A001175 -id:A001175 - cp's OEIS frontend

A010077 a(n) = sum of digits of a(n-1) + sum of digits of a(n-2); a(0) = 0, a(1) = 1.

0, 1, 1, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10

Offset: 0

N. J. A. Sloane, Leonid Broukhis

The digital sum analog (in base 10) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007
a(n) and Fibonacci(n) = A000045(n) are congruent modulo 9 which implies that (a(n) mod 9) is equal to (Fibonacci(n) mod 9) = A007887(n). Thus (a(n) mod 9) is periodic with the Pisano period A001175(9)=24. - Hieronymus Fischer, Jun 27 2007
a(n) == A004090(n) (mod 9) (A004090(n) = digital sum of Fibonacci(n)). - Hieronymus Fischer, Jun 27 2007
For general bases p > 2, we have the inequality 2 <= a(n) <= 2p-3 (for n > 2). Actually, a(n) <= 17 = A131319(10) for the base p=10. - Hieronymus Fischer, Jun 27 2007

David A. Corneth, Table of n, a(n) for n = 0..9999
Index entries for Colombian or self numbers and related sequences

Cf. A007953, A007612, A065076.
Cf. A000045, A010073, A010074, A010075, A010076.
Cf. A131294, A131295, A131296, A131297, A131318, A131319, A131320.

a[0] = 0; a[1] = 1; a[n_] := a[n] = Apply[ Plus, IntegerDigits[ a[n - 1] ]] + Apply[ Plus, IntegerDigits[ a[n - 2] ]]; Table[ a[n], {n, 0, 100} ]
nxt[{a_,b_}]:={b, Total[IntegerDigits[a]]+Total[IntegerDigits[b]]}; NestList[ nxt,{0,1},80][[All,1]] (* Harvey P. Dale, Apr 15 2018 *)

first(n) = {n = max(n, 2); my(res = vector(n)); res[2] = 1; for(i = 3, n, res[i] = sumdigits(res[i-1]) + sumdigits(res[i-2]) ); res } \\ David A. Corneth, May 26 2021

Periodic from n=3 with period 24. - Franklin T. Adams-Watters, Mar 13 2006
a(n) = A030132(n-4) + A030132(n-3) for n>3. - Reinhard Zumkeller, Jul 04 2007
a(n) = a(n-1) + a(n-2) - 9*(floor(a(n-1)/10) + floor(a(n-2)/10)). - Hieronymus Fischer, Jun 27 2007
a(n) = floor(a(n-1)/10) + floor(a(n-2)/10) + (a(n-1) mod 10) + (a(n-2) mod 10). - Hieronymus Fischer, Jun 27 2007
a(n) = A059995(a(n-1)) + A059995(a(n-2)) + A010879(a(n-1)) + A010879(a(n-2)). - Hieronymus Fischer, Jun 27 2007
a(n) = Fibonacci(n) - 9*Sum_{k=2..n-1} Fibonacci(n-k+1)*floor(a(k)/10) where Fibonacci(n) = A000045(n). - Hieronymus Fischer, Jun 27 2007

A106535 Numbers k such that the smallest x > 1 for which Fibonacci(x) == 0 mod k is x = k - 1.

Original entry on oeis.org

11, 19, 31, 59, 71, 79, 131, 179, 191, 239, 251, 271, 311, 359, 379, 419, 431, 439, 479, 491, 499, 571, 599, 631, 659, 719, 739, 751, 839, 971, 1019, 1039, 1051, 1091, 1171, 1259, 1319, 1399, 1439, 1451, 1459, 1499, 1531, 1559, 1571, 1619, 1759, 1811, 1831

