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 10 results.

A081264 Odd Fibonacci pseudoprimes: odd composite numbers k such that either (1) k divides Fibonacci(k-1) if k == +-1 (mod 5) or (2) k divides Fibonacci(k+1) if k == +-2 (mod 5).

Original entry on oeis.org

323, 377, 1891, 3827, 4181, 5777, 6601, 6721, 8149, 10877, 11663, 13201, 13981, 15251, 17119, 17711, 18407, 19043, 23407, 25877, 27323, 30889, 34561, 34943, 35207, 39203, 40501, 50183, 51841, 51983, 52701, 53663, 60377, 64079, 64681
Offset: 1

Views

Author

T. D. Noe, Mar 15 2003, Jun 09 2008

Keywords

Comments

Lehmer shows that there are an infinite number of Fibonacci pseudoprimes (FPPs). In particular, the number Fibonacci(2p) is an FPP for all primes p > 5. Anderson lists over 5000 FPPs, while Jacobsen lists over 170000. The sequences A069106 and A069107 give k such that k divides Fibonacci(k-1) and k divides Fibonacci(k+1), respectively. See A141137 for even FPPs.

References

  • R. Crandall and C. Pomerance, Prime Numbers: A Computational Perspective, Springer, 2002, p. 131.
  • Paulo Ribenboim, The New Book of Prime Number Records, Springer, 1995, p. 127.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 104.
  • A. Witno, Theory of Numbers, BookSurge, North Charleston, SC; see p. 83.

Crossrefs

Programs

  • Maple
    filter:= proc(n) local M,r;
       uses LinearAlgebra:-Modular;
       if isprime(n) then return false fi;
       M:= Mod(n, [[1,1],[1,0]],float[8]);
       if n^2 mod 5 = 1 then r:= n-1 else r:= n+1 fi;
       M:= MatrixPower(n,M,r);
       M[1,2] = 0
    end proc:select(filter, [2*i+1 $ i=1..10^5]); # Robert Israel, Aug 05 2015
  • Mathematica
    lst={}; f0=0; f1=1; Do[f2=f1+f0; If[n>1&&!PrimeQ[n], If[MemberQ[{1, 4}, Mod[n, 5]], If[Mod[f0, n]==0, AppendTo[lst, n]]]; If[MemberQ[{2, 3}, Mod[n, 5]], If[Mod[f2, n]==0, AppendTo[lst, n]]]]; f0=f1; f1=f2, {n, 100000}]; lst
    ocnQ[n_]:=CompositeQ[n]&&Which[Mod[n,5]==1,Divisible[Fibonacci[ n-1], n],Mod[n,5] == 4,Divisible[ Fibonacci[n-1],n],Mod[n,5]==2,Divisible[ Fibonacci[n+1],n], Mod[n,5]==3,Divisible[Fibonacci[n+1],n],True,False]; Select[Range[1,65001,2],ocnQ] (* Harvey P. Dale, Aug 23 2017 *)
  • Perl
    use ntheory ":all"; foroddcomposites { $e = (0,-1,1,1,-1)[$%5]; say unless $e==0 || (lucas_sequence($, 1, -1, $+$e))[0] } 1e10; # _Dana Jacobsen, Aug 05 2015

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

Views

Author

Peter K. Pearson (ppearson+att(AT)spamcop.net), May 06 2005

Keywords

Comments

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)

Crossrefs

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).

Programs

  • GAP
    Filtered([2..2000], n -> Fibonacci(n-1) mod n = 0 and Filtered( [2..n-2], x -> Fibonacci(x) mod n = 0 ) = [] );
    
  • Maple
    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
  • Mathematica
    f[n_] := Block[{x = 2}, While[Mod[Fibonacci@ x, n] != 0, x++]; x];Select[Range@ 1860, f@ # == # - 1 &] (* Michael De Vlieger, Oct 13 2015 *)
  • PARI
    isok(n) = {x = 2; while(fibonacci(x) % n, x++); x == n-1;} \\ Michel Marcus, Oct 20 2015

Formula

{n: A001177(n) = n-1}. - R. J. Mathar, Jul 09 2012

Extensions

Corrected by T. D. Noe, Oct 25 2006

A123976 Numbers k such that Fibonacci(k-1) is divisible by k.

