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 1-5 of 5 results.

A280687 a(n) = n / A280686(n); n divided by its largest Fibonacci proper divisor, a(1) = 1.

Original entry on oeis.org

1, 2, 3, 2, 5, 2, 7, 4, 3, 2, 11, 4, 13, 7, 3, 2, 17, 6, 19, 4, 7, 11, 23, 3, 5, 2, 9, 14, 29, 6, 31, 4, 11, 17, 7, 12, 37, 19, 3, 5, 41, 2, 43, 22, 9, 23, 47, 6, 49, 10, 17, 4, 53, 18, 11, 7, 19, 29, 59, 12, 61, 31, 3, 8, 5, 22, 67, 2, 23, 14, 71, 9, 73, 37, 15, 38, 77, 6, 79, 10, 27, 41, 83, 4, 17, 43, 29, 11, 89, 18, 7, 46, 31, 47, 19, 12, 97, 49, 33, 20
Offset: 1

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Author

Antti Karttunen, Jan 11 2017

Keywords

Comments

For n > 1, a(n) = n divided by the greatest Fibonacci number that divides n and is less than n.

Crossrefs

Programs

Formula

a(n) = n / A280686(n).
a(A000045(n)) = A280690(n).

A280688 a(n) = A000045(A054576(n))= A280686(A105800(n)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 5, 1, 1, 1, 8, 1, 1, 2, 13, 1, 5, 1, 21, 1, 1, 1, 34, 1, 1, 1, 55, 1, 13, 1, 89, 5, 1, 1, 144, 1, 5, 1, 233, 1, 34, 1, 377, 1, 1, 1, 610, 1, 1, 13, 987, 1, 89, 1, 1597, 1, 13, 1, 2584, 1, 1, 5, 4181, 1, 233, 1, 6765, 34, 1, 1, 10946, 1, 1, 1, 17711, 1, 610, 1, 28657, 1, 1, 1, 46368, 1, 13, 89, 75025, 1, 1597
Offset: 1

Views

Author

Antti Karttunen, Jan 11 2017

Keywords

Crossrefs

Formula

a(n) = A000045(A054576(n)).
a(n) = A280686(A105800(n)).
a(n) = A105800(A032742(n)).

A105800 Greatest Fibonacci number that is a proper divisor of the n-th Fibonacci number; a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 2, 5, 1, 8, 1, 13, 5, 21, 1, 34, 1, 55, 13, 89, 1, 144, 5, 233, 34, 377, 1, 610, 1, 987, 89, 1597, 13, 2584, 1, 4181, 233, 6765, 1, 10946, 1, 17711, 610, 28657, 1, 46368, 13, 75025, 1597, 121393, 1, 196418, 89, 317811, 4181, 514229, 1, 832040, 1
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 20 2005

Keywords

Comments

a(A001605(n)) = 1.

Crossrefs

Cf. A046022 (gives the positions of ones).

Programs

  • Mathematica
    nn=70;Join[{1,1},With[{fibs=Fibonacci[Range[nn]]},Table[ Max[ Intersection[ Most[Divisors[fibs[[n]]]],fibs]],{n,3,nn}]]] (* Harvey P. Dale, Apr 10 2012 *)

Formula

From Antti Karttunen, Jan 11 2017: (Start)
a(n) = A280686(A000045(n)).
a(n) = A000045(A032742(n)). [Because A000045 is a divisibility sequence.]
a(A032742(n)) = A280688(n).
(End)

A054494 Largest Fibonacci factor of n.

Original entry on oeis.org

1, 2, 3, 2, 5, 3, 1, 8, 3, 5, 1, 3, 13, 2, 5, 8, 1, 3, 1, 5, 21, 2, 1, 8, 5, 13, 3, 2, 1, 5, 1, 8, 3, 34, 5, 3, 1, 2, 13, 8, 1, 21, 1, 2, 5, 2, 1, 8, 1, 5, 3, 13, 1, 3, 55, 8, 3, 2, 1, 5, 1, 2, 21, 8, 13, 3, 1, 34, 3, 5, 1, 8, 1, 2, 5, 2, 1, 13, 1, 8, 3, 2, 1, 21, 5, 2, 3, 8, 89, 5, 13, 2, 3, 2, 5, 8, 1, 2, 3, 5
Offset: 1

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Author

Henry Bottomley, Apr 04 2000

Keywords

Examples

			a(10)=5 because 1, 2 and 5 are the Fibonacci numbers which divide 10 and 5 is the largest.
		

Crossrefs

Sequences with similar definitions: A047930 (smallest Fibonacci multiple), A280686 (restricted to proper divisors), A280694 (equivalent for Lucas numbers).
Positions of 1's: A147956.

Programs

  • Mathematica
    With[{fibs=Fibonacci[Range[20]]},Table[Max[Select[fibs,Divisible[ n,#]&]],{n,100}]] (* Harvey P. Dale, Jul 17 2012 *)
  • PARI
    A010056(n)=my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8))
    a(n)=fordiv(n,d,if(A010056(n/d), return(n/d))) \\ Charles R Greathouse IV, Nov 05 2014
    
  • Python
    from sympy import divisors
    from sympy.ntheory.primetest import is_square
    def A054494(n): return next(d for d in sorted(divisors(n,generator=True),reverse=True) if is_square(m:=5*d**2-4) or is_square(m+8)) # Chai Wah Wu, May 06 2024

Formula

a(n) = n/A054495(n).

Extensions

Corrected by Harvey P. Dale, Jul 17 2012

A280696 Largest Lucas proper divisor of n, a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 4, 3, 2, 1, 4, 1, 7, 3, 4, 1, 3, 1, 4, 7, 11, 1, 4, 1, 2, 3, 7, 1, 3, 1, 4, 11, 2, 7, 18, 1, 2, 3, 4, 1, 7, 1, 11, 3, 2, 1, 4, 7, 2, 3, 4, 1, 18, 11, 7, 3, 29, 1, 4, 1, 2, 7, 4, 1, 11, 1, 4, 3, 7, 1, 18, 1, 2, 3, 4, 11, 3, 1, 4, 3, 2, 1, 7, 1, 2, 29, 11, 1, 18, 7, 4, 3, 47, 1, 4, 1, 7, 11, 4, 1, 3, 1, 4, 7, 2, 1, 18, 1, 11, 3, 7, 1, 3, 1
Offset: 1

Views

Author

Antti Karttunen, Jan 11 2017

Keywords

Comments

For n > 1, a(n) = greatest Lucas number (A000032) that divides n and is less than n.

Crossrefs

Cf. A057854 (gives the positions n > 1 where this sequence and A280694 obtain equal values).

Programs

  • Scheme
    ;; A stand-alone program:
    (define (A280696 n) (let loop ((l1 1) (l2 3) (lpd 1)) (cond ((>= l1 n) (if (and (= 1 lpd) (even? n) (> n 2)) 2 lpd)) ((zero? (modulo n l1)) (loop l2 (+ l1 l2) l1)) (else (loop l2 (+ l1 l2) lpd)))))

Formula

a(n) = n / A280697(n).
Other identities. For all n >= 1:
a(A057854(n)) = A280694(A057854(n)).
a(A000204(n)) = A280698(n).
Showing 1-5 of 5 results.