A001177 -id:A001177 - cp's OEIS frontend

A053032 Odd primes p with one zero in Fibonacci numbers mod p.

11, 19, 29, 31, 59, 71, 79, 101, 131, 139, 151, 179, 181, 191, 199, 211, 229, 239, 251, 271, 311, 331, 349, 359, 379, 419, 431, 439, 461, 479, 491, 499, 509, 521, 541, 571, 599, 619, 631, 659, 691, 709, 719, 739, 751, 809, 811, 839, 859, 911, 919, 941, 971

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Also, odd primes that divide Lucas numbers of odd index. - T. D. Noe, Jul 25 2003
From Charles R Greathouse IV, Dec 14 2016: (Start)
It seems that this sequence contains about 1/3 of the primes. In particular, members of this sequence constitute:
        35 of the first 10^2 primes
       330 of the first 10^3 primes
      3328 of the first 10^4 primes
     33371 of the first 10^5 primes
    333329 of the first 10^6 primes
   3333720 of the first 10^7 primes
  33333463 of the first 10^8 primes
etc. (End)
Of the Fibonacci-like sequences modulo a prime p that are not A000004, one of them has a period length less than A001175(p) if and only if p = 5 or p is in this sequence. - Isaac Saffold, Dec 18 2018
Odd primes in A053031. - Jianing Song, Jun 19 2019

			From _Michael B. Porter_, Jan 25 2019: (Start)
The Fibonacci numbers (mod 7) repeat the pattern 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1. Since there are two zeros, 7 is not in the sequence.
The Fibonacci numbers (mod 11) repeat the pattern 0, 1, 1, 2, 3, 5, 8, 2, 10, 1 which has only one zero, so 11 is in the sequence.
(End)

T. D. Noe, Table of n, a(n) for n = 1..1000
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B1).
Nicholas Bragman and Eric Rowland, Limiting density of the Fibonacci sequence modulo powers of p, arXiv:2202.00704 [math.NT], 2022.
M. Renault, Fibonacci sequence modulo m

Cf. A001175, A001177. See A112860 for another version.
Cf. A000204 (Lucas numbers), A001602 (index of the smallest Fibonacci number divisible by prime(n)).
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |    m=1     |   m=2   |   m=3
-----------------------------+------------+---------+---------
The sequence {x(n)}          | A000045    | A000129 | A006190
The sequence {w(k)}          | A001176    | A214027 | A322906
Primes p such that w(p) = 1  | A112860*   | A309580 | A309586
Primes p such that w(p) = 2  | A053027**  | A309581 | A309587
Primes p such that w(p) = 4  | A053028*** | A261580 | A309588
Numbers k such that w(k) = 1 | A053031    | A309583 | A309591
Numbers k such that w(k) = 2 | A053030    | A309584 | A309592
Numbers k such that w(k) = 4 | A053029    | A309585 | A309593
* and also this sequence U {2}
** also primes dividing Lucas numbers of even index
*** also primes dividing no Lucas number

Prime@ Rest@ Position[Table[Count[Drop[NestWhile[Append[#, Mod[Total@ Take[#, -2], n]] &, {1, 1}, If[Length@ # < 3, True, Take[#, -2] != {1, 1}] &], -2], 0], {n, Prime@ Range@ 168}], 1][[All, 1]] (* Michael De Vlieger, Aug 08 2018 *)

fibmod(n,m)=(Mod([1, 1; 1, 0], m)^n)[1, 2]
is(n)=my(k=n+[0, -1, 1, 1, -1][n%5+1]); k>>=valuation(k,2)-1; fibmod(k,n)==0 && fibmod(k/2,n) && isprime(n) \\ Charles R Greathouse IV, Dec 14 2016

A prime p = prime(i) is in this sequence if p > 2 and A001602(i)/2 is odd. - T. D. Noe, Jul 25 2003

A053027 Odd primes p with 2 zeros in Fibonacci numbers mod p.

