n and 3 for odd n.
pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n, p), so pi(n,p^e) = 4p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1), pi(n,m_2)) if gcd(m_1,m_2) = 1.
If n != 2, the largest possible value of pi(n,m)/m is 4 for even n and 6 for odd n. For even n, pi(n,p^e) = 4p^e; for odd n, pi(n,2p^e) = 12p^e, where p is any odd prime factor of n^2 + 4. For n = 2 it is 8/3, obtained by m = 3^e.
Let z(n,m) be the number of zeros in a period of A(n, k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: z(n,p) = 4 for odd primes p that are divisible by n^2 + 4. For other odd primes p, z(n,p) = 4 if E(n,p) is odd; 1 if E(n,p) is even but not divisible by 4; 2 if E(n,p) is divisible by 4; see the table below. z(n,2) = z(n,4) = 1.
Among all values of z(n,p) when p runs through all odd primes that are not divisible by n^2 + 4, we have:
((n^2+4)/p)...p mod 8....proportion of 1.....proportion of 2.....proportion of 4
......1..........1......1/6 (conjectured)...2/3 (conjectured)...1/6 (conjectured)*
......1..........5......1/2 (conjectured)...........0...........1/2 (conjectured)*
......1.........3,7.............1...................0...................0
.....-1.........1,5.............0...................0...................1
.....-1.........3,7.............0...................1...................0
* The result is that among all odd primes that are not divisible by n^2 + 4, 7/24 of them are with z(n,p) = 1, 5/12 are with z(n,p) = 2 and 7/24 are with z(n,p) = 4 if n^2 + 4 is a twice a square; 1/3 of them are with z(n,p) = 1, 1/3 are with z(n,p) = 2 and 1/3 are with z(n,p) = 4 otherwise. [Corrected by Jianing Song, Jul 06 2019]
z(n,p^e) = z(n,p) for all odd primes p; z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.
(End)
From Michael A. Allen, Mar 06 2023: (Start)
Removing the first (n=0) row of A352361 gives this sequence.
Row n is the n-metallonacci sequence.
A(n,k) is (for k>0) the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

			The array, A(n, k), starts in row n = 1 with columns k >= 0 as
  0      1      1      2      3      5      8
  0      1      2      5     12     29     70
  0      1      3     10     33    109    360
  0      1      4     17     72    305   1292
  0      1      5     26    135    701   3640
  0      1      6     37    228   1405   8658
  0      1      7     50    357   2549  18200
  0      1      8     65    528   4289  34840
  0      1      9     82    747   6805  61992
  0      1     10    101   1020  10301 104030
  0      1     11    122   1353  15005 166408
Antidiagonal triangle, T(n, k), begins as:
  0;
  0, 1;
  0, 1, 1;
  0, 1, 2,  2;
  0, 1, 3,  5,   3;
  0, 1, 4, 10,  12,   5;
  0, 1, 5, 17,  33,  29,    8;
  0, 1, 6, 26,  72, 109,   70,   13;
  0, 1, 7, 37, 135, 305,  360,  169,  21;
  0, 1, 8, 50, 228, 701, 1292, 1189, 408, 34;

Jianing Song, Antidiagonals n = 1..100, flattened (the first 30 antidiagonals from Vincenzo Librandi)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.

Rows n include: A000045 (n=1), A000129 (n=2), A006190 (n=3), A001076 (n=4), A052918 (n=5), A005668 (n=6), A054413 (n=7), A041025 (n=8), A099371 (n=9), A041041 (n=10), A049666 (n=11), A041061 (n=12), A140455 (n=13), A041085 (n=14), A154597 (n=15), A041113 (n=16), A178765 (n=17), A041145 (n=18), A243399 (n=19), A041181 (n=20). (Note that there are offset shifts for rows n = 5, 7, 8, 10, 12, 14, 16..20.)
Columns k include: A000004 (k=0), A000012 (k=1), A000027 (k=2), A002522 (k=3), A054602 (k=4), A057721 (k=5), A124152 (k=6).
Entry points for A(n,k) modulo m: A001177 (n=1), A214028 (n=2), A322907 (n=3).
Pisano period for A(n,k) modulo m: A001175 (n=1), A175181 (n=2), A175182 (n=3), A175183 (n=4), A175184 (n=5), A175185 (n=6).
Number of zeros in a period for A(n,k) modulo m: A001176 (n=1), A214027 (n=2), A322906 (n=3).
Cf. A084844, A084845, A118243, A157103, A271782, A316269.
Sums include: A304357, A304359.
Similar to: A073133.

