cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 13 results. Next

A366951 a(n) = 2*(p_n - 1)/A060305(n) iff p_n == +/- 1 (mod 5), 2*(p_n + 1)/A060305(n) iff p_n == +/- 2 (mod 5), 0 iff p_n = 5.

Original entry on oeis.org

2, 1, 0, 1, 2, 1, 1, 2, 1, 4, 2, 1, 2, 1, 3, 1, 2, 2, 1, 2, 1, 2, 1, 4, 1, 4, 1, 3, 2, 3, 1, 2, 1, 6, 2, 6, 1, 1, 1, 1, 2, 4, 2, 1, 1, 18, 10, 1, 1, 4, 9, 2, 2, 2, 1, 3, 2, 2, 1, 10, 1, 1, 7, 2, 1, 1, 6, 1, 3, 4, 3, 2, 1, 1, 2, 1, 2, 1, 4, 2, 2, 10, 2, 1, 2, 1
Offset: 1

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Author

A.H.M. Smeets, Oct 29 2023

Keywords

Crossrefs

Formula

a(n) == 0 (mod 2) for prime(n) == +/- 1 (mod 5) and n > 2.
a(n) == 1 (mod 2) for Prime(n) == +/- 2 (mod 5) and n > 2.
a(n) = 1 iff prime(n) in A071774.
a(n) = 2 iff prime(n) in ({2} union A003147)/{5}.
a(n) = 3 iff prime(n) in A308784.
a(n) = 4 iff prime(n) in A308787.
a(n) = 6 iff prime(n) in A308788.
a(n) = 7 iff prime(n) in A308785.
a(n) = 8 iff prime(n) in A308789.
a(n) = 9 iff prime(n) in A308786.
a(n) = 10 iff prime(n) in A308790.
a(n) = 12 iff prime(n) in A308791.
a(n) = 14 iff prime(n) in A308792.
a(n) = 16 iff prime(n) in A308793.
a(n) = 18 iff prime(n) in A308794.
a(n) = A296240(n) iff prime(n) == +/- 2 (mod 5) and n > 3.
a(n) = 2*A296240(n) iff prime(n) == +/- 1 (mod 5) and n > 3.
a(n) in {2^k: k > 1} iff prime(n) in {A047650}.
a(n) == 3 (mod 6) iff prime(n) in {A124096}.
a(n) == 6 (mod 12) iff prime(n) in {A046652}.
a(n) == 0 (mod 14) iff prime(n) in {A125252}.

A001175 Pisano periods (or Pisano numbers): period of Fibonacci numbers mod n.

Original entry on oeis.org

1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, 16, 30, 48, 24, 100, 84, 72, 48, 14, 120, 30, 48, 40, 36, 80, 24, 76, 18, 56, 60, 40, 48, 88, 30, 120, 48, 32, 24, 112, 300, 72, 84, 108, 72, 20, 48, 72, 42, 58, 120, 60, 30, 48, 96, 140, 120, 136
Offset: 1

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Author

Keywords

Comments

These numbers might also have been called Fibonacci periods.
Also, number of perfect multi-Skolem type sequences of order n.
Index the Fibonacci numbers so that 3 is the fourth number. If the modulo base is a Fibonacci number (>=3) with an even index, the period is twice the index. If the base is a Fibonacci number (>=5) with an odd index, the period is 4 times the index. - Kerry Mitchell, Dec 11 2005
Each row of the image represents a different modulo base n, from 1 at the bottom to 24 at the top. The columns represent the Fibonacci numbers mod n, from 0 mod n at the left to 59 mod n on the right. In each cell, the brightness indicates the value of the residual, from dark for 0 to near-white for n-1. Blue squares on the left represent the first period; the number of blue squares is the Pisano number. - Kerry Mitchell, Feb 02 2013
a(n) = least positive integer k such that F(k) == 0 (mod n) and F(k+1) == 1 (mod n), where F = A000045 is the Fibonacci sequence. a(n) exists for all n by Dirichlet's box principle and the fact that the positive integers are well-ordered. Cf. Saha and Karthik (2011). - L. Edson Jeffery, Feb 12 2014

Examples

			For n=4, take the Fibonacci sequence (A000045), 1, 1, 2, 3, 5, 8, 13, 21, ... (mod 4), which gives 1, 1, 2, 3, 1, 0, 1, 1, .... This repeats a pattern of length 6, so a(4) = 6. - _Michael B. Porter_, Jul 19 2016
		