Original entry on oeis.org

3, 7, 23, 41, 43, 47, 67, 83, 103, 107, 127, 163, 167, 223, 227, 241, 263, 281, 283, 307, 347, 367, 383, 401, 409, 443, 449, 463, 467, 487, 503, 523, 547, 563, 569, 587, 601, 607, 641, 643, 647, 683, 727, 743, 769, 787, 823, 827, 863, 881, 883, 887, 907, 929

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Also, odd primes that divide Lucas numbers of even index. - T. D. Noe, Jul 25 2003
Primes in A053030. - Jianing Song, Jun 19 2019
From Jianing Song, Jun 16 2024: (Start)
Primes p such that A001176(p) = 2.
For p > 2, p is in this sequence if and only if 8 divides of A001175(p), and if and only if 4 divides A001177(p). For a proof of the equivalence between A001176(p) = 2 and 4 dividing A001177(p), see Section 2 of my link below.
This sequence contains all primes congruent to 3, 7 (mod 20). This corresponds to case (2) for k = 3 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. (End) [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

T. D. Noe, Table of n, a(n) for n=1..1000
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B3).
Nicholas Bragman and Eric Rowland, Limiting density of the Fibonacci sequence modulo powers of p, arXiv:2202.00704 [math.NT], 2022.
M. Renault, Fibonacci sequence modulo m
Jianing Song, Lucas sequences and entry point modulo p

Cf. A001175, A001177.
Cf. A000204 (Lucas numbers), A001602 (index of the smallest Fibonacci number divisible by prime(n)).
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |    m=1    |   m=2   |   m=3
-----------------------------+-----------+---------+---------
The sequence {x(n)}          | A000045   | A000129 | A006190
The sequence {w(k)}          | A001176   | A214027 | A322906
Primes p such that w(p) = 1  | A112860*  | A309580 | A309586
Primes p such that w(p) = 2  | this seq  | A309581 | A309587
Primes p such that w(p) = 4  | A053028** | A261580 | A309588
Numbers k such that w(k) = 1 | A053031   | A309583 | A309591
Numbers k such that w(k) = 2 | A053030   | A309584 | A309592
Numbers k such that w(k) = 4 | A053029   | A309585 | A309593
* and also A053032 (primes dividing Lucas numbers of odd index) U {2}
** also primes dividing no Lucas number

A prime p = prime(i) is in this sequence if p > 2 and A001602(i)/2 is even. - T. D. Noe, Jul 25 2003

A214027 The number of zeros in the fundamental Pisano period of the sequence A000129 mod n.

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 1, 1, 2, 2, 2, 2, 4, 1, 2, 1, 2, 2, 2, 1, 2, 2, 1, 1, 4, 2, 2, 1, 4, 2, 1, 1, 2, 2, 2, 2, 4, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 4, 2, 2, 1, 2, 2, 2, 2, 4, 1, 2, 1, 4, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 2, 2, 2, 2, 2, 1

Offset: 1

Art DuPre, Jul 04 2012

nonn

This is intimately connected with A175181 and A214028, much as A001176 is intimately connected with A001175 and A001177. In fact, A175181(n)/a(n) = A214028(n). This is the same divisibility relation that holds between A001175, A001176 and A001177.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A175181, A214028.

Similar sequences: A001176, A322906.

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

A000129(m) = ([2, 1; 1, 0]^m)[2, 1]
a(n) = my(i=1); while(A000129(i)%n!=0, i++); znorder(Mod(A000129(i+1), n)) \\ Jianing Song, Aug 10 2019

From Jianing Song, Sep 12 2018: (Start)
For odd primes p, a(p^e) = 4 if A214028(p) is odd; 1 if A214028(p) is even but not divisible by 4; 2 if A214028(p) is divisible by 4.
a(n) = 2 for n == 3 (mod 8). For primes p, a(p^e) = 1 if p == 7 (mod 8), 4 if p == 5 (mod 8). Conjecture: 1/6 of the primes congruent to 1 mod 8 satisfy a(p^e) = 1, 2/3 of them satisfy a(p^e) = 2 and 1/6 of them satisfy a(p^e) = 4.
(End)

A213648 The minimum number of 1's in the relation n*[n,1,1,...,1,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

2, 3, 5, 4, 11, 7, 5, 11, 14, 9, 11, 6, 23, 19, 11, 8, 11, 17, 29, 7, 29, 23, 11, 24, 20, 35, 23, 13, 59, 29, 23, 19, 8, 39, 11, 18, 17, 27, 29, 19, 23, 43, 29, 59, 23, 15, 11, 55, 74, 35, 41, 26, 35, 9, 23, 35, 41, 57, 59, 14, 29, 23, 47, 34, 59, 67

