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.

A094394 Odd composites m that divide Fibonacci(m)-1.

Original entry on oeis.org

323, 2737, 4181, 6479, 6721, 7743, 11663, 13201, 15251, 18407, 19043, 23407, 27071, 34561, 34943, 35207, 39203, 44099, 47519, 51841, 51983, 53663, 54839, 64079, 64681, 65471, 67861, 68251, 72831, 78089, 79547, 82983, 86063, 90061, 94667
Offset: 1

Views

Author

Eric Rowland, May 01 2004

Keywords

Comments

No terms satisfy the Fermat criterion 2^(a(n)-1) mod a(n) = 1. - Gary Detlefs, May 25 2014
For each prime p, Fibonacci(p) = 5^((p-1)/2) mod p, so p divides Fibonacci(p) - 1 for each prime p=10k+-1. Hence it is interesting to seek also nonprimes with the same property, a motivation for this sequence. - Robert FERREOL, Jul 14 2015

Crossrefs

Programs

  • Maple
    with(combinat):test:=n->(fibonacci(n)-1) mod n= 0:
    select(test and not isprime ,[seq(2*k+1,k=1..10000)]); # Robert FERREOL, Jul 14 2015
  • Mathematica
    Select[Range[2, 50000], OddQ[#] && ! PrimeQ[#] && Mod[Fibonacci[#] - 1, #] == 0 &]
  • PARI
    main(m)=forcomposite(n=1,m,if(((n%2==1)&&(fibonacci(n)-1)%n==0),print1(n,", "))); \\ Anders Hellström, Aug 12 2015

Extensions

Offset corrected by Giovanni Resta, Jul 20 2013

A094401 Composite n such that n divides both Fibonacci(n-1) and Fibonacci(n) - 1.

Original entry on oeis.org

2737, 4181, 6721, 13201, 15251, 34561, 51841, 64079, 64681, 67861, 68251, 90061, 96049, 97921, 118441, 146611, 163081, 179697, 186961, 194833, 197209, 219781, 252601, 254321, 257761, 268801, 272611, 283361, 302101, 303101, 327313, 330929
Offset: 1

Views

Author

Eric Rowland, May 01 2004

Keywords

Comments

Composite n such that Q^(n-1) = I (mod n), where Q is the Fibonacci matrix {{1,1},{1,0}} and I is the identity matrix. The identity is also true for the primes congruent to 1 or 4 (mod 5), which is sequence A045468. The period of Q^k (mod n) is the same as the period of the Fibonacci numbers F(k) (mod n), A001175. Hence the terms in this sequence are the composite n such that A001175(n) divides n-1. [T. D. Noe, Jan 09 2009]

Crossrefs

Programs

  • Mathematica
    Select[Range[2, 50000], ! PrimeQ[ # ] && Mod[Fibonacci[ # - 1], # ] == 0 && Mod[Lucas[ # ] - 1, # ] == 0 &]

Extensions

More terms from Ryan Propper, Sep 24 2005

A094402 Numbers k that divide both Fibonacci(k+1) and Lucas(k) + 1.

Original entry on oeis.org

1, 2, 323, 6479, 11663, 18407, 19043, 23407, 34943, 35207, 39203, 44099, 47519, 51983, 53663, 65471, 78089, 79547, 82983, 86063, 94667, 104663, 109871, 121103, 139359, 142883, 157079, 168299, 195227, 196559, 200147, 233519
Offset: 1

Views

Author

Eric Rowland, May 01 2004

Keywords

Crossrefs

Programs

  • Mathematica
    Select[Range[10^5],Divisible[Fibonacci[#+1],#]&&Divisible[LucasL[#]+1,#]&] (* Giorgos Kalogeropoulos, Aug 20 2021 *)
  • PARI
    is(n)=(Mod([0,1;1,1],n)^(n+1)*[0;1])[1,1]==0 && (Mod([0,1;1,1],n)^n*[2;1])[1,1]==-1 \\ Charles R Greathouse IV, Nov 04 2016

Extensions

More terms from Emeric Deutsch, Apr 13 2005

A095401 number of steps required to reach 1 if the following modified juggler map is iterated: a[n]=(1-Mod[n, 2])*Floor[n^(3/4)]+Mod[n, 2]*Floor[n^(4/3)]; original exponents {1/2, 3/2} are replaced with {3/4, 4/3}.

Original entry on oeis.org

0, 1, 3, 2, 4, 4, 8, 3, 5, 5, 7, 5, 7, 9, 11, 4, 8, 4, 6, 6, 12, 6, 18, 6, 14, 8, 12, 6, 14, 6, 18, 8, 18, 10, 12, 10, 16, 12, 14, 12, 20, 5, 7, 9, 11, 9, 13, 5, 9, 5, 9, 7, 13, 7, 11, 7, 11, 13, 19, 13, 15, 7, 9, 7, 17, 19, 21, 19, 23, 7, 11, 7, 13, 15, 17, 15, 19, 9, 11, 9, 11, 13, 15, 13, 19
Offset: 1

Views

Author

Labos Elemer, Jun 18 2004

Keywords

Examples

			n=101: the trajectory is {101, 470, 100, 31, 97, 445, 3397, 51065, 1894513, 234421146, 1894512, 51064, 3396, 444, 96, 30, 12, 6, 3, 4, 2, 1}, number of required steps is a[101]=22-1=21.
		

Crossrefs

Programs

  • Mathematica
    e[x_]:=e[x]=(1-Mod[x, 2])*Floor[N[x^(3/4), 50]] +Mod[x, 2]*Floor[N[x^(4/3), 50]];e[1]=1; fe[x_]:=Delete[FixedPointList[e, x], -1]; Table[ -1+Length[fe[w]], {w, 1, 150}]
Showing 1-4 of 4 results.