Offset: 1

Peter K. Pearson (ppearson+att(AT)spamcop.net), May 06 2005

nonn

This is a sister sequence to A000057, because this sequence, since {k : A001177(k) = k-1}, might be called a subdiagonal sequence of A001177, and {k : A001177(k) = k+1}, which might be called a superdiagonal sequence of A001177. Sequences A000057 and A106535 are disjoint. Is this sequence the set of all divisors of some family of sequences, like A000057 is? - Art DuPre, Jul 11 2012
Are all members of this sequence prime? Using A069106, any composite members must exceed 89151931. - Robert Israel, Oct 13 2015
From Jianing Song, Jul 02 2019: (Start)
Yes, all terms are primes. See a brief proof below.
Also, if p == 1 (mod 4) then b(p) divides (p-Legendre(p,5))/2. So terms in this sequence are congruent to 11 or 19 modulo 20.
Primes p such that ord(-(3+sqrt(5))/2,p) = p-1, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer. (End)
Comments from Amiram Eldar, Jan 30 2022 (Start)
Sequence A003147, "Primes p with a Fibonacci primitive root", is defined in the paper: Daniel Shanks, Fibonacci primitive roots, Fibonacci Quarterly, Vol.  10, No. 2 (1972), pp. 163-168, and 181.
A second paper on this subject Daniel Shanks and Larry Taylor, An Observation of Fibonacci Primitive Roots, Fibonacci Quarterly, Vol. 11, No. 2 (1973), pp. 159-160,
deals with terms p == 3 (mod 4) of A003147, i.e., the intersection of A003147 and A002145 (or A004767).
It states that if g is a Fibonacci primitive root of a prime p such that p == 3 (mod 4) then g-1 and g-2 are also primitive roots of p.
The first 2000 terms of (A003147 intersect A002145) agree with the present sequence, although the definitions are quite different. Are these two sequences the same? (End)

Robert Israel, Table of n, a(n) for n = 1..2000
Alfred Brousseau, Primes which are factors of all Fibonacci sequences, Fib. Quart., 2 (1964), 33-38.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972.
Jianing Song, Proof that all terms are prime

Cf. A000057, A001175, A069106.

Similar sequences that give primes p such that A001177(p) = (p-1)/s: this sequence (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7),A308801 (s=8), A308802 (s=9).

Filtered([2..2000], n -> Fibonacci(n-1) mod n = 0 and Filtered( [2..n-2], x -> Fibonacci(x) mod n = 0 ) = [] );

A106535 := proc(n)
        option remember;
        if n = 1 then
                11;
        else
                for a from procname(n-1)+1 do
                        if A001177(a) = a-1 then
                                return a;
                        end if;
                end do:
        end if;
end proc: # R. J. Mathar, Jul 09 2012
# Alternative:
fmod:= proc(a,b) local A;
  uses LinearAlgebra[Modular];
  A:= Mod(b, <<1,1>|<1,0>>,integer[8]);
  MatrixPower(b,A,a)[1,2];
end proc:
filter:= proc(n)
  local cands;
  if fmod(n-1,n) <> 0 then return false fi;
  cands:= map(t -> (n-1)/t, numtheory:-factorset(n-1));
  andmap(c -> (fmod(c,n) > 0), cands);
end proc:
select(filter, [$2..10^4]); # Robert Israel, Oct 13 2015

f[n_] := Block[{x = 2}, While[Mod[Fibonacci@ x, n] != 0, x++]; x];Select[Range@ 1860, f@ # == # - 1 &] (* Michael De Vlieger, Oct 13 2015 *)

isok(n) = {x = 2; while(fibonacci(x) % n, x++); x == n-1;} \\ Michel Marcus, Oct 20 2015

{n: A001177(n) = n-1}. - R. J. Mathar, Jul 09 2012

Corrected by T. D. Noe, Oct 25 2006

A248218 Period in residues modulo n in iteration of x^2 + 1 starting at 0.