A172236:= func< n,k | k le 1 select k else Evaluate(DicksonSecond(k-1,-1), n-k) >;
[A172236(n,k): k in [0..n-1], n in [1..13]]; // G. C. Greubel, Sep 29 2024

A172236[n_,k_]:=Fibonacci[k, n-k];
Table[A172236[n, k], {n,15}, {k,0,n-1}]//Flatten

A(n, k) = if (k==0, 0, if (k==1, 1, n*A(n, k-1) + A(n, k-2)));
tabl(nn) = for(n=1, nn, for (k=0, nn, print1(A(n, k), ", ")); print); \\ Jianing Song, Jul 14 2018 (program from Michel Marcus; see also A316269)

A(n, k) = ([n, 1; 1, 0]^k)[2, 1] \\ Jianing Song, Nov 10 2018

def A172236(n,k): return sum(binomial(k-j-1,j)*(n-k)^(k-2*j-1) for j in range(1+(k-1)//2))
flatten([[A172236(n,k) for k in range(n)] for n in range(1,14)]) # G. C. Greubel, Sep 29 2024

A(n,k) = (((n + sqrt(n^2 + 4))/2)^k - ((n-sqrt(n^2 + 4))/2)^k)/sqrt(n^2 + 4), n >= 1, k >= 0. - Jianing Song, Jun 27 2018
For n >= 1, Sum_{i=1..k} 1/A(n,2^i) = ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/A(n,2^k)), where u = (n + sqrt(n^2 + 4))/2, v = (n - sqrt(n^2 + 4))/2 are the two roots of the polynomial x^2 - n*x - 1. As a result, Sum_{i>=1} 1/A(n,2^i) = (n^2 + 4 - n*sqrt(n^2 + 4))/(2*n). - Jianing Song, Apr 21 2019
From G. C. Greubel, Sep 29 2024: (Start)
A(n, k) = F_{k}(n) (Fibonacci polynomials F_{n}(x)) (array).
T(n, k) = F_{k}(n-k) (antidiagonal triangle).
Sum_{k=0..n-1} T(n, k) = A304357(n) - (1-(-1)^n)/2.
Sum_{k=0..n-1} (-1)^k*T(n, k) = (-1)*A304359(n) + (1-(-1)^n)/2.
T(2*n, n) = A084844(n).
T(2*n+1, n+1) = A084845(n). (End)

More terms from Jianing Song, Jul 14 2018

A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).

Original entry on oeis.org

1, 5, 26, 135, 701, 3640, 18901, 98145, 509626, 2646275, 13741001, 71351280, 370497401, 1923838285, 9989688826, 51872282415, 269351100901, 1398627786920, 7262490035501, 37711077964425, 195817879857626

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

A087130(n)^2 - 29*a(n-1)^2 = 4*(-1)^n, n >= 1. - Gary W. Adamson, Jul 01 2003, corrected Oct 07 2008, corrected by Jianing Song, Feb 01 2019
a(p-1) == 29^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
For more information about this type of recurrence, follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A015523. - Johannes W. Meijer, Aug 01 2010
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 5's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,5} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 5-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 5 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 901
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
Milan Janjić, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,1).

Row 5 of A073133, A172236, and A352361.