References

  • B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers. Report ORNL-4261, Oak Ridge National Laboratory, Oak Ridge, Tennessee, Jun 1968.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 162.
  • J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, pp. 304 - 309.
  • 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).
  • S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989. See p. 89. - N. J. A. Sloane, Feb 20 2013

Crossrefs

Cf. A060305 (Fibonacci period mod prime(n)), A003893.
Cf. A001178 (Fibonacci frequency), A001179 (Leonardo logarithm), A235702 (fixed points), A066853 (number of elements in the set of residua), A222413, A296240 (Pisano quotients for primes).

Programs

  • Haskell
    a001175 1 = 1
    a001175 n = f 1 ps 0 where
       f 0 (1 : xs) pi = pi
       f _ (x : xs) pi = f x xs (pi + 1)
       ps = 1 : 1 : zipWith (\u v -> (u + v) `mod` n) (tail ps) ps
    -- Reinhard Zumkeller, Jan 15 2014
    
  • Maple
    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, 1];
             for k do f:=[f[2], f[1]+f[2] mod n];
                      if f=[0, 1] then break fi
             od; k
          fi
        end:
    seq(a(n), n=1..100);  # Alois P. Heinz, Sep 18 2013
  • Mathematica
    Table[a={1, 0}; a0=a; k=0; While[k++; s=Mod[Plus@@a, n]; a=RotateLeft[a]; a[[2]]=s; a!=a0]; k, {n, 2, 100}] (* T. D. Noe, Jul 19 2005 *)
    a[1]=1; a[n_]:= For[k = 1, True, k++, If[Mod[Fibonacci[k], n]==0 && Mod[ Fibonacci[k+1], n]==1, Return[k]]]; Table[a[n], {n, 100}] (* Jean-François Alcover, Feb 11 2015 *)
    test[{0, 1, }]:= False; test[]:= True;
    nest[k_][{a_, b_, c_}]:= {Mod[b, k], Mod[a+b, k], c+1};
    A001175[1] := 1;
    A001175[k_]:= NestWhile[nest[k], {1, 1, 1}, test][[3]];
    Table[A001175[n], {n, 100}] (* Leo C. Stein, Nov 08 2019 *)
    Module[{nn=1000,fibs},fibs=Fibonacci[Range[nn]];Table[Length[ FindTransientRepeat[ Mod[fibs,n],2][[2]]],{n,70}]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Nov 02 2020 *)
  • PARI
    minWSS=2^64; \\ PrimeGrid search
    fibmod(n,m)=((Mod([1,1;1,0],m))^n)[1,2]
    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]>=minWSS, 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, Feb 13 2014; updated Dec 14 2016; updated Aug 24 2021; updated Jul 08 2024
    
  • Python
    from functools import reduce
    from sympy import factorint, lcm
    def A001175(n):
        if n == 1:
            return 1
        f = factorint(n)
        if len(f) > 1:
            return reduce(lcm, (A001175(a**f[a]) for a in f))
        else:
            k,x = 1, [1,1]
            while x != [0,1]:
                k += 1
                x = [x[1], (x[0]+x[1]) % n]
            return k # Chai Wah Wu, Jul 17 2019
  • Sage
    def a(n): return BinaryRecurrenceSequence(1,1).period(n) # Ralf Stephan, Jan 23 2014
    

Formula

Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)). - T. D. Noe, May 02 2005
a(n) = n-1 if n is a prime > 5 included in A003147 ( n = 11, 19, 31, 41, 59, 61, 71, 79, 109...). - Benoit Cloitre, Jun 04 2002
K. S. Brown shows that a(n)/n <= 6 for all n and a(n)=6n if and only if n has the form 2*5^k.
a(n) = A001177(n)*A001176(n) for n >= 1. - Henry Bottomley, Dec 19 2001
a(n) <= 2*n+2 if n is a prime > 5, by Wall's Theorems 6 and 7; see A060305, A222413, A296240. - Jonathan Sondow, Dec 10 2017
a(2^k) = 3*2^(k-1) for k > 0. In general, if a(p) != a(p^2) for p prime, then a(p^k) = p^(k-1)*a(p) [Wall, 1960]. - Chai Wah Wu, Feb 25 2022

A001177 Fibonacci entry points: a(n) = least k >= 1 such that n divides Fibonacci number F_k (=A000045(k)).

