A001175 -id:A001175 - cp's OEIS frontend

A030132 Digital root of Fibonacci(n).

0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8

Offset: 0

youngelder(AT)webtv.net (Ana)

Any initial pair (a(0), a(1)) of nonzero single-digit numbers enters a cycle of length 24, except for the 8 cases where 3 divides both a(0), a(1) and (a(0), a(1)) != (9, 9), which enter a cycle of length 8 and (9, 9), which is immediately periodic of period length 1. - Jonathan Vos Post, Dec 29 2005 [Corrected by Jianing Song, Apr 17 2021]
First term that differs from A004090 is a(10). In general, all terms of A004090 having one digit are the same in this sequence. - Alonso del Arte, Sep 16 2012
Decimal expansion of 12484270798876404618091 / 1111111111111111111111110 = 0.0[112358437189887641562819] (periodic). - Daniel Forgues, Feb 27 2017

			a(10) = 1 because F(10) = 55, and since 5 + 5 = 10 and 1 + 0 = 1 is the digital root of 55.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Robert Bruce Gray, Another context for the first 48 terms of this sequence, May 08 2025
S. Marivani and others, Digital Roots of Fibonacci Numbers: Problem 10974, Amer. Math. Monthly, 111 (No. 7, 2004), 628.
Colm Mulcahy, Gibonacci Bracelets.
Marc Renault, The Fibonacci sequence modulo m
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

Cf. A000045 (Fibonacci numbers), A010888 (digital roots), A004090, A007953, A030133.

Cf. A001175, A010077, A017641, A065076.

a030132 n = a030132_list !! n
a030132_list =
   0 : 1 : map a007953 (zipWith (+) a030132_list (tail a030132_list))
-- Reinhard Zumkeller, Aug 20 2011

digitalRoot[n_Integer?Positive] := FixedPoint[Plus@@IntegerDigits[#]&, n]; Table[If[n == 0, 0, digitalRoot[Fibonacci[n]]], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, May 02 2011 *)
Table[NestWhile[Total[IntegerDigits[#]]&, Fibonacci[n], # > 9 &], {n, 0, 90}] (* Harvey P. Dale, May 07 2012 *)
PadRight[{0},120,{9,1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1}] (* Harvey P. Dale, Jul 20 2024 *)

a(n)=if(n,(fibonacci(n)-1)%9+1,0) \\ Charles R Greathouse IV, Jan 23 2013

a(n + 1) = sum of digits of (a(n) + a(n - 1)).
Periodic with period 24 = A001175(9) given by {1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9}.
a(n) + a(n + 1) = A010077(n + 4); a(A017641(n)) = 9. - Reinhard Zumkeller, Jul 04 2007
G.f.: x*( -1 -x -2*x^2 -3*x^3 -5*x^4 -8*x^5 -4*x^6 -3*x^7 -7*x^8 -x^9 -8*x^10 -9*x^11 -8*x^12 -8*x^13 -7*x^14 -6*x^15 -4*x^16 -x^17 -5*x^18 -6*x^19 -2*x^20 -8*x^21 -x^22 -9*x^23 ) / ( (x-1) *(1+x+x^2) *(1+x) *(1-x+x^2) *(1+x^2) *(x^4-x^2+1) *(1+x^4) *(x^8-x^4+1) ). - R. J. Mathar, Feb 08 2013

Entry revised by N. J. A. Sloane, Aug 29 2004

A046738 Period of Fibonacci 3-step sequence A000073 mod n.

Original entry on oeis.org

1, 4, 13, 8, 31, 52, 48, 16, 39, 124, 110, 104, 168, 48, 403, 32, 96, 156, 360, 248, 624, 220, 553, 208, 155, 168, 117, 48, 140, 1612, 331, 64, 1430, 96, 1488, 312, 469, 360, 2184, 496, 560, 624, 308, 440, 1209, 2212, 46, 416, 336, 620, 1248, 168

Offset: 1

David W. Wilson

Could also be called the tribonacci Pisano periods. [Carl R. White, Oct 05 2009]
Klaska notes that n=208919=59*3541 satisfies a(n) = a(n^2). - Michel Marcus, Mar 03 2016
39, 78, 273, 546 also satisfy a(n) = a(n^2). - Michel Marcus, Mar 07 2016