Cf. A087130, A099365 (squares), A100237, A175184 (Pisano periods), A201005 (prime subsequence).

a:=[1,5];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Oct 16 2019

I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 23 2013

R:=PowerSeriesRing(Integers(), 22); Coefficients(R!( 1/(1 - 5*x - x^2) )); // Marius A. Burtea, Oct 16 2019

spec := [S,{S=Sequence(Union(Z,Z,Z,Z,Z,Prod(Z,Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006
with(combinat):a:=n->fibonacci(n,5):seq(a(n),n=1..30); # Zerinvary Lajos, Dec 07 2008

LinearRecurrence[{5, 1}, {1, 5}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
Table[Fibonacci[n+1, 5], {n,0,30}] (* Vladimir Reshetnikov, May 08 2016 *)

Vec(1/(1-5*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011

[lucas_number1(n,5,-1) for n in range(1, 22)] # Zerinvary Lajos, Apr 24 2009

G.f.: 1/(1 - 5*x - x^2).
a(3n) = A041047(5n), a(3n+1) = A041047(5n+3), a(3n+2) = 2*A041047(5n+4). - Henry Bottomley, May 10 2000
a(n) = Sum_{alpha=RootOf(-1+5*z+z^2)} (1/29)*(5+2*alpha)*alpha^(-1-n).
a(n-1) = (((5 + sqrt(29))/2)^n - ((5 - sqrt(29))/2)^n)/sqrt(29). - Gary W. Adamson, Jul 01 2003
a(n) = U(n, 5*i/2)*(-i)^n with i^2 = -1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See triangle A049310.
Let M = {{0, 1}, {1, 5}}, then a(n) is the lower-right term of M^n. - Roger L. Bagula, May 29 2005
a(n) = F(n, 5), the n-th Fibonacci polynomial evaluated at x = 5. - T. D. Noe, Jan 19 2006
a(n) = denominator of n-th convergent to [1, 4, 5, 5, 5, ...], for n > 0. Continued fraction [1, 4, 5, 5, 5, ...] = 0.807417596..., the inradius of a right triangle with legs 2 and 5. n-th convergent = A100237(n)/A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135, 566/701, ... - Gary W. Adamson, Dec 21 2007
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 5*A097781(n), a(2n) = A097835(n).
Limit_{k->oo} a(n+k)/a(k) = (A087130(n) + a(n-1)*sqrt(29))/2.
Limit_{n->oo} A087130(n)/a(n-1) = sqrt(29). (End)
From L. Edson Jeffery, Jan 07 2012: (Start)
Define the 2 X 2 matrix A = {{1, 1}, {5, 4}}. Then:
a(n) is the upper-left term of (1/5)*(A^(n+2) - A^(n+1));
a(n) is the upper-right term of A^(n+1);
a(n) is the lower-left term of (1/5)*A^(n+1);
a(n) is the lower-right term of (Sum_{k=0..n} A^k). (End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(29) - 5)/2. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 5*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 5 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

A175181 Pisano period of the 2-Fibonacci numbers A000129.

Original entry on oeis.org

1, 2, 8, 4, 12, 8, 6, 8, 24, 12, 24, 8, 28, 6, 24, 16, 16, 24, 40, 12, 24, 24, 22, 8, 60, 28, 72, 12, 20, 24, 30, 32, 24, 16, 12, 24, 76, 40, 56, 24, 10, 24, 88, 24, 24, 22, 46, 16, 42, 60, 16, 28, 108, 72, 24, 24, 40, 20, 40, 24, 124, 30, 24, 64, 84, 24, 136

Offset: 1

R. J. Mathar, Mar 01 2010

nonn

Period of the sequence defined by reading A000129 modulo n.

T. D. Noe, Table of n, a(n) for n = 1..1000
Sergio Falcon and Ángel Plaza, k-Fibonacci sequences modulo m, Chaos, Solit. Fractals 41 (2009), 497-504.
Eric Weisstein's World of Mathematics, Pisano period.
Wikipedia, Pisano period.