Offset: 2

Art DuPre, Jun 17 2012

nonn

Multiplying n by a simple continued fraction with an increasing number of 1's sandwiched between n generates fractions that have a leading term x in their continued fraction, where x is obviously > n^2. We increase the number of 1's until the first and the last term in the simple terminating continued fraction of n*[n,1,...,1,n] =[x,...,x] is the same, x, and set a(n) to the count of these 1's.
Conjecture: the fixed points of this sequence are in A000057.
We have [n,1,1,...,1,n] = n + (n*Fib(m)+Fib(m-1))/(n*Fib(m+1)+Fib(m)) and n*[n,1,1,...,1,n] = n^2 + 1 + (n^2-n-1)*Fib(m)/(n*Fib(m+1)+Fib(m)), where m is the number of 1's. - Max Alekseyev, Aug 09 2012
The analog sequence with 11 instead of 1, A213900, seems to have the same fixed points, while other variants (A262212 - A262220, A262211) have other fixed points (A213891 - A213899,  A261311). - M. F. Hasler, Sep 15 2015

			3* [3,1,1,1,3] = [10,1,10],so a(3)=3
4* [4,1,1,1,1,1,4] = [18,2,18],so a(4)=5
5* [5,1,1,1,1,5] = [28,28],so a(5)=4
6* [6,1,1,1,1,1,1,1,1,1,1,1,6] = [39,1,2,2,2,1,39], so a(6)=11
7* [7,1,1,1,1,1,1,1,7] = [53,3,53], so a(7)=7

A. Hurwitz, Über die Kettenbrüche, deren Teilnenner arithmetische Reihen bilden, Vierteljahrsschrift der Naturforschenden Gesellschaft in Zürich, Jahrg XLI, 1896, Jubelband II, S. 34-64.

Bill Gosper, Appendix 2 Continued Fraction Arithmetic

Cf. A000057, A262212 - A262220.

A213648 := proc(n)
        local h,ins,c ;
        for ins from 1 do
                c := [n,seq(1,i=1..ins),n] ;
                h := numtheory[cfrac](n*simpcf(c),quotients) ;
                if op(1,h) = op(-1,h) then
                        return ins;
                end if;
        end do:
end proc: # R. J. Mathar, Jul 06 2012

f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; f[1, #] & /@ Range[2, 67] (* Michael De Vlieger, Sep 16 2015 *)

{a(n) = local(t, m=1); if( n<2, 0, while( t = contfracpnqn( concat( [n, vector(m, i, 1 ), n])), t = contfrac( n * t[1, 1] / t[2, 1]); if( t[1] < n^2 || t[#t] < n^2, m++, break)); m)} /* Michael Somos, Jun 17 2012 */

{a(n) = local(t, m=0); if( n<2, 0, until(t[1]==t[#t], m++; t = contfrac(n^2 + 1 + (n^2-n-1)*fibonacci(m)/(n*fibonacci(m+1)+fibonacci(m))); ); m )} /* Max Alekseyev, Aug 09 2012 */

Conjecture: a(n)=A001177(n)-1.

A322906 The number of zeros in the fundamental Pisano period of the 3-Fibonacci numbers A006190 modulo n.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 2, 2, 1, 4, 2, 1, 4, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 4, 4, 1, 2, 4, 2, 2, 2, 2, 2, 2, 1, 4, 2, 2, 2, 4, 2, 1, 2, 2, 1, 2, 2, 2, 4, 2, 2, 1, 1, 2, 2, 2, 4, 2, 2, 1, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 4, 4, 2, 2, 2, 2, 1, 2, 1, 4, 2, 2, 2, 1, 2

Offset: 1

Jianing Song, Jan 05 2019

nonn

a(n) is the multiplicative order of A006190(A322907(n)+1) modulo n.
a(n) has value 1, 2 or 4. This is because A006190(k,m+1)^4 == 1 (mod A006190(k,m)).
Conjecture: For primes p == 1, 9, 17, 25, 49, 81 (mod 104), the probability of a(p^e) taking on the value 1, 2, 4 is 1/6, 2/3, 1/6, respectively; for primes p == 29, 53, 61, 69, 77, 101 (mod 104), the probability of a(p^e) taking on the value 1, 4 is 1/2, 1/2, respectively.

Jianing Song, Table of n, a(n) for n = 1..1000

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = k*x(n+1) + x(n). Then the periods, ranks and the ratios of the periods to the ranks modulo a given integer n are given by:
k = 1: A001175 (periods), A001177 (ranks), A001176 (ratios).
k = 2: A175181 (periods), A214028 (ranks), A214027 (ratios).
k = 3: A175182 (periods), A322907 (ranks), this sequence (ratios).