T. D. Noe [1..1000] + Jean-François Alcover [1001..2000] + Zhong Ziqian [2001..20000], Table of n, a(n) for n = 1..20000
Jirí Klaška, A search for Tribonacci-Wieferich primes, Acta Mathematica Universitatis Ostraviensis, vol. 16 (2008), issue 1, pp. 15-20.
Jirí Klaška, On Tribonacci-Wieferich primes, Fibonacci Quart. 46/47 (2008/2009), no. 4, 290-297.
Jirí Klaška, Tribonacci partition formulas modulo m, Acta Mathematica Sinica, English Series, March 2010, Volume 26, Issue 3, pp 465-476.
M. E. Waddill, Some properties of a generalized Fibonacci sequence modulo m, The Fibonacci Quarterly, vol. 16, no. 4, pp. 344-353 (1978).

Cf. A106302.

Cf. A001175.

a:= proc(n) local f, k, l; l:= ifactors(n)[2];
      if nops(l)<>1 then ilcm(seq(a(i[1]^i[2]), i=l))
    else f:= [0, 0, 1];
         for k do f:=[f[2], f[3], f[1]+f[2]+f[3] mod n];
                  if f=[0, 0, 1] then break fi
         od; k
      fi
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 27 2023

Table[a = {0, 1, 1}; a = a0 = Mod[a, n]; k = 0; While[k++; s = a[[3]] + a[[2]] + a[[1]]; a = RotateLeft[a]; a[[-1]] = Mod[s, n]; a != a0]; k, {n, 100}] (* T. D. Noe, Aug 28 2012 *)

from itertools import count
def A046738(n):
    a = b = (0,0,1%n)
    for m in count(1):
        b = b[1:] + (sum(b) % n,)
        if a == b:
            return m # Chai Wah Wu, Feb 27 2022

a(3^k) = 13*3^(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 [Waddill, 1978]. - Chai Wah Wu, Feb 25 2022

Let the prime factorization of n be p1^e1*...*pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)) [Waddill, 1978]. - Avery Diep, Aug 26 2025

A089911 a(n) = Fibonacci(n) mod 12.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1

Offset: 0

Casey Mongoven, Nov 14 2003

From Reinhard Zumkeller, Jul 05 2013: (Start)
Sequence has been applied by several composers to 12-tone equal temperament pitch structure. The complete Fibonacci mod 12 system (a set of 10 periodic sequences) exhausts all possible ordered dyads; that is, every possible combination of two pitches is found in these sets.
a(A008594(n)) = 0;
a(A227144(n)) = 1;
a(3*A047522(n)) = 2;
a(A017569(n)) = a(2*A016933(n)) = a(4*A016777(n)) = 3;
a(2*A017629(n)) = a(3*A017137(n)) = a(6*A004767(n)) = 4;
a(A227146(n)) = 5;
a(nonexistent) = 6;
a(2*A017581(n)) = 7;
a(2*A017557(n)) = a(4*A016813(n)) = 8;
a(A017617(n)) = a(2*A016957(n)) = a(4*A016789(n)) = 9;
a(3*A047621(n)) = 10;
a(2*A017653(n)) = 11. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1199
Ron Knott, Fibonacci Numbers and the Golden Section
M. Renault, The Fibonacci Sequence Modulo M
A. P. Shah, Fibonacci Sequence Modulo m, Fibonacci Quarterly, Vol.6, No.2 (1968), 139-141.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Index entries for linear recurrences with constant coefficients, signature (1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1).

Cf. A000045, A001175, A003893, A010881.

a089911 n = a089911_list !! n
a089911_list = 0 : 1 : zipWith (\u v -> (u + v) `mod` 12)
                       (tail a089911_list) a089911_list
-- Reinhard Zumkeller, Jul 01 2013

[Fibonacci(n) mod 12: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

with(combinat,fibonacci); A089911 := proc(n) fibonacci(n) mod 12; end;

Table[Mod[Fibonacci[n], 12], {n, 0, 100}] (* Vincenzo Librandi, Feb 04 2014 *)

a(n)=fibonacci(n)%12 \\ Charles R Greathouse IV, Feb 03 2014

Has period of 24, restricted period 12 and multiplier 5.

a(n) = (a(n-1) + a(n-2)) mod 12, a(0) = 0, a(1) = 1.

More terms from Ray Chandler, Nov 15 2003

A001060 a(n) = a(n-1) + a(n-2) with a(0)=2, a(1)=5. Sometimes called the Evangelist Sequence.