Original entry on oeis.org

1, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, 211, 229, 239, 241, 251, 269, 271, 281, 311, 331, 349, 359, 379, 389, 401, 409, 419, 421, 431, 439, 442, 449, 461, 479, 491, 499, 509, 521, 541, 569, 571, 599, 601
Offset: 1

Views

Author

Tanya Khovanova, Oct 30 2006

Keywords

Comments

a(n) is a union of {1}, A069106(n) and A045468(n). Composite a(n) are listed in A069106(n) = {442, 1891, 2737, 4181, 6601, 6721, 8149, ...}. Prime a(n) are listed in A045468(n) = {11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, ...} Primes congruent to {1, 4} mod 5. - Alexander Adamchuk, Nov 02 2006

Examples

			Fibonacci(10) = 55, is divisible by 11.
		

Crossrefs

Programs

  • Haskell
    import Data.List (elemIndices)
    a123976 n = a123976_list !! (n-1)
    a123976_list = map (+ 1) $ elemIndices 0 $ zipWith mod a000045_list [1..]
    -- Reinhard Zumkeller, Oct 13 2011
    
  • Mathematica
    Select[Range[1000], IntegerQ[Fibonacci[ # - 1]/# ] &]
  • PARI
    is(n)=((Mod([1,1;1,0],n))^n)[2,2]==0 \\ Charles R Greathouse IV, Feb 03 2014

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

A094400 Odd n dividing Fibonacci(n)-1 but neither Fibonacci(n-1) nor Fibonacci(n+1).

Original entry on oeis.org

7743, 27071, 54839, 72831, 217257, 388367, 417601, 575599, 670879, 691447, 701569, 809999, 850541, 881011, 1274897, 1365407, 1383249, 1464449, 1504097, 1653751, 1922817, 2106017, 2276351, 2385811, 2474047, 2556553, 2628879, 2697899, 2804543, 3017729, 3352049
Offset: 1

Views

Author

Eric Rowland, May 01 2004

Keywords

Crossrefs

Programs

  • Mathematica
    Select[Range[50000], OddQ[ # ] && Mod[Fibonacci[ # ] - 1, # ] == 0 && ! Mod[Fibonacci[ # - 1], # ] == 0 && ! Mod[Fibonacci[ # + 1], # ] == 0 &]

Extensions

Offset corrected by and a(15)-a(31) from Giovanni Resta, Jul 20 2013

A100992 Indices k of Fibonacci numbers F(k) (A000045) that are divisible by k+1.

Original entry on oeis.org

10, 18, 28, 30, 40, 58, 60, 70, 78, 88, 100, 108, 130, 138, 148, 150, 178, 180, 190, 198, 210, 228, 238, 240, 250, 268, 270, 280, 310, 330, 348, 358, 378, 388, 400, 408, 418, 420, 430, 438, 441, 448, 460, 478, 490, 498, 508, 520, 540, 568, 570, 598, 600, 618
Offset: 1

Views

Author

Ron Knott, Nov 25 2004

Keywords

Comments

When k+1 is prime, it is in A045468; when k+1 is composite (such as 442), it is in A069106. - T. D. Noe, Dec 13 2004

Examples

			18 is a term because F(18) = 2584 = 2*2*2*17*19 is divisible by 19, one more than its index number 18.
		

Crossrefs

Programs

  • Mathematica
    Select[ Range[ 620], Mod[ Fibonacci[ # ], # + 1] == 0 &] (* Robert G. Wilson v, Nov 26 2004 *)

Extensions

More terms from Robert G. Wilson v, Nov 26 2004

A094409 Odd numbers k that divide Fibonacci(k) - 1 but not Fibonacci(k-1).

Original entry on oeis.org

323, 6479, 7743, 11663, 18407, 19043, 23407, 27071, 34943, 35207, 39203, 44099, 47519, 51983, 53663, 54839, 65471, 72831, 78089, 79547, 82983, 86063, 94667, 104663, 109871, 121103, 139359, 142883, 157079, 168299, 195227, 196559
Offset: 1

Views

Author

Eric Rowland, May 01 2004

Keywords

Crossrefs

Programs

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

Extensions

Offset corrected by Giovanni Resta, Jul 20 2013

A094410 Numbers k that divide Fibonacci(k-1) but do not divide Fibonacci(k) - 1.