Cf. A000129, A001175, A175182, A175183, A175184, A175185.

F := proc(k,n) option remember; if n <= 1 then n; else k*procname(k,n-1)+procname(k,n-2) ; end if; end proc:
Pper := proc(k,m) local cha, zer,n,fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k,n) mod m ; cha := [op(cha),fmodm] ; if fmodm = 0 then zer := [op(zer),n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2],cha) ] = [ op(zer[2]+1..zer[3],cha) ] and [op(1..zer[2],cha)] = [ op(zer[3]+1..zer[4],cha) ] and [op(1..zer[2],cha)] = [ op(zer[4]+1..zer[5],cha) ] then return zer[2] ; elif [op(1..zer[3],cha) ] = [ op(zer[3]+1..zer[5],cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
k := 2 ; seq( Pper(k,m),m=1..80) ;

Table[s = t = Mod[{0, 1}, n]; cnt = 1; While[tmp = Mod[2*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s != t, cnt++]; cnt, {n, 100}] (* T. D. Noe, Jul 09 2012 *)

A175182 Pisano period of the 3-Fibonacci numbers A006190.

Original entry on oeis.org

1, 3, 2, 6, 12, 6, 16, 12, 6, 12, 8, 6, 52, 48, 12, 24, 16, 6, 40, 12, 16, 24, 22, 12, 60, 156, 18, 48, 28, 12, 64, 48, 8, 48, 48, 6, 76, 120, 52, 12, 28, 48, 42, 24, 12, 66, 96, 24, 112, 60, 16, 156, 26, 18, 24, 48, 40, 84, 24, 12, 30, 192, 48, 96, 156, 24, 136, 48, 22, 48, 144

Offset: 1

R. J. Mathar, Mar 01 2010

Period of the sequence defined by reading A006190 modulo n.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Sergio Falcon and Ángel Plaza, k-Fibonacci sequences modulo m, Chaos, Solit. Fractals 41 (2009), 497-504.
Eric Weisstein's World of Mathematics, Pisano period.
Wikipedia, Pisano period.

Cf. A001175, A006190, A175181, A175183, A175184, A175185.

F := proc(k,n) option remember; if n <= 1 then n; else k*procname(k,n-1)+procname(k,n-2) ; end if; end proc:
Pper := proc(k,m) local cha, zer,n,fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k,n) mod m ; cha := [op(cha),fmodm] ; if fmodm = 0 then zer := [op(zer),n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2],cha) ] = [ op(zer[2]+1..zer[3],cha) ] and [op(1..zer[2],cha)] = [ op(zer[3]+1..zer[4],cha) ] and [op(1..zer[2],cha)] = [ op(zer[4]+1..zer[5],cha) ] then return zer[2] ; elif [op(1..zer[3],cha) ] = [ op(zer[3]+1..zer[5],cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
k := 3 ; seq( Pper(k,m),m=1..80) ;

Table[s = t = Mod[{0, 1}, n]; cnt = 1; While[tmp = Mod[3*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s!= t, cnt++]; cnt, {n, 100}] (* Vincenzo Librandi, Dec 20 2012, T. D. Noe *)

A175185 Pisano period of the 6-Fibonacci numbers A005668.

Original entry on oeis.org

1, 2, 2, 4, 20, 2, 16, 8, 6, 20, 24, 4, 6, 16, 20, 16, 36, 6, 8, 20, 16, 24, 48, 8, 100, 6, 18, 16, 60, 20, 30, 32, 24, 36, 80, 12, 12, 8, 6, 40, 40, 16, 42, 24, 60, 48, 96, 16, 112, 100, 36, 12, 26, 18, 120, 16, 8, 60, 40, 20, 124, 30, 48, 64, 60, 24, 22, 36, 48, 80, 70, 24, 148

Offset: 1

R. J. Mathar, Mar 01 2010

Period of the sequence defined by reading A005668 modulo n.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Sergio Falcon and Ángel Plaza, k-Fibonacci sequences modulo m, Chaos, Solit. Fractals 41 (2009), 497-504.
Eric Weisstein's World of Mathematics, Pisano period.
Wikipedia, Pisano period.