Original entry on oeis.org

2, 5, 7, 12, 19, 31, 50, 81, 131, 212, 343, 555, 898, 1453, 2351, 3804, 6155, 9959, 16114, 26073, 42187, 68260, 110447, 178707, 289154, 467861, 757015, 1224876, 1981891, 3206767, 5188658, 8395425, 13584083, 21979508, 35563591, 57543099, 93106690, 150649789

Offset: 0

N. J. A. Sloane

Literally the same as A013655(n+1), since A001060(-1) = A013655(0) = 3. - Eric W. Weisstein, Jun 30 2017
Used by the Sofia Gubaidulina and other composers. - Ian Stewart, Jun 07 2012
From a(2) on, sums of five consecutive Fibonacci numbers; the subset of primes is essentially in A153892. - R. J. Mathar, Mar 24 2010
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (is this A001175?). - R. J. Mathar, Aug 10 2012
Also the number of independent vertex sets and vertex covers in the (n+1)-pan graph. - Eric W. Weisstein, Jun 30 2017
From Wajdi Maaloul, Jun 10 2022: (Start)
For n > 0, a(n) is the number of ways to tile the figure below with squares and dominoes (a strip of length n+1 that contains a vertical strip of height 3 in its second tile). For instance, a(4) is the number of ways to tile this figure (of length 5) with squares and dominoes.
     _
    |_|
   ||_______
  |||_|||_|
(End)

R. V. Jean, Mathematical Approach to Pattern and Form in Plant Growth, Wiley, 1984. See p. 5.
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).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Alfred Brousseau, Seeking the lost gold mine or exploring Fibonacci factorizations, Fib. Quart., 3 (1965), 129-130.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 52.
Paul Coleman, An Introduction to the Music of Sofia Gubaidulina
Tanya Khovanova, Recursive Sequences
Casey Mongoven, Fibonacci Pitch Sequences. - _Ian Stewart_, Jun 07 2012
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
Eric Weisstein's World of Mathematics, Independent Vertex Set
Eric Weisstein's World of Mathematics, Pan Graph
Eric Weisstein's World of Mathematics, Vertex Cover
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000032, A000045, A000285, A104449.

Apart from initial term, same as A013655.

F:=Fibonacci;; List([0..40], n-> F(n+4) - F(n-1) ); # G. C. Greubel, Sep 19 2019

I:=[2,5]; [n le 2 select I[n] else Self(n-1)+Self(n-2): n in [1..50]]; // Vincenzo Librandi, Jan 16 2012

a0:=2; a1:=5; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..35]]; // Bruno Berselli, Feb 12 2013

with(combinat): a:= n-> 2*fibonacci(n)+fibonacci(n+3): seq(a(n), n=0..40); # Zerinvary Lajos, Oct 05 2007
A001060:=-(2+3*z)/(-1+z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

Table[Fibonacci[n+4] -Fibonacci[n-1], {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
LinearRecurrence[{1,1}, {2,5}, 50] (* Vincenzo Librandi, Jan 16 2012 *)
Table[Fibonacci[n+2] + LucasL[n+1], {n, 0, 40}] (* Eric W. Weisstein, Jun 30 2017 *)
CoefficientList[Series[(2+3x)/(1-x-x^2), {x, 0, 40}], x] (* Eric W. Weisstein, Sep 22 2017 *)

a(n)=6*fibonacci(n)+fibonacci(n-3) \\ Charles R Greathouse IV, Jul 14 2017

a(n)=([0,1; 1,1]^n*[2;5])[1,1] \\ Charles R Greathouse IV, Jul 14 2017

f=fibonacci; [f(n+4) - f(n-1) for n in (0..40)] # G. C. Greubel, Sep 19 2019

a(n) = 2*Fibonacci(n) + Fibonacci(n+3). - Zerinvary Lajos, Oct 05 2007
a(n) = Fibonacci(n+4) - Fibonacci(n-1) for n >= 1. - Ian Stewart, Jun 07 2012
a(n) = Fibonacci(n) + 2*Fibonacci(n+2) = 5*Fibonacci(n) + 2*Fibonacci(n-1). The ratio r(n) := a(n+2)/a(n) satisfies the recurrence r(n+1) = (2*r(n) - 1)/(r(n) - 1). If M denotes the 2 X 2 matrix [2, -1; 1, -1] then [a(n+2), a(n)] = M^n[2, -1]. - Peter Bala, Dec 06 2013
a(n) = 6*F(n) + F(n-3), for F(n)=A000045. - J. M. Bergot, Jul 14 2017
a(n) = -(-1)^n*A000285(-2-n) = -(-1)^n*A104449(-1-n) for all n in Z. - Michael Somos, Oct 28 2018
E.g.f.: 2*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 4*sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, May 26 2025

More terms from James Sellers, May 04 2000

A036284 Periodic vertical binary vectors of Fibonacci numbers.