Original entry on oeis.org

1, 2, 1, 2, 3, 2, 1, 2, 3, 6, 2, 2, 4, 2, 3, 2, 6, 6, 1, 6, 1, 2, 2, 2, 3, 4, 3, 2, 2, 6, 1, 2, 2, 6, 3, 6, 1, 2, 4, 6, 7, 2, 1, 2, 3, 2, 4, 2, 6, 6, 6, 4, 2, 6, 6, 2, 1, 2, 3, 6, 10, 2, 3, 2, 12, 2, 2, 6, 2, 6, 11, 6, 6, 2, 3, 2, 2, 4, 4, 6, 9, 14, 5, 2, 6

Offset: 1

Vaclav Kotesovec, Oct 04 2014

nonn

a(n) is a period in the sequence A003095 modulo n.
For n <= 10000 is the maximal period a(7921) = 1232.
For n <= 100000 is the maximal period a(73205) = 7260.
For n <= 500000 is the maximal period a(357911) = 54670.
From Hermann Stamm-Wilbrandt, Jun 21 2021: (Start)
357911 = 71^3; a(71^2) = 770; a(71^3) = 71 * a(71^2); a(71^4) = 71 * a(71^3); a(71^5) = 71 * a(71^4); a(71^6) = 71 * a(71^5). 770/71^2 = 0.15274747073993255306, so cycle length is linear in n for these composite numbers. a(71^6) = 19566994370.
Let A(n) be number of start values that end on same cycle as start value 0. A(71^2) = 3711; A(71^3) = 71 * A(71^2); A(71^4) = 71 * A(71^3); A(71^5) = 71 * A(71^4). 3711/71^2 = 0.73616345963102559016, so majority of start values end on start value 0 cycle. (End)
Linear cycle length for a(71^i) with 2 <= i <= 5 sounds bad for runtime of Pollard-Rho factorization algorithm (heuristic claim assumes square root cycle length). The opposite is true, every value on start value 0 cycle has same remainder mod 71 as the value after applying "x -> (x^2 + 1) mod n" 11 times, so factorization completes quickly. - Hermann Stamm-Wilbrandt, Jun 29 2021

			n=5, residues are 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, ..., period is 3, a(5)=3.
n=7, residues are 1, 2, 5, 5, 5, 5, 5, ..., final period is 1, therefore a(7)=1.
n=10, residues are 1, 2, 5, 6, 7, 0, 1, 2, 5, 6, 7, 0, 1, 2, ..., a(10)=6.
n=43, residues are 1, 2, 5, 26, 32, 36, 7, 7, 7, 7, ..., a(43) = 1.
n=229, residues are 1, 2, 5, 26, 219, 101, 126, 76, 52, 186, 18, 96, 57, 44, 105, 34, 12, 145, 187, 162, 139, 86, 69, 182, 149, 218, 122, 0, 1, 2, 5, 26, 219, 101, 126, 76, 52, 186, 18, 96, 57, 44, 105, 34, 12, 145, 187, 162, 139, 86, 69, 182, 149, 218, 122, 0, 1, 2, 5, 26, ..., period is 28, a(229)=28.
This program is for experiments (n<100): Rest[NestList[Mod[#^2+1, n] &, 0, 100]]

Vaclav Kotesovec, Table of n, a(n) for n = 1..100000
Hermann Stamm-Wilbrandt, analyze.c
Hermann Stamm-Wilbrandt, period.c
Wikipedia, Pollard's Rho algorithm.

Cf. A248219, A256342-A256349, A003095, A247981, A001175.

/* See analyze.c in the Links section. This program computes a(n) for n < 2^31, all periods for any starting value. See also period.c which only computes period length, but with arbitrary precision gmplib. This allowed to compute a(71^6). - Hermann Stamm-Wilbrandt, Jun 22 2021 */

Table[m=Rest[NestList[Mod[#^2+1,n]&,0,1000]]; period=0; j=1; While[j<=Length[m] && period==0,If[m[[Length[m]-j]]==m[[Length[m]]],period=j]; j++]; period,{n,1,1000}]

A248218(m,t=0,u=[t])=until(#Set(u=concat(u,t=(t^2+1)%m))<#u,);for(i=1,#u,t==u[#u-i]&&return(i)) \\ M. F. Hasler, Mar 25 2015

a(LCM(i,j)) = LCM(a(i),a(j)). - Robert Israel, Mar 08 2021

A015134 Consider Fibonacci-type sequences b(0)=X, b(1)=Y, b(k)=b(k-1)+b(k-2) mod n; all are periodic; sequence gives number of distinct periods.

