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 1-4 of 4 results.

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

A139044 Smallest prime divisor of the Fibonacci numbers > 1.

Original entry on oeis.org

2, 3, 5, 2, 13, 3, 2, 5, 89, 2, 233, 13, 2, 3, 1597, 2, 37, 3, 2, 89, 28657, 2, 5, 233, 2, 3, 514229, 2, 557, 3, 2, 1597, 5, 2, 73, 37, 2, 3, 2789, 2, 433494437, 3, 2, 139, 2971215073, 2, 13, 5, 2, 3, 953, 2, 5, 3, 2, 59, 353, 2, 4513, 557, 2, 3, 5, 2, 269, 3, 2, 5, 6673, 2
Offset: 1

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Author

Omar E. Pol, Apr 23 2008

Keywords

Comments

Fibonacci number > 1, divided by its largest proper divisor.

Crossrefs

Programs

  • Magma
    [Minimum(PrimeDivisors(Fibonacci(n+2))): n in [1..70]]; // Vincenzo Librandi, Dec 24 2016
  • Maple
    with(numtheory): with(combinat): a:=proc(n) options operator, arrow: op(2, divisors(fibonacci(n))) end proc: seq(a(n),n=3..60); # Emeric Deutsch, May 02 2008
  • Mathematica
    First[First[FactorInteger[ # ]]]&/@Fibonacci[Range[3,40]] (* Harvey P. Dale, Apr 30 2008 *)
  • PARI
    a(n) = factor(fibonacci(n+2))[1,1]; \\ Michel Marcus, Nov 15 2014
    

Formula

a(n) = A000045(n+2)/A032742(A000045(n+2)) = A000045(n+2)/A139045(n).
a(n) = A020639(A000045(n+2)). - Michel Marcus, Nov 15 2014
a(n) = A060383(n+2). - Alois P. Heinz, Oct 11 2015

Extensions

More terms from Emeric Deutsch and Harvey P. Dale, May 02 2008
More terms from Vincenzo Librandi, Dec 24 2016

A139045 Largest proper divisor of the Fibonacci numbers > 1.

Original entry on oeis.org

1, 1, 1, 4, 1, 7, 17, 11, 1, 72, 1, 29, 305, 329, 1, 1292, 113, 2255, 5473, 199, 1, 23184, 15005, 521, 98209, 105937, 1, 416020, 2417, 726103, 1762289, 3571, 1845493, 7465176, 330929, 1056437, 31622993, 34111385, 59369, 133957148, 1, 233802911, 567451585
Offset: 3

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Author

Omar E. Pol, Apr 23 2008

Keywords

Comments

See the list of divisors of positive Fibonacci numbers in the triangle A133021.
See the largest proper divisor of n in A032742.
Fibonacci(1)=Fibonacci(2)=1 do not have proper divisors. - Emeric Deutsch, May 18 2008

Examples

			a(9) = 17 because the 9th Fibonacci number is 34 and the divisors of 34 are 1, 2, 17, 34, then the largest proper divisor of 34 is 17.
		

Crossrefs

Programs

  • Maple
    with(combinat): with(numtheory): a:=proc(n) options operator, arrow: op(tau(fibonacci(n))-1, divisors(fibonacci(n))) end proc: seq(a(n),n=3..40); # Emeric Deutsch, May 18 2008
    # second Maple program:
    a:= n-> (f-> f/min(numtheory[factorset](f)))((<<0|1>, <1|1>>^n)[1, 2]):
    seq(a(n), n=3..47);  # Alois P. Heinz, Sep 03 2019
  • Mathematica
    lpd[n_]:=Divisors[n][[-2]]; lpd/@(Fibonacci[Range[3,40]]) (* Harvey P. Dale, Mar 29 2015 *)

Formula

a(n) = A032742(A000045(n)).
a(n) = A000045(n)/A060383(n). - Alois P. Heinz, Sep 03 2019

Extensions

More terms from Emeric Deutsch, May 18 2008

A364820 a(n) is the smallest prime factor of the n-th Pell number A000129(n).

Original entry on oeis.org

2, 5, 2, 29, 2, 13, 2, 5, 2, 5741, 2, 33461, 2, 5, 2, 137, 2, 37, 2, 5, 2, 229, 2, 29, 2, 5, 2, 44560482149, 2, 61, 2, 5, 2, 13, 2, 593, 2, 5, 2, 1746860020068409, 2, 11437, 2, 5, 2, 3761, 2, 13, 2, 5, 2, 68480406462161287469, 2, 29, 2, 5, 2
Offset: 2

Views

Author

Sean A. Irvine, Oct 21 2023

Keywords

Crossrefs

Cf. A246556 (smallest primitive factor), A264137, A280104, A060383, A020639.

Programs

  • Mathematica
    FactorInteger[#][[1, 1]] & /@ LinearRecurrence[{2, 1}, {2, 5}, 57] (* Amiram Eldar, Oct 21 2023 *)

Formula

a(n) = A020639(A000129(n)).
Showing 1-4 of 4 results.