Original entry on oeis.org

6, 24, 1440, 5728448, 92568198012160, 26494530374406845814111659520, 2095920895719545919920115988669687683503034097906010941440, 13128614603426246034591796912897206548807135027496968025827278400248602613784037111736380004928525614173642247188480

Offset: 0

Antti Karttunen, Nov 01 1998

The sequence can be also computed with a recurrence that does not explicitly refer to Fibonacci numbers. See the given Maple and C programs.

Conjecture: For n>=1, each term a(n), when considered as a GF(2)[X]-polynomial, is divisible by GF(2)[X] -polynomial (x^3 + 1) ^ A000225(n-1). If this holds, then for n>=1, a(n) = A048720bi(A136380(n),A048723bi(9,A000225(n-1))). Conjecture 2: there is also one extra (x^1 + 1) factor present, see A136384.

			When Fibonacci numbers are written in binary (see A004685), under each other as:
0000000 (0)
0000001 (1)
0000001 (1)
0000010 (2)
0000011 (3)
0000101 (5)
0001000 (8)
0001101 (13)
0010101 (21)
0100010 (34)
0110111 (55)
1011001 (89)
it can be seen that the bits in the n-th column from right repeat after a period of A007283(n): 3, 6, 12, 24, ... (See also A001175). This sequence is formed from those bits: 011, reversed is 110, is binary for 6, thus a(0) = 6. 000110, reversed is 11000, is binary for 24, thus a(1) = 24, 000001011010, reversed is 10110100000, is binary for 1440, thus a(2) = 1440.

A. Karttunen, Table of n, a(n) for n = 0..10
A. Karttunen, C program for computing this sequence

Same sequence in octal base: A036285. Bits reversed: A036286. See also A136378, A136379, A136380, A136382, A136384, A037096, A037093, A000045.

A036284:=proc(n) option remember; local a, b, c, i, j, k, l, s, x, y, z; if (0 = n) then (6) else a := 0; b := 0; s := 0; x := 0; y := 0; k := 3*(2^(n-1)); l := 3*(2^n); j := 0; for i from 0 to l do z := bit_i(A036284(n-1),(j)); c := (a + b + (`if`((x = y),x,(z+1))) mod 2); if(c <> 0) then s := s + (2^i); fi; a := b; b := c; x := y; y := z; j := j + 1; if(j = k) then j := 0; fi; od; RETURN(s); fi; end:
bit_i := (x,i) -> `mod`(floor(x/(2^i)),2);

a[n_] := Sum[Mod[Fibonacci[k]/2^n // Floor, 2]* 2^k, {k, 0, 3*2^n - 1}]; Table[a[n], {n, 0, 7}] (* Jean-François Alcover, Mar 04 2016 *)

a(n) = Sum_{k=0..A007283(n)-1} ([A000045(k)/(2^n)] mod 2) * 2^k, where [] stands for floor function, i.e. Sum (bit n of Fibonacci(k))*(2^k), k = 0 ... (3*(2^n))-1.

Entry revised Dec 29 2007

A214028 Entry points for the Pell sequence: smallest k such that n divides A000129(k).

Original entry on oeis.org

1, 2, 4, 4, 3, 4, 6, 8, 12, 6, 12, 4, 7, 6, 12, 16, 8, 12, 20, 12, 12, 12, 22, 8, 15, 14, 36, 12, 5, 12, 30, 32, 12, 8, 6, 12, 19, 20, 28, 24, 10, 12, 44, 12, 12, 22, 46, 16, 42, 30, 8, 28, 27, 36, 12, 24, 20, 10, 20, 12, 31, 30, 12, 64, 21, 12, 68, 8, 44, 6, 70, 24, 36, 38

Offset: 1

Art DuPre, Jul 04 2012

nonn

Conjecture: A175181(n)/A214027(n) = a(n). This says that the zeros appear somewhat uniformly in a period. The second zero in a period is exactly where n divides the first Lucas number, so this relationship is not really surprising.
From Jianing Song, Aug 29 2018: (Start)
The comment above is correct, since n divides A000129(k*a(n)) for all integers k and clearly a(n) divides A175181(n), so the zeros appear uniformly.
a(n) <= 4*n/3 for all n, where the equality holds iff n is a power of 3.
(End)

			11 first divides the term A000129(12) = 13860 = 2*3*5*7*11.