Original entry on oeis.org

442, 1891, 6601, 8149, 13981, 17119, 17711, 30889, 40501, 52701, 68101, 78409, 88601, 88831, 115231, 138601, 145351, 150121, 153781, 191351, 199801, 218791, 265881, 266071, 316561, 332949, 342271, 348517, 407353, 453151, 456301, 464101
Offset: 1

Views

Author

Eric Rowland, May 01 2004

Keywords

Crossrefs

Programs

  • Mathematica
    Select[Range[50000], ! Mod[Fibonacci[ # ] - 1, # ] == 0 && Mod[Fibonacci[ # - 1], # ] == 0 &]
  • PARI
    fibmod(n,m)=(Mod([0,1;1,1],m)^n*[0;1])[1,1]
    is(n)=fibmod(n-1,n)==0 && fibmod(n,n)!=1 \\ Charles R Greathouse IV, Nov 04 2016

Extensions

More terms from David Wasserman, May 08 2007

A177086 Semiprimes k that divide Fibonacci(k-1).

Original entry on oeis.org

1891, 4181, 8149, 13201, 15251, 17711, 40501, 51841, 64079, 64681, 67861, 68251, 78409, 88601, 88831, 90061, 96049, 97921, 115231, 118441, 145351, 146611, 153781, 191351, 197209, 218791, 219781, 254321, 272611, 302101, 303101
Offset: 1

Views

Author

Jonathan Vos Post, Dec 09 2010

Keywords

Comments

This is the semiprime (A001358) analog of A045468. Now A045468 has a very simple characterization: it consists of the primes ending in 1 or 9. Can one say anything about the present sequence?

Examples

			46368/23 = 2016 = 2^5 * 3^2 * 7 so (24-1) | Fibonacci(24) but 24 is not semiprime, so is not in the sequence.
a(1) = 1891 = 31 * 61 is not in the sequence because 1891 divides Fibonacci(1891-1) = Fibonacci(1890).
a(21) = 146611 = 271 * 541 because 146611 | Fibonacci(146610).
		

Crossrefs

Cf. A000040, A000045, A001358, A069106, A045468, A003631, A064739, A081264 (Fibonacci pseudoprimes).
Cf. A177745 (semiprimes k that divide Fibonacci(k+1)).

Programs

  • Mathematica
    Select[Range[310000],PrimeOmega[#]==2 && Divisible[Fibonacci[#-1],#]&] (* Harvey P. Dale, May 02 2016 *)

Formula

{k: k is in A001358 and k|A000045(k-1)} = A069106 INTERSECTION A001358.

A177745 Semiprimes k that divide Fibonacci(k+1).

Original entry on oeis.org

323, 377, 3827, 5777, 10877, 11663, 18407, 19043, 23407, 25877, 27323, 34943, 39203, 51983, 53663, 60377, 75077, 86063, 94667, 100127, 113573, 121103, 121393, 161027, 162133, 182513, 195227, 200147, 231703, 240239, 250277, 294527, 306287, 345913, 381923, 429263, 430127, 454607, 500207, 507527, 548627, 569087, 600767, 635627, 636707, 685583, 697883, 736163, 753377, 775207, 828827, 851927, 948433, 983903
Offset: 1

Views

Author

Jonathan Vos Post, Dec 12 2010

Keywords

Comments

Data from T. D. Noe.

Examples

			a(1) = 323 = 17 * 19 because it is semiprime (product of two prime A000040), and 323 divides F(324) = 23041483585524168262220906489642018075101617466780496790573690289968, with dividend 2^4 * 3^5 * 53 * 107 * 109 * 2269 * 3079 * 4373 * 5779 * 19441 * 11128427 * 62650261 * 1828620361 * 6782976947987.
		

Crossrefs

Cf. A177086, A000045, A001358, A069106, A045468, A003631, A064739, A081264 (Fibonacci pseudoprimes).

Programs

  • Mathematica
    With[{semis=Select[Range[1000000],PrimeOmega[#]==2&]},Select[semis, Divisible[Fibonacci[#+1],#]&]] (* Harvey P. Dale, Aug 20 2012 *)

Formula

{k: k is in A001358 and k|A000045(k+1)}.
Showing 1-10 of 10 results.