A006190(m) = ([3, 1; 1, 0]^m)[2, 1]
a(n) = my(i=1); while(A006190(i)%n!=0, i++); znorder(Mod(A006190(i+1), n))

For n > 2, T(n,k) = 4 iff A322907(n) is odd; 1 iff A322907(n) is even but not divisible by 4; 2 iff A322907(n) is divisible by 4.
For primes p == 3, 23, 27, 35, 43, 51 (mod 52), a(p^e) = 1.
For primes p == 5, 21, 33, 37, 41, 45 (mod 52), a(p^e) = 4.
For primes p == 7, 11, 15, 19, 31, 47 (mod 52), a(p^e) = 2.
a(13^e) = 4. a(2^e) = 1 if e = 1, 2 and 2 if e >= 3.
a(n) = A175182(n)/A322907(n).

A053031 Numbers with 1 zero in Fibonacci numbers mod m.

Original entry on oeis.org

1, 2, 4, 11, 19, 22, 29, 31, 38, 44, 58, 59, 62, 71, 76, 79, 101, 116, 118, 121, 124, 131, 139, 142, 151, 158, 179, 181, 191, 199, 202, 209, 211, 229, 236, 239, 242, 251, 262, 271, 278, 284, 302, 311, 316, 319, 331, 341, 349, 358, 359, 361, 362, 379, 382, 398

Offset: 1

Henry Bottomley, Feb 23 2000

nonn

Conjecture: m is on this list iff m is an odd number all of whose factors are on this list or m is 2 or 4 times such an odd number.
A001176(a(n)) = A128924(a(n),1) = 1. - Reinhard Zumkeller, Jan 16 2014
Also numbers n such that A001175(n) = A001177(n). - Daniel Suteu, Aug 08 2018

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Brennan Benfield and Michelle Manes, The Fibonacci Sequence is Normal Base 10, arXiv:2202.08986 [math.NT], 2022.
Brennan Benfield and Oliver Lippard, Connecting Zeros in Pisano Periods to Prime Factors of K-Fibonacci Numbers, arXiv:2407.20048 [math.NT], 2024.
M. Renault, Fibonacci sequence modulo m

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1    |   m=2   |   m=3
-----------------------------+----------+---------+---------
The sequence {x(n)}          | A000045  | A000129 | A006190
The sequence {w(k)}          | A001176  | A214027 | A322906
Primes p such that w(p) = 1  | A112860* | A309580 | A309586
Primes p such that w(p) = 2  | A053027  | A309581 | A309587
Primes p such that w(p) = 4  | A053028  | A261580 | A309588
Numbers k such that w(k) = 1 | this seq | A309583 | A309591
Numbers k such that w(k) = 2 | A053030  | A309584 | A309592
Numbers k such that w(k) = 4 | A053029  | A309585 | A309593
* and also A053032 U {2}

a053031 n = a053031_list !! (n-1)
a053031_list = filter ((== 1) . a001176) [1..]
-- Reinhard Zumkeller, Jan 16 2014

With[{s = {1}~Join~Table[Count[Drop[NestWhile[Append[#, Mod[Total@ Take[#, -2], n]] &, {1, 1}, If[Length@ # < 3, True, Take[#, -2] != {1, 1}] &], -2], 0], {n, 2, 400}]}, Position[s, 1][[All, 1]] ] (* Michael De Vlieger, Aug 08 2018 *)

entryp(p)=my(k=p+[0, -1, 1, 1, -1][p%5+1], f=factor(k)); for(i=1, #f[, 1],for(j=1, f[i, 2], if((Mod([1, 1; 1, 0], p)^(k/f[i, 1]))[1, 2], break); k/=f[i, 1])); k
entry(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e14, entryp(f[i, 1]^f[i, 2]), entryp(f[i, 1])*f[i, 1]^(f[i, 2]-1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<Charles R Greathouse IV, Dec 14 2016

A112860 2 together with A053032.

Original entry on oeis.org

2, 11, 19, 29, 31, 59, 71, 79, 101, 131, 139, 151, 179, 181, 191, 199, 211, 229, 239, 251, 271, 311, 331, 349, 359, 379, 419, 431, 439, 461, 479, 491, 499, 509, 521, 541, 571, 599, 619, 631, 659, 691, 709, 719, 739, 751, 809, 811, 839, 859, 911, 919, 941, 971

Offset: 1

N. J. A. Sloane, Nov 30 2007

nonn

Consists of the primes that are in neither A053027 nor A053028.
From Jianing Song, Jun 16 2024: (Start)
Primes p such that A001176(p) = 1.
For p > 2, p is in this sequence if and only if A001175(p) == 2 (mod 4), and if and only if A001177(p) == 2 (mod 4). For a proof of the equivalence between A001176(p) = 1 and A001177(p) == 2 (mod 4), see Section 2 of my link below.
This sequence contains all primes congruent to 11, 19 (mod 20). This corresponds to case (3) for k = 3 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. (End) [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]