G. C. Greubel, Table of n, a(n) for n = 1..10000
Bernadette Faye and Florian Luca, Pell Numbers whose Euler Function is a Pell Number, arXiv:1508.05714 [math.NT], 2015 (called z(n)).
N. Robbins, On Pell numbers of the form p*x^2, where p is prime, Fib. Quart. 22 (4) (1984) 340-348, definition 1.

Cf. A001175, A001176, A001177, A214027, A175181.

A214028 := proc(n)
    local a000129,k ;
    a000129 := [1,2,5] ;
    for k do
        if modp(a000129[1],n) = 0 then
            return k;
        end if;
        a000129[1] := a000129[2] ;
        a000129[2] := a000129[3] ;
        a000129[3] := 2*a000129[2]+a000129[1] ;
    end do:
end proc:
seq(A214028(n),n=1..40); # R. J. Mathar, May 26 2016

a[n_] := With[{s = Sqrt@ 2}, ((1 + s)^n - (1 - s)^n)/(2 s)] // Simplify; Table[k = 1; While[Mod[a[k], n] != 0, k++]; k, {n, 80}] (* Michael De Vlieger, Aug 25 2015, after Michael Somos at A000129 *)
Table[k = 1; While[Mod[Fibonacci[k, 2], n] != 0, k++]; k, {n, 100}] (* G. C. Greubel, Aug 10 2018 *)

pell(n) = polcoeff(Vec(x/(1-2*x-x^2) + O(x^(n+1))), n);
a(n) = {k=1; while (pell(k) % n, k++); k;} \\ Michel Marcus, Aug 25 2015

If p^2 does not divide A000129(a(p)) (that is, p is not in A238736) then a(p^e) = a(p)*p^(e - 1). If gcd(m, n) = 1 then a(mn) = lcm(a(m), a(n)). - Jianing Song, Aug 29 2018

A022097 Fibonacci sequence beginning 1, 7.

Original entry on oeis.org

1, 7, 8, 15, 23, 38, 61, 99, 160, 259, 419, 678, 1097, 1775, 2872, 4647, 7519, 12166, 19685, 31851, 51536, 83387, 134923, 218310, 353233, 571543, 924776, 1496319, 2421095, 3917414, 6338509, 10255923, 16594432, 26850355, 43444787, 70295142, 113739929

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(7;n-1-k,k) with n>=1, a(-1)=6. These are the SW-NE diagonals in P(7;n,k), the (7,1) Pascal triangle A093564. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (perhaps the same as A001175). - R. J. Mathar, Aug 10 2012
For n >= 1, a(n) is the number of edge covers of the tadpole graph T_{4,n-1} with T_{4,0} interpreted as just the cycle graph C_4. Example: If n=2, we have C_4 and path P_1 joined by a bridge. This is the cycle with a pendant and has 7 edge covers. - Feryal Alayont, Sep 22 2024

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Tadpole Graph.
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000045, A001175, A093564, A101220, A109754, A118654, A131778.

a0:=1; a1:=7; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

First /@ NestList[{Last@ #, Total@ #} &, {1, 7}, 36] (* or *)
CoefficientList[Series[(1 + 6 x)/(1 - x - x^2), {x, 0, 36}], x] (* Michael De Vlieger, Feb 20 2017 *)
LinearRecurrence[{1,1},{1,7},40] (* Harvey P. Dale, May 17 2018 *)

a(n)=([0,1; 1,1]^n*[1;7])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

A022097=BinaryRecurrenceSequence(1,1,1,7)
print([A022097(n) for n in range(41)]) # G. C. Greubel, Jun 03 2025

a(n) = a(n-1) + a(n-2) for n>=2, a(0)=1, a(1)=7, a(-1):=6.
G.f.: (1+6*x)/(1-x-x^2).
a(n) = A109754(6, n+1).
a(n) = A118654(3, n).
a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(-13 + sqrt(5)) + (1 + sqrt(5))^n*(13 + sqrt(5))))/sqrt(5). - Herbert Kociemba
a(n) = 6*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 8*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
From Aamen Muharram, Aug 05 2022: (Start)
a(n) = F(n-4) + F(n-1) + F(n+4),
a(n) = F(n) + F(n+4) - F(n-3),
where F(n) = A000045(n) is the Fibonacci numbers. (End)

A106291 Period of the Lucas sequence A000032 mod n.