Original entry on oeis.org

1, 3, 4, 6, 5, 12, 8, 6, 12, 15, 10, 12, 7, 24, 20, 12, 9, 12, 18, 30, 8, 30, 24, 12, 25, 21, 36, 24, 14, 60, 30, 24, 20, 9, 40, 12, 19, 18, 28, 30, 20, 24, 44, 30, 60, 24, 16, 12, 56, 75, 36, 42, 27, 36, 10, 24, 36, 42, 58, 60, 15, 30, 24, 48, 35, 60, 68, 18, 24, 120
Offset: 1

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Author

Keywords

Comments

In the formula, the relation a(p^e) = p^(e-1)*a(p) is called Wall's conjecture, which has been verified for primes up to 10^14. See A060305. Primes for which this relation fails are called Wall-Sun-Sun primes. - T. D. Noe, Mar 03 2009
All solutions to F_m == 0 (mod n) are given by m == 0 (mod a(n)). For a proof see, e.g., Vajda, p. 73. [Old comment changed by Wolfdieter Lang, Jan 19 2015]
If p is a prime of the form 10n +- 1 then a(p) is a divisor of p-1. If q is a prime of the form 10n +- 3 then a(q) is a divisor of q+1. - Robert G. Wilson v, Jul 07 2007
Definition 1 in Riasat (2011) calls this k(n), or sometimes just k. Corollary 1 in the same paper, "every positive integer divides infinitely many Fibonacci numbers," demonstrates that this sequence is infinite. - Alonso del Arte, Jul 27 2013
If p is a prime then a(p) <= p+1. This is because if p is a prime then exactly one of the following Fibonacci numbers is a multiple of p: F(p-1), F(p) or F(p+1). - Dmitry Kamenetsky, Jul 23 2015
From Renault 1996:
1. a(lcm(n,m)) = lcm(a(n), a(m)).
2. if n|m then a(n)|a(m).
3. if m has prime factorization m=p1^e1 * p2^e2 * ... * pn^en then a(m) = lcm(a(p1^e1), a(p2^e2), ..., a(pn^en)). - Dmitry Kamenetsky, Jul 23 2015
a(n)=n if and only if n=5^k or n=12*5^k for some k >= 0 (see Marques 2012). - Dmitry Kamenetsky, Aug 08 2015
Every positive integer (except 2) eventually appears in this sequence. This is because every Fibonacci number bigger than 1 (except Fibonacci(6)=8 and Fibonacci(12)=144) has at least one prime factor that is not a factor of any earlier Fibonacci number (see Knott reference). Let f(n) be such a prime factor for Fibonacci(n); then a(f(n))=n. - Dmitry Kamenetsky, Aug 08 2015
We can reconstruct the Fibonacci numbers from this sequence using the formula Fibonacci(n+2) = 1 + Sum_{i: a(i) <= n} phi(i)*floor(n/a(i)), where phi(n) is Euler's totient function A000010 (see the Stroinski link). For example F(6) = 1 + phi(1)*floor(4/a(1)) + phi(2)*floor(4/a(2)) + phi(3)*floor(4/a(4)) = 1 + 1*4 + 1*1 + 2*1 = 8. - Peter Bala, Sep 10 2015
Conjecture: Sum_{d|n} phi(d)*a(d) = A232656(n). - Logan J. Kleinwaks, Oct 28 2017
a(F_m) = m for all m > 1. Indeed, let (b(j)) be defined by b(1)=b(2)=1, and b(j+2) = (b(j) + b(j+1)) mod n. Then a(n) equals the index of the first occurrence of 0 in (b(j)). Example: if n=4 then b = A079343 = 1,1,2,3,1,0,1,1,..., so a(4)=6. If n is a Fibonacci number n=F_m, then obviously a(n)=m. Note that this gives a simple proof of the fact that all integers larger than 2 occur in (a(n)). - Michel Dekking, Nov 10 2017