Cf. A001175, A005668, A175181, A175182, A175183, A175184.

F := proc(k,n) option remember; if n <= 1 then n; else k*procname(k,n-1)+procname(k,n-2) ; end if; end proc:
Pper := proc(k,m) local cha, zer,n,fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k,n) mod m ; cha := [op(cha),fmodm] ; if fmodm = 0 then zer := [op(zer),n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2],cha) ] = [ op(zer[2]+1..zer[3],cha) ] and [op(1..zer[2],cha)] = [ op(zer[3]+1..zer[4],cha) ] and [op(1..zer[2],cha)] = [ op(zer[4]+1..zer[5],cha) ] then return zer[2] ; elif [op(1..zer[3],cha) ] = [ op(zer[3]+1..zer[5],cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
k := 6 ; seq( Pper(k,m),m=1..80) ;

Table[s = t = Mod[{0, 1}, n]; cnt = 1; While[tmp = Mod[6*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s!= t, cnt++]; cnt, {n, 100}] (* Vincenzo Librandi, Dec 20 2012, after T. D. Noe *)

A175183 Pisano period of the 4-Fibonacci numbers A001076.

Original entry on oeis.org

1, 2, 8, 2, 20, 8, 16, 4, 8, 20, 10, 8, 28, 16, 40, 8, 12, 8, 6, 20, 16, 10, 16, 8, 100, 28, 24, 16, 14, 40, 10, 16, 40, 12, 80, 8, 76, 6, 56, 20, 40, 16, 88, 10, 40, 16, 32, 8, 112, 100, 24, 28, 36, 24, 20, 16, 24, 14, 58, 40, 20, 10, 16, 32, 140, 40, 136, 12, 16, 80, 70, 8, 148, 76

Offset: 1

R. J. Mathar, Mar 01 2010

Period of the sequence defined by reading A001076 modulo n.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Sergio Falcon and Ángel Plaza, k-Fibonacci sequences modulo m, Chaos, Solit. Fractals 41 (2009), 497-504.
Eric Weisstein's World of Mathematics, Pisano period.
Wikipedia, Pisano period.

Cf. A001076, A001175, A175181, A175182, A175184, A175185.

F := proc(k,n) option remember; if n <= 1 then n; else k*procname(k,n-1)+procname(k,n-2) ; end if; end proc:
Pper := proc(k,m) local cha, zer,n,fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k,n) mod m ; cha := [op(cha),fmodm] ; if fmodm = 0 then zer := [op(zer),n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2],cha) ] = [ op(zer[2]+1..zer[3],cha) ] and [op(1..zer[2],cha)] = [ op(zer[3]+1..zer[4],cha) ] and [op(1..zer[2],cha)] = [ op(zer[4]+1..zer[5],cha) ] then return zer[2] ; elif [op(1..zer[3],cha) ] = [ op(zer[3]+1..zer[5],cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
k := 4 ; seq( Pper(k,m),m=1..80) ;

Table[s = t = Mod[{0, 1}, n]; cnt=1; While[tmp = Mod[4*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s!= t, cnt++]; cnt, {n, 100}] (* Vincenzo Librandi, Dec 20 2012, after T. D. Noe *)

cp's OEIS Frontend

A172236 Array A(n,k) = n*A(n,k-1) + A(n,k-2) read by upward antidiagonals, starting A(n,0) = 0, A(n,1) = 1.

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A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).

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A175181 Pisano period of the 2-Fibonacci numbers A000129.

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A175182 Pisano period of the 3-Fibonacci numbers A006190.

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A175185 Pisano period of the 6-Fibonacci numbers A005668.

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A175183 Pisano period of the 4-Fibonacci numbers A001076.

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A175289 Pisano period of A002605 modulo n.

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