Original entry on oeis.org

1, 2, 2, 4, 3, 4, 4, 8, 5, 6, 14, 10, 7, 8, 12, 16, 9, 16, 22, 16, 29, 28, 12, 30, 13, 14, 14, 22, 63, 24, 34, 32, 39, 34, 30, 58, 19, 86, 32, 52, 43, 58, 22, 78, 39, 46, 70, 102, 25, 26, 42, 40, 27, 52, 160, 74, 63, 126, 62, 70, 63, 134, 104, 64, 57, 78, 34, 132, 101, 60, 74, 222

Offset: 1

Phil Carmody

nonn

b(k) >= k/4 (by counting zeros). - R. C. Johnson (bob.johnson(AT)dur.ac.uk), Nov 20 2003

Peter McAndrew, Table of n, a(n) for n = 1..10000 (terms 1..1000 from David Radcliffe)
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.
Jesse Fischer, Number Necklace Generator.
R. C. Johnson, Fibonacci Numbers and Resources.

Cf. A015135 (number of different orbit lengths of the 2-step recursion mod n), A106306 (primes that yield a simple orbit structure in 2-step recursions).

a(2^n) = 2^n. - Thomas Anton, Apr 16 2023
Conjectures from Stephen Bartell, Aug 20 2023: (Start)
a(3 * 2^n) = 3*(3 * 2^n - 14), n >= 3.
a(3^n) = (3^n + 1)/2, n >= 0.
a(2 * 3^n) = 2*(3^n - 1), n >= 1.
a(5 * 3^n) = (3^(n+2) - 3)/2, n >= 0.
a(10 * 3^n) = 6(3^(n+1) - 5), n >= 1.
a(5^n) = (5^n + 1)/2, n >= 0.
a(2 * 5^n) = 5^n + 1, n >= 0.
a(3 * 5^n) = (5^(n+1) - 1)/2, n >= 0.
a(4 * 5^n) = 3 * 5^n + 1, n >= 0.
a(6 * 5^n) = 5^(n+1) - 1, n >= 0.
a(7^n) = (7^n + 1)/2, n >= 0.
a(2 * 7^n) = 7^n + 1, n >= 0. (End)
Conjectures from Stephen Bartell, Aug 16 2024: (Start)
a(11^n) = (6*11^n - 1)/5 + n, n >= 0.
a(13^n) = (13^n + 1)/2, n >= 0.
a(17^n) = (17^n + 1)/2, n >= 0.
a(19^n) = (10*19^n - 1)/9 + n, n >= 0.
a(23^n) = (23^n + 1)/2, n >= 0.
a(29^n) = (15*29^n - 8)/7 + 2n, n >= 0.
For prime p not in A053032, a(p^n) = 1 + ((p+1)(p^n-1))/(A001175(p)) [except for p = 5].
For prime p in A053032, a(p^n) = 1 + ((p+1)(p^n-1)+n(p-1))/(A001175(p)). (End)

More terms from Larry Reeves (larryr(AT)acm.org), Jan 06 2005

A060305 Pisano periods for primes: period of Fibonacci numbers mod prime(n).

Original entry on oeis.org

3, 8, 20, 16, 10, 28, 36, 18, 48, 14, 30, 76, 40, 88, 32, 108, 58, 60, 136, 70, 148, 78, 168, 44, 196, 50, 208, 72, 108, 76, 256, 130, 276, 46, 148, 50, 316, 328, 336, 348, 178, 90, 190, 388, 396, 22, 42, 448, 456, 114, 52, 238, 240, 250, 516, 176, 268, 270, 556

Offset: 1

Louis Mello (mellols(AT)aol.com), Mar 26 2001

nonn

Assuming Wall's conjecture (which is still open) allows one to calculate A001175(m) when m is a prime power since for any k >= 1: A001175(prime(n)^k) = a(n)*prime(n)^(k-1). For example: A001175(2^k) = 3*2^(k-1) = A007283(k-1).