Original entry on oeis.org

1, 3, 8, 6, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, 16, 30, 48, 24, 20, 84, 72, 48, 14, 24, 30, 48, 40, 36, 16, 24, 76, 18, 56, 12, 40, 48, 88, 30, 24, 48, 32, 24, 112, 60, 72, 84, 108, 72, 20, 48, 72, 42, 58, 24, 60, 30, 48, 96, 28, 120, 136, 36, 48, 48

Offset: 1

T. D. Noe, May 02 2005

nonn

This sequence differs from the Fibonacci periods (A001175) only when n is a multiple of 5, which can be traced to 5 being the discriminant of the characteristic polynomial x^2-x-1.

This sequence coincides with the Fibonacci periods (A001175) if n is a multiple of 5^j and the following conditions apply: n contains at least one prime factor of the form p = 10*k+1 (A030430) which occurs in Fibonacci(m) or Lucas(m) as prime factor, where m must be the smallest possible index containing p and a factor 5^i and j <= i. If n contains several prime factors from A030430 that satisfy the above conditions, the largest applicable i is decisive. - Klaus Purath, Apr 26 2019

			From _Klaus Purath_, Jul 10 2019: (Start)
n = 3*5*31 = 465, j = 1; L(15) is the smallest Lucas number with prime factor 31; 15 = 3*5, i = 1 = j. Hence Lucas period (mod 465) = Fibonacci period (mod 465) = 120, but if n = 3*5^2*31 = 2325, j = 2 > i. Hence Lucas period (mod 2325) = 120 < Fibonacci period (mod 2325) = 600.
n = 5*701 = 3505, j = 1; F(175) is the smallest Fibonacci number with prime factor 701; 175 = 7*5^2, i = 2 > j. Therefore Lucas period (mod 3505) = Fibonacci period (mod 3505) = 700, but if n = 5^3*701 = 87625, j = 3 > i. Therefore Lucas period (mod 87625) = 700 < Fibonacci period (mod 87625) = 3500.
n = 5^2*11*101 = 27775, j =2; L(5) is the smallest Lucas number with prime factor 11, i = 1; L(25) = is the smallest Lucas number with prime factor 101; 25 = 5^2, i = 2 ( decisive); j = i. Hence Lucas period (mod 27775) = Fibonacci period (mod 27775) = 100, but if n = 5^3*11*101 = 138875, j = 3 > i. Hence Lucas period (mod 138875) = 100 < Fibonacci period (mod 138875) = 500. (End)

S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989. See p. 89. - From N. J. A. Sloane, Feb 20 2013

G. C. Greubel and D. Turner, Table of n, a(n) for n = 1..10000
Brennan Benfield and Oliver Lippard, Fixed points of K-Fibonacci sequences, arXiv:2404.08194 [math.NT], 2024. See p. 11.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number

Cf. A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).

Cf. A000032, A001175, A030430.

n=2; 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, 70}]

from math import lcm
from functools import lru_cache
from sympy import factorint
@lru_cache(maxsize=None)
def A106291(n):
    if n < 3:
        return (n<<1)-1
    f = factorint(n).items()
    if len(f) > 1:
        return lcm(*(A106291(a**b) for a,b in f))
    else:
        k,x = 1, (1,3)
        while x != (2,1):
            k += 1
            x = (x[1], (x[0]+x[1]) % n)
        return k # Chai Wah Wu, Apr 25 2025

def a(n): return BinaryRecurrenceSequence(1, 1, 2, 1).period(n)
[a(n) for n in (1..100)] # G. C. Greubel, Apr 27 2019

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

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 *)

A003147 Primes p with a Fibonacci primitive root g, i.e., such that g^2 = g + 1 (mod p).