Examples

			a(4) = 6 because the smallest Fibonacci number that 4 divides is F(6) = 8.
a(5) = 5 because the smallest Fibonacci number that 5 divides is F(5) = 5.
a(6) = 12 because the smallest Fibonacci number that 6 divides is F(12) = 144.
From _Wolfdieter Lang_, Jan 19 2015: (Start)
a(2) = 3, hence 2 | F(m) iff m = 2*k, for k >= 0;
a(3) = 4, hence 3 | F(m) iff m = 4*k, for k >= 0;
etc. See a comment above with the Vajda reference.
(End)
		

References

  • A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 25.
  • B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers. Report ORNL-4261, Oak Ridge National Laboratory, Oak Ridge, Tennessee, June 1968.
  • Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Afterword by Herbert A. Hauptman, Nobel Laureate, 2. 'The Minor Modulus m(n)', Prometheus Books, NY, 2007, page 329-342.
  • 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).
  • S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
  • N. N. Vorob'ev, Fibonacci numbers, Blaisdell, NY, 1961.

Crossrefs

Cf. A000045, A001175, A001176, A060383, A001602. First occurrence of k is given in A131401. A233281 gives such k that a(k) is a prime.
From Antti Karttunen, Dec 21 2013: (Start)
Various derived sequences:
A047930(n) = A000045(a(n)).
A037943(n) = A000045(a(n))/n.
A217036(n) = A000045(a(n)-1) mod n.
A132632(n) = a(n^2).
A132633(n) = a(n^3).
A214528(n) = a(n!).
A215011(n) = a(A000217(n)).
A215453(n) = a(n^n).
Analogous sequence for the tribonacci numbers: A046737, for Lucas numbers: A223486, for Pell numbers: A214028.

Programs

  • Haskell
    a001177 n = head [k | k <- [1..], a000045 k `mod` n == 0]
    -- Reinhard Zumkeller, Jan 15 2014
  • Maple
    A001177 := proc(n)
            for k from 1 do
                    if combinat[fibonacci](k) mod n = 0 then
                            return k;
                    end if;
            end do:
    end proc: # R. J. Mathar, Jul 09 2012
    N:= 1000: # to get a(1) to a(N)
    L:= ilcm($1..N):
    count:= 0:
    for n from 1 while count < N do
      fn:= igcd(L,combinat:-fibonacci(n));
      divs:= select(`<=`,numtheory:-divisors(fn),N);
      for d in divs do if not assigned(A[d]) then count:= count+1; A[d]:= n fi od:
    od:
    seq(A[n],n=1..N); # Robert Israel, Oct 14 2015
  • Mathematica
    fibEntry[n_] := Block[{k = 1}, While[ Mod[ Fibonacci@k, n] != 0, k++ ]; k]; Array[fibEntry, 74] (* Robert G. Wilson v, Jul 04 2007 *)
  • PARI
    a(n)=if(n<0,0,s=1;while(fibonacci(s)%n>0,s++);s) \\ Benoit Cloitre, Feb 10 2007
    
  • PARI
    ap(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
    a(n)=if(n==1,return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i,1]>1e14,ap(f[i,1]^f[i,2]), ap(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, May 08 2017
    
  • Scheme
    (define (A001177 n) (let loop ((k 1)) (cond ((zero? (modulo (A000045 k) n)) k) (else (loop (+ k 1)))))) ;; Antti Karttunen, Dec 21 2013
    

Formula

A001175(n) = A001176(n) * a(n) for n >= 1.
a(n) = n if and only if n is of form 5^k or 12*5^k (proved in Marques paper), a(n) = n - 1 if and only if n is in A106535, a(n) = n + 1 if and only if n is in A000057, a(n) = n + 5 if and only if n is in 5*A000057, ... - Benoit Cloitre, Feb 10 2007
a(1) = 1, a(2) = 3, a(4) = 6 and for e > 2, a(2^e) = 3*2^(e-2); a(5^e) = 5^e; and if p is an odd prime not 5, then a(p^e) = p^max(0, e-s)*a(p) where s = valuation(A000045(a(p)), p) (Wall's conjecture states that s = 1 for all p). If (m, n) = 1 then a(m*n) = lcm(a(m), a(n)). See Posamentier & Lahmann. - Robert G. Wilson v, Jul 07 2007; corrected by Max Alekseyev, Oct 19 2007, Jun 24 2011
Apparently a(n) = A213648(n) + 1 for n >= 2. - Art DuPre, Jul 01 2012
a(n) < n^2. [Vorob'ev]. - Zak Seidov, Jan 07 2016
a(n) < n^2 - 3n + 6. - Jinyuan Wang, Oct 13 2018
a(n) <= 2n [Salle]. - Jon Maiga, Apr 25 2019

Extensions

Definition corrected by Wolfdieter Lang, Jan 19 2015

A071774 Related to Pisano periods: integers k such that the period of Fibonacci numbers mod k equals 2*(k+1).