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014), # 14.8.5.
A. Elsenhans and J. Jahnel, The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 10^14, (2004).
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Wikipedia, Pisano period.

Cf. A001175, A000961, A071774, A003147, A296240.

a:= proc(n) option remember; local F, k, p;
      F:=[1,1]; p:=ithprime(n);
      for k while F<>[0,1] do
        F:=[F[2], irem(F[1]+F[2],p)]
      od: k
    end:
seq(a(n), n=1..70);  # Alois P. Heinz, Oct 16 2015

Table[p=Prime[n]; a={1,0}; a0=a; k=0; While[k++; s=Mod[Plus@@a,p];a=RotateLeft[a]; a[[2]]=s; a!=a0]; k, {n,100}] (* T. D. Noe, Jun 12 2006 *)

for(n=1,100,s=1; while(sum(i=n,n+s,abs(fibonacci(i)%prime(n)-fibonacci(i+s)%prime(n)))+sum(i=n+1,n+1+s,abs(fibonacci(i)%prime(n)-fibonacci(i+s)%prime(n)))>0,s++); print1(s,","))

from itertools import count
from sympy import prime
def A060305(n):
    x, p = (1,1), prime(n)
    for k in count(1):
        if x == (0,1):
            return k
        x = (x[1], (x[0]+x[1]) % p) # Chai Wah Wu, May 31 2022

a(n) = A001175(prime(n)). - Jonathan Sondow, Dec 09 2017

a(n) = (3 - L(p))/2 * (p - L(p)) / A296240(n) for n >= 4, where p = prime(n) and L(p) = Legendre(p|5); so a(n) <= p-1 if p == +- 1 mod 5, and a(n) <= 2*p+2 if p == +- 2 mod 5. See Wall's Theorems 6 and 7. - Jonathan Sondow, Dec 10 2017

Corrected by Benoit Cloitre, Jun 04 2002

Name clarified by Jonathan Sondow, Dec 09 2017

A131294 a(n)=ds_3(a(n-1))+ds_3(a(n-2)), a(0)=0, a(1)=1; where ds_3=digital sum base 3.

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Jun 27 2007

The digital sum analog (in base 3) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(2)=3.
a(n) and Fib(n)=A000045(n) are congruent modulo 2 which implies that (a(n) mod 2) is equal to (Fib(n) mod 2)=A011655(n). Thus (a(n) mod 2) is periodic with the Pisano period A001175(2)=3 too.
For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=3=A131319(3) for the base p=3.

			a(5)=3, since a(3)=2, ds_3(2)=2, a(4)=3=10(base 3),
ds_3(3)=1 and so a(5)=2+1.

Index entries for Colombian or self numbers and related sequences
Index entries for linear recurrences with constant coefficients, signature (0, 0, 1).

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131295, A131296, A131297, A131318, A131319, A131320.

nxt[{a_,b_}]:={b,Total[IntegerDigits[a,3]]+Total[IntegerDigits[b,3]]}; Transpose[NestList[nxt,{0,1},100]][[1]] (* Harvey P. Dale, Aug 02 2016 *)

a(n) = a(n-1)+a(n-2)-2*(floor(a(n-1)/3)+floor(a(n-2)/3)).
a(n) = floor(a(n-1)/3)+floor(a(n-2)/3)+(a(n-1)mod 3)+(a(n-2)mod 3).
a(n) = A002264(a(n-1))+A002264(a(n-2))+A010872(a(n-1))+A010872(a(n-2)).
a(n) = Fib(n)-2*sum{1A000045(n).