Original entry on oeis.org

5, 11, 19, 31, 41, 59, 61, 71, 79, 109, 131, 149, 179, 191, 239, 241, 251, 269, 271, 311, 359, 379, 389, 409, 419, 431, 439, 449, 479, 491, 499, 569, 571, 599, 601, 631, 641, 659, 701, 719, 739, 751, 821, 839, 929, 971, 1019, 1039, 1051, 1091, 1129, 1171, 1181, 1201, 1259, 1301

Offset: 1

N. J. A. Sloane

Primes p with a primitive root g such that g^2 = g + 1 (mod p).
Not the same as primes with a Fibonacci number as primitive root; cf. A083701. - Jonathan Sondow, Feb 17 2013
For all except the initial term 5, these are numbers such that the Pisano period equals 1 less than the Pisano number, i.e., where A001175(n) = n-1. - Matthew Goers, Sep 20 2013
As shown in the paper by Brison, these are also the primes p such that there is a Fibonacci-type sequence (mod p) that begins with (1,b) and encounters all numbers less than p in the first p-1 iterations (for some b). - T. D. Noe, Feb 26 2014
Shanks (1972) conjectured that the relative asymptotic density of this sequence in the sequence of primes is 27*c/38 = 0.2657054465..., where c is Artin's constant (A005596). The conjecture was proved on the assumption of a generalized Riemann hypothesis by Lenstra (1977) and Sander (1990). - Amiram Eldar, Jan 22 2022

			3 is a primitive root mod 5, and 3^2 = 3 + 1 mod 5, so 5 is a member. - _Jonathan Sondow_, Feb 17 2013

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1000 terms from Noe)
Bob Bastasz, Lyndon words of a second-order recurrence, Fibonacci Quarterly, Vol. 58, No. 5 (2020), pp. 25-29.
Owen J. Brison, Complete Fibonacci sequences in finite fields, Fibonacci Quarterly, Vol. 30, No. 4 (1992), pp. 295-304.
Alexandru Gica, Quadratic Residues in Fibonacci Sequences, Fibonacci Quart., Vol. 46/47, No. 1 (2008/2009), pp. 68-72. See Theorem 5.1.
Liang-Chung Hsia, Hua-Chieh Li, and Wei-Liang Sun, Certain Diagonal Equations and Conflict-Avoiding Codes of Prime Lengths, arXiv:2302.00920 [math.NT], 2023.
H. W. Lenstra, Jr., On Artin's conjecture and Euclid's algorithm in global fields, Invent. Math., Vol. 42 (1977), pp. 202-224; alternative link.
J. W. Sander, On Fibonacci primitive roots, Fibonacci Quarterly, Vol. 28, No. 1 (1990), pp. 79-80.
Daniel Shanks, Fibonacci primitive roots, end of article, Fibonacci Quarterly, Vol. 10, No. 2 (1972), pp. 163-168, 181.
Daniel Shanks and Larry Taylor, An Observation of Fibonacci Primitive Roots, Fibonacci Quarterly, Vol. 11, No. 2 (1973), pp. 159-160.
Index entries for primes by primitive root.

Subsequence of A038872.
Cf. A001175, A005596, A083701.
See also A106535.

filter:=proc(n) local g,r;
if not isprime(n) then return false fi;
r:= [msolve(g^2 -g - 1, n)][1];
numtheory:-order(rhs(op(r)),n) = n-1
end proc:
select(filter, [5,seq(seq(10*i+j,j=[1,9]),i=1..1000)]); # Robert Israel, May 22 2015

okQ[p_] := AnyTrue[PrimitiveRootList[p], Mod[#^2, p] == Mod[#+1, p]&]; Select[Prime[Range[300]], okQ] (* Jean-François Alcover, Jan 04 2016 *)

is(n)=if(kronecker(5,n)<1||!isprime(n), return(n==5)); my(s=sqrt(Mod(5,n))); znorder((1+s)/2)==n-1 || znorder((1-s)/2)==n-1 \\ Charles R Greathouse IV, May 22 2015

More terms from David W. Wilson
Cross-reference from Charles R Greathouse IV, Nov 05 2009
Definition clarified by M. F. Hasler, Jun 05 2018

cp's OEIS Frontend

A030132 Digital root of Fibonacci(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Formula

Extensions

A046738 Period of Fibonacci 3-step sequence A000073 mod n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A089911 a(n) = Fibonacci(n) mod 12.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

Formula

Extensions

A001060 a(n) = a(n-1) + a(n-2) with a(0)=2, a(1)=5. Sometimes called the Evangelist Sequence.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Magma

Magma

Maple

Mathematica

PARI

PARI

Sage

Formula

Extensions

A036284 Periodic vertical binary vectors of Fibonacci numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A214028 Entry points for the Pell sequence: smallest k such that n divides A000129(k).

Views

Author

Keywords

Comments

Examples