T. D. Noe, Table of n, a(n) for n=1..1001
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B1).
Jianing Song, Lucas sequences and entry point modulo p

Cf. A001175, A001177.
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
                             |   m=1     |   m=2   |   m=3
-----------------------------+-----------+---------+---------
The sequence {x(n)}          | A000045   | A000129 | A006190
The sequence {w(k)}          | A001176   | A214027 | A322906
Primes p such that w(p) = 1  | this seq* | A309580 | A309586
Primes p such that w(p) = 2  | A053027   | A309581 | A309587
Primes p such that w(p) = 4  | A053028   | A261580 | A309588
Numbers k such that w(k) = 1 | A053031   | A309583 | A309591
Numbers k such that w(k) = 2 | A053030   | A309584 | A309592
Numbers k such that w(k) = 4 | A053029   | A309585 | A309593
* and also A053032 U {2}

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

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

A104714 Greatest common divisor of a Fibonacci number and its index.

Original entry on oeis.org

0, 1, 1, 1, 1, 5, 2, 1, 1, 1, 5, 1, 12, 1, 1, 5, 1, 1, 2, 1, 5, 1, 1, 1, 24, 25, 1, 1, 1, 1, 10, 1, 1, 1, 1, 5, 36, 1, 1, 1, 5, 1, 2, 1, 1, 5, 1, 1, 48, 1, 25, 1, 1, 1, 2, 5, 7, 1, 1, 1, 60, 1, 1, 1, 1, 5, 2, 1, 1, 1, 5, 1, 72, 1, 1, 25, 1, 1, 2, 1, 5, 1, 1, 1, 12, 5, 1, 1, 1, 1, 10, 13, 1, 1, 1, 5, 96, 1

Offset: 0

Harmel Nestra (harmel.nestra(AT)ut.ee), Apr 23 2005

Considering this sequence is a natural sequel to the investigation of the problem when F_n is divisible by n (the numbers occurring in A023172). This sequence has several nice properties. (1) n | m implies a(n) | a(m) for arbitrary naturals n and m. This property is a direct consequence of the analogous well-known property of Fibonacci numbers. (2) gcd (a(n), a(m)) = a(gcd(n, m)) for arbitrary naturals n and m. Also this property follows directly from the analogous (perhaps not so well-known) property of Fibonacci numbers. (3) a(n) * a(m) | a(n * m) for arbitrary naturals n and m. This property is remarkable especially in the light that the analogous proposition for Fibonacci numbers fails if n and m are not relatively prime (e.g. F_3 * F_3 does not divide F_9). (4) The set of numbers satisfying a(n) = n is closed w.r.t. multiplication. This follows easily from (3).

			The natural numbers:    0 1 2 3 4 5 6  7  8  9 10 11  12 ...
The Fibonacci numbers:  0 1 1 2 3 5 8 13 21 34 55 89 144 ...
The corresponding GCDs: 0 1 1 1 1 5 2  1  1  1  5  1  12 ...

Alois P. Heinz, Table of n, a(n) for n = 0..20000 (first 1001 terms from T. D. Noe)
Paolo Leonetti, Carlo Sanna, On the greatest common divisor of n and the nth Fibonacci number, arXiv:1704.00151 [math.NT], 2017.
Carlo Sanna, Emanuele Tron, The density of numbers n having a prescribed G.C.D. with the nth Fibonacci number, arXiv:1705.01805 [math.NT], 2017.

Cf. A023172, A000045, A001177, A001175, A001176. a(n) = gcd(A000045(n), A001477(n)). a(n) = n iff n occurs in A023172 iff n | A000045(n).

Cf. A074215 (a(n)==1).

let fibs@(_ : fs) = 0 : 1 : zipWith (+) fibs fs in 0 : zipWith gcd [1 ..] fs

b:= proc(n) option remember; local r, M, p; r, M, p:=
      <<1|0>, <0|1>>, <<0|1>, <1|1>>, n;
      do if irem(p, 2, 'p')=1 then r:= r.M mod n fi;
         if p=0 then break fi; M:= M.M mod n
      od; r[1, 2]
    end:
a:= n-> igcd(n, b(n)):
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 05 2017

Table[GCD[Fibonacci[n],n],{n,0,97}] (* Alonso del Arte, Nov 22 2010 *)

a(n)=if(n,gcd(n,lift(Mod([1,1;1,0],n)^n)[1,2]),0) \\ Charles R Greathouse IV, Sep 24 2013

a(n) = gcd(F(n), n).

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