Incorrect comment removed by Michel Marcus, Apr 29 2018

A131295 a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3

Offset: 0

Hieronymus Fischer, Jun 27 2007

The digital sum analog (in base 4) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(3)=8.
Also a(n)==A004090(n) modulo 3 (A004090(n)=digital sum of Fib(n)).
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(4) for the base p=4.
a(n) and Fib(n)=A000045(n) are congruent modulo 3 which implies that (a(n) mod 3) is equal to (Fib(n) mod 3)=A082115(n-1) (for n>0). Thus (a(n) mod 3) is periodic with the Pisano period = A001175(3)=8 too. - Hieronymus Fischer

			a(8)=3, since a(6)=5=11(base 4), ds_4(5)=2,
a(7)=4=10(base 4), ds_4(4)=1 and so a(8)=2+1.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131296, A131297, A131318, A131319, A131320.

nxt[{a_,b_}]:={b,Total[IntegerDigits[a,4]]+Total[IntegerDigits[b,4]]}; NestList[ nxt,{0,1},110][[All,1]] (* Harvey P. Dale, Jul 30 2018 *)

a(n)=a(n-1)+a(n-2)-3*(floor(a(n-1)/4)+floor(a(n-2)/4)).
a(n)=floor(a(n-1)/4)+floor(a(n-2)/4)+(a(n-1)mod 4)+(a(n-2)mod 4).
a(n)=A002265(a(n-1))+A002265(a(n-2))+A010873(a(n-1))+A010873(a(n-2)).
a(n)=Fib(n)-3*sum{1A000045(n).

A010074 a(n) = sum of base-7 digits of a(n-1) + sum of base-7 digits of a(n-2).

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7

Offset: 0

N. J. A. Sloane, Leonid Broukhis

The digital sum analog (in base 7) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007
a(n) and Fib(n)=A000045(n) are congruent modulo 6 which implies that (a(n) mod 6) is equal to (Fib(n) mod 6) = A082117(n-1) (for n>0). Thus (a(n) mod 6) is periodic with the Pisano period A001175(6)=24. - Hieronymus Fischer, Jun 27 2007
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds (for n>2). Actually, a(n)<=11=A131319(7) for the base p=7. - Hieronymus Fischer, Jun 27 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

Cf. A000045, A010073, A010075, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

nxt[{a_,b_}]:={b,Total[IntegerDigits[a,7]]+Total[IntegerDigits[b,7]]}; Transpose[NestList[nxt,{0,1},80]][[1]] (* Harvey P. Dale, Oct 12 2013 *)

Periodic from n=3 with period 24. - Franklin T. Adams-Watters, Mar 13 2006
From Hieronymus Fischer, Jun 27 2007: (Start)
a(n) = a(n-1)+a(n-2)-6*(floor(a(n-1)/7)+floor(a(n-2)/7)).
a(n) = floor(a(n-1)/7)+floor(a(n-2)/7)+(a(n-1)mod 7)+(a(n-2)mod 7).
a(n) = (a(n-1)+a(n-2)+6*(A010876(a(n-1))+A010876(a(n-2))))/7.
a(n) = Fib(n)-6*sum{1A000045(n). (End)

Incorrect comment removed by Michel Marcus, Apr 29 2018

A010075 a(n) = sum of base-8 digits of a(n-1) + sum of base-8 digits of a(n-2).

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8

Offset: 0

N. J. A. Sloane, Leonid Broukhis

The digital sum analog (in base 8) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007
a(n) and Fib(n)=A000045(n) are congruent modulo 7 which implies that (a(n) mod 7) is equal to (Fib(n) mod 7). Thus (a(n) mod 7) is periodic with the Pisano period A001175(7)=16. - Hieronymus Fischer, Jun 27 2007
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=11=A131319(8) for the base p=8. - Hieronymus Fischer, Jun 27 2007

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

Cf. A000045, A010073, A010074, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

nxt[{a_,b_}]:={b,Total[IntegerDigits[a,8]]+Total[IntegerDigits[b,8]]}; NestList[ nxt,{0,1},100][[All,1]] (* or *) PadRight[{0,1,1},100,{7,8,8,2,3,5,8,6,7,13,13,12,11,9,6,8}] (* Harvey P. Dale, Apr 19 2020 *)