Original entry on oeis.org

3, 7, 13, 17, 23, 37, 43, 53, 67, 73, 83, 97, 103, 127, 137, 157, 163, 167, 173, 193, 197, 223, 227, 257, 277, 283, 293, 313, 317, 337, 367, 373, 383, 397, 433, 443, 457, 463, 467, 487, 503, 523, 547, 577, 587, 593, 607, 613, 617, 643, 647, 653, 673, 683, 727
Offset: 1

Views

Author

Benoit Cloitre, Jun 04 2002

Keywords

Comments

Terms are primes with final digit 3 or 7.
Apparently these are the primes given in A003631 without 2 and A216067. - Klaus Purath, Dec 11 2020
If k is a term, then for m=5*k the period of Fibonacci numbers mod m equals 2*(m+5). - Matthew Goers, Jan 13 2021

Crossrefs

Programs

  • Mathematica
    Select[Prime@ Range[129], Function[n, Mod[Last@ NestWhile[{Mod[#2, n], Mod[#1 + #2, n], #3 + 1} & @@ # &, {1, 1, 1}, #[[1 ;; 2]] != {0, 1} &], n] == Mod[2 (n + 1), n] ]] (* Michael De Vlieger, Mar 31 2021, after Leo C. Stein at A001175 *)
  • PARI
    for(n=2,5000,t=2*(n+1);good=1;if(fibonacci(t)%n==0, for(s=0,t,if(fibonacci(t+s)%n!=fibonacci(s)%n,good=0;break); if(s>1&&s
    				
  • PARI
    forprime(p=3,3000,if(p%5==2||p%5==3,a=1;b=0;c=1;while(a!=0||b!=1,c++;d=a;a=b;a=(a+d)%p;b=d%p);if(c==(2*(p+1)),print1(p",")))) /* V. Raman, Nov 22 2012 */

Extensions

More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Dec 21 2004

A296240 Pisano quotients: a(n) = (p-1)/k(p) if p == +- 1 mod 5, = (2*p+2)/k(p) if p == +- 2 mod 5, where p = prime(n) and k(p) = Pisano period(p).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 9, 5, 1, 1, 2, 9, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 7, 1, 1, 1, 3, 1, 3, 2, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 5, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 2, 20, 1, 6, 1, 9, 3, 1, 1, 1, 1, 1, 1
Offset: 4

Views

Author

Jonathan Sondow, Dec 09 2017

Keywords

Comments

Wall (1960) in Theorems 6 and 7 proved that a(n) is an integer for n >= 4. Jarden (1946) proved that the sequence is unbounded. See Elsenhans and Jahnel (2010), pp. 1-2.

Crossrefs

Programs

  • Mathematica
    With[{p = Prime[n]}, T = Table[a = {1, 0}; a0 = a; k = 0; While[k++; s = Mod[Plus @@ a, p]; a = RotateLeft[a]; a[[2]] = s; a != a0]; k, {n, 1, 130}]; Table[L = KroneckerSymbol[p, 5]; (3 - L)/2 (p - L)/T[[n]], {n, 4, 130}]] (* after T. D. Noe *)

Formula

a(n) = (3 - L(p))/2 * (p - L(p)) / k(p), where p = prime(n), L(p) = Legendre(p|5), and k(p) = Pisano period(p) = A001175(p).
a(n) > 1 if and only if prime(n) is in A222413.

A222413 All primes p > 5 such that A001175(p) is smaller than the maximal value permitted by Wall's Theorems 6 and 7.