Periodic from n=3 with period 16. - Franklin T. Adams-Watters, Mar 13 2006
From Hieronymus Fischer, Jun 27 2007: (Start)
a(n) = a(n-1)+a(n-2)-7*(floor(a(n-1)/8)+floor(a(n-2)/8)).
a(n) = floor(a(n-1)/8)+floor(a(n-2)/8)+(a(n-1)mod 8)+(a(n-2)mod 8).
a(n) = (a(n-1)+a(n-2)+7*(A010877(a(n-1))+A010877(a(n-2))))/8.
a(n) = Fib(n)-7*sum{1A000045(n). (End)

Incorrect comment removed by Michel Marcus, Apr 29 2018

A066853 Number of different remainders (or residues) for the Fibonacci numbers (A000045) when divided by n (i.e., the size of the set of F(i) mod n over all i).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 6, 9, 10, 7, 11, 9, 14, 15, 11, 13, 11, 12, 20, 9, 14, 19, 13, 25, 18, 27, 21, 10, 30, 19, 21, 19, 13, 35, 15, 29, 13, 25, 30, 19, 18, 33, 20, 45, 21, 15, 15, 37, 50, 35, 30, 37, 29, 12, 25, 33, 20, 37, 55, 25, 21, 23, 42, 45, 38, 51, 20, 29, 70, 44, 15, 57

Offset: 1

Reiner Martin, Jan 21 2002

nonn

The Fibonacci numbers mod n for any n are periodic - see A001175 for period lengths. - Ron Knott, Jan 05 2005

a(n) = number of nonzeros in n-th row of triangle A128924. - Reinhard Zumkeller, Jan 16 2014

			a(8)=6 since the Fibonacci numbers, 0,1,1,2,3,5,8,13,21,34,55,89,144,... when divided by 8 have remainders 0,1,1,2,3,5,0,5,5,2,7,1 (repeatedly) which only contains the remainders 0,1,2,3,5 and 7, i.e., 6 remainders, so a(8)=6.
a(11)=7 since Fibonacci numbers reduced modulo 11 are {0, 1, 2, 3, 5, 8, 10}.

T. D. Noe, Table of n, a(n) for n = 1..10000
Casey Mongoven, Sonification of multiple Fibonacci-related sequences, Annales Mathematicae et Informaticae, 41 (2013) pp. 175-192.

Cf. A001175, A079002.

a066853 1 = 1
a066853 n = f 1 ps [] where
   f 0 (1 : xs) ys = length ys
   f _ (x : xs) ys = if x `elem` ys then f x xs ys else f x xs (x:ys)
   ps = 1 : 1 : zipWith (\u v -> (u + v) `mod` n) (tail ps) ps
-- Reinhard Zumkeller, Jan 16 2014

a[n_] := Module[{v = {1, 2}}, If[n<8, n, While[v[[-1]] != 1 || v[[-2]] != 0, AppendTo[v, Mod[v[[-1]] + v[[-2]], n]]]; v // Union // Length]]; Array[a, 100] (* Jean-François Alcover, Feb 15 2018, after Charles R Greathouse IV *)

a(n)=if(n<8, return(n)); my(v=List([1,2])); while(v[#v]!=1 || v[#v-1]!=0, listput(v, (v[#v]+v[#v-1])%n)); #Set(v) \\ Charles R Greathouse IV, Jun 19 2017

cp's OEIS Frontend

A010077 a(n) = sum of digits of a(n-1) + sum of digits of a(n-2); a(0) = 0, a(1) = 1.

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A106535 Numbers k such that the smallest x > 1 for which Fibonacci(x) == 0 mod k is x = k - 1.

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A248218 Period in residues modulo n in iteration of x^2 + 1 starting at 0.

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A015134 Consider Fibonacci-type sequences b(0)=X, b(1)=Y, b(k)=b(k-1)+b(k-2) mod n; all are periodic; sequence gives number of distinct periods.

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A060305 Pisano periods for primes: period of Fibonacci numbers mod prime(n).

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A131294 a(n)=ds_3(a(n-1))+ds_3(a(n-2)), a(0)=0, a(1)=1; where ds_3=digital sum base 3.

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A131295 a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4.

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