Original entry on oeis.org

29, 47, 89, 101, 107, 113, 139, 151, 181, 199, 211, 229, 233, 263, 281, 307, 331, 347, 349, 353, 401, 421, 461, 509, 521, 541, 557, 563, 619, 661, 677, 691, 709, 743, 761, 769, 797, 809, 811, 829, 859, 881, 911, 919, 941, 953, 967, 977, 991, 1009, 1021, 1031, 1049, 1061, 1069, 1087, 1097, 1103, 1109, 1151, 1217, 1223, 1229, 1231, 1249, 1277
Offset: 1

Views

Author

N. J. A. Sloane, Feb 28 2013

Keywords

Comments

Included because A001175 is still a mystery (as are many sequences of the same type).
A222414 gives the corresponding values of A001175(a(n)).
The maximal value for a prime p > 5 is p-1 if p == 1 or 9 (mod 10) and 2*(p+1) if p == 3 or 7 (mod 10). See Wall's Theorems 6 and 7. These values are given in A253806. - Wolfdieter Lang, Jan 16 2015
Prime(n) is a member if and only if A296240(n) > 1. - Jonathan Sondow, Dec 10 2017

Examples

			From _Wolfdieter Lang_, Jan 16 2015: (Start)
a(1) = 29 because A001175(29) = 14 but the maximal value is 29 - 1 = 28.
a(2) = 47 because A001175(47) = 32 but the maximal value is 2*(47 + 1) = 96.
All other primes p > 5 have A001175(p) = maximal value for p.
E.g., p = 11 has  A001175(11) = 11-1 = 10 and  p = 7 has A001175(7) = 2*(7 + 1) = 16. (End)
		

Crossrefs

Extensions

Name corrected by Wolfdieter Lang, Jan 16 2015

A271953 a(n) is the period of A000930 modulo n.

Original entry on oeis.org

1, 7, 8, 14, 31, 56, 57, 28, 24, 217, 60, 56, 168, 399, 248, 56, 288, 168, 381, 434, 456, 420, 528, 56, 155, 168, 72, 798, 840, 1736, 930, 112, 120, 2016, 1767, 168, 342, 2667, 168, 868, 1723, 3192, 1848, 420, 744, 3696, 46, 56, 399, 1085, 288, 168, 468, 504, 1860, 1596, 3048, 840, 3541, 1736, 1240, 6510
Offset: 1

Views

Author

Joerg Arndt, Apr 17 2016

Keywords

Crossrefs

Cf. A000930, A271901 (periods mod primes), A001175 (periods of A000045 modulo n).

Programs

  • Mathematica
    minlen = 100; maxlen = 2*10^4;
    per[lst_] := FindTransientRepeat[lst, 2] // Last // Length;
    a[n_] := Module[{p0=0, len=minlen}, While[p0 = Mod[LinearRecurrence[{1, 0, 1}, {1, 1, 1}, len], n] // per; p0<=1 && len<=maxlen, len = 2 len]; p0];
    Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 100}] (* Jean-François Alcover, Jul 21 2018 *)
  • PARI
    per(n, S, R) = {  \\ S[]: leading terms, R[]: recurrence
        if ( n==1, return( 1 ) );
        my ( r = #R );
        if ( r != #S , error("Mismatch in length of S[] and R[]") );
        S = vector(#S, j, Mod(S[j], n) );
        R = vector(#S, j, Mod(R[j], n) );
        my( T = S );
        my( j = 0 );
        until ( 0,  \\ forever
            j += 1;
            my( t = sum(i=1, r, R[i] * T[r+1-i] ) );  \\ next term
            for (k=1, r-1, T[k] = T[k+1] );
            T[r] = t;
            if ( T == S , return(j) );
        );
    }
    \\vector(66, n, per(n, [0,1], [1,1]) )  \\ A001175
    \\vector(66, n, per(prime(n), [0,1], [1,1]) )  \\ A060305
    vector(66, n, per(n, [0,0,1], [1,0,1]) )  \\ A271953
    \\vector(66, n, per(prime(n), [0,0,1], [1,0,1]) )  \\ A271901
    \\vector(66, n, per(n, [0,0,1], [0,1,1]) )  \\ A104217
    /* Joerg Arndt, Apr 17 2016 */

Formula

Let the prime factorization of n be p1^e1*...*pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)) [Engstrom]. - N. J. A. Sloane, Feb 18 2017

A216067 Prime numbers p such that p is odd and is congruent to 2 (mod 5) or 3 (mod 5), but the period of the irreducible polynomial x^2-x-1 in GF(p^2) is not 2*(p+1).

Original entry on oeis.org

47, 107, 113, 233, 263, 307, 347, 353, 557, 563, 677, 743, 797, 953, 967, 977, 1087, 1097, 1103, 1217, 1223, 1277, 1307, 1427, 1483, 1523, 1553, 1597, 1733, 1823, 1877, 1913, 1973, 2027, 2207, 2237, 2243, 2267, 2333, 2417, 2447, 2663, 2687, 2753, 2777
Offset: 1

Views

Author

V. Raman, Sep 01 2012

Keywords

Examples

			47 is in the sequence because the period of the Fibonacci / Lucas numbers (mod 47) = 32, is not 2*(47+1) = 96.
		

Crossrefs

Programs

  • PARI
    forprime(p=3,3000,if(p%5==2||p%5==3,a=1;b=0;c=1;while(a!=0||b!=1,c++;d=a;a=b;a=(a+d)%p;b=d%p);if(c!=(2*(p+1)),print1(p",")))) \\ V. Raman, Nov 22 2012

Extensions

Definition corrected by V. Raman, Nov 22 2012

A116515 a(n) = the period of the Fibonacci numbers modulo p divided by the smallest m such that p divides Fibonacci(m), where p is the n-th prime.

Original entry on oeis.org

1, 2, 4, 2, 1, 4, 4, 1, 2, 1, 1, 4, 2, 2, 2, 4, 1, 4, 2, 1, 4, 1, 2, 4, 4, 1, 2, 2, 4, 4, 2, 1, 4, 1, 4, 1, 4, 2, 2, 4, 1, 1, 1, 4, 4, 1, 1, 2, 2, 1, 4, 1, 2, 1, 4, 2, 4, 1, 4, 2, 2, 4, 2, 1, 4, 4, 1, 4, 2, 1, 4, 1, 2, 4, 1, 2, 4, 4, 2, 2, 1, 4, 1, 4, 1, 2, 2, 4, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 4, 2, 2, 1
Offset: 1

Views

Author

Nick Krempel, Mar 24 2006

Keywords

Comments

Conditions on p_n mod 4 and mod 5 restrict possible values of a(n). The unknown (?) case is p = 1 mod 4 and (5|p) = 1, equivalently, p = 1 or 9 mod 20, where {1, 2, 4} all occur.
Number of zeros in fundamental period of Fibonacci numbers mod prime(n). [From T. D. Noe, Jan 14 2009]

Examples

			a(4) = 2, as 7 is the 4th prime, the Fibonacci numbers mod 7 have period 16, the first Fibonacci number divisible by 7 is F(8) = 21 = 3*7 and 16 / 8 = 2.
One period of the Fibonacci numbers mod 7 is 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, which has two zeros. Hence a(4)=2. [From _T. D. Noe_, Jan 14 2009]
		

Crossrefs

Cf. A112860, A053027, A053028 (primes producing 1, 2 and 4 zeros) [From T. D. Noe, Jan 14 2009]

Formula

a(n) = A060305(n) / A001602(n). a(n) is always one of {1, 2, 4}.
a(n) = A001176(prime(n)) [From T. D. Noe, Jan 14 2009]

A253246 Pisano period of A006190 to mod prime(n).

Original entry on oeis.org

3, 2, 12, 16, 8, 52, 16, 40, 22, 28, 64, 76, 28, 42, 96, 26, 24, 30, 136, 144, 148, 26, 168, 180, 196, 50, 102, 106, 20, 112, 126, 10, 92, 138, 300, 304, 156, 328, 336, 86, 178, 180, 190, 388, 396, 198, 30, 448, 456, 460, 116, 160, 484, 250, 128, 262, 268, 544, 138, 564
Offset: 1

Views

Author

Eric Chen, Apr 11 2015

Keywords

Comments

If the generalized Wall's conjecture to A006190 is true, then we can calculate A175182(m) when m is a prime power since for any k>=1 : A175182(prime(n)^k)=a(n)*prime(n)^(k-1). For example: A175182(2^k)=3*2^(k-1)=A007283(k-1).
In fact, the conjecture fails on p=241, and this is the only counterexample below 10^8.

Crossrefs

Programs

  • Mathematica
    Table[s = t = Mod[{0, 1}, Prime[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}]
  • PARI
    fibmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
    entry(p)=my(k=1, c=Mod(1, p), o); while(c, [o, c]=[c, 3*c+o]; k++); ka(n)=entry(prime(n))

Formula

a(n) = A175182(A000040(n)).
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