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 1-6 of 6 results.

A023172 Self-Fibonacci numbers: numbers k that divide Fibonacci(k).

Original entry on oeis.org

1, 5, 12, 24, 25, 36, 48, 60, 72, 96, 108, 120, 125, 144, 168, 180, 192, 216, 240, 288, 300, 324, 336, 360, 384, 432, 480, 504, 540, 552, 576, 600, 612, 625, 648, 660, 672, 684, 720, 768, 840, 864, 900, 960, 972, 1008, 1080, 1104, 1152, 1176, 1200, 1224, 1296, 1320
Offset: 1

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Author

Keywords

Comments

Sequence contains all powers of 5, infinitely many multiples of 12 and other numbers (including some factors of Fibonacci(5^j), e.g., 75025).
If m is in this sequence then 5*m is (since 5*m divides 5*F(m) which in turn divides F(5*m)). Also, if m is in this sequence then F(m) is in this sequence (since if gcd(F(m),m)=m then gcd(F(F(m)),F(m)) = F(gcd(F(m),m)) = F(m)). - Max Alekseyev, Sep 20 2009
From Max Alekseyev, Nov 29 2010: (Start)
Every term greater than 1 is a multiple of 5 or 12.
Proof. Let n>1 divide Fibonacci number F(n) and let p be the smallest prime divisor of n.
If p=2, then 3|n implying further that 4|n. Hence, 12|n.
If p=5, then 5|n.
If p is different from 2 and 5, then p divides either F(p+1) or F(p-1) and thus p divides either F(gcd(n,p+1)) or F(gcd(n,p-1)). Minimality of p implies that gcd(n,p-1)=1 and gcd(n,p+1)=1 (notice that p+1 being prime implies p=2 which is not the case). Therefore, p divides F(1)=1, a contradiction to the existence of such p. (End)
Luca & Tron give an upper bound, see links. - Charles R Greathouse IV, Aug 04 2021

References

  • S. Wolfram, "A new kind of science", p. 891

Crossrefs

Cf. A000350. See A127787 for an essentially identical sequence.
Cf. A128974 (12n does not divide Fibonacci(12n)). - Zak Seidov, Jan 10 2016

Programs

  • Haskell
    import Data.List (elemIndices)
    a023172 n = a023172_list !! (n-1)
    a023172_list =
       map (+ 1) $ elemIndices 0 $ zipWith mod (tail a000045_list) [1..]
    -- Reinhard Zumkeller, Oct 13 2011
    
  • Magma
    [n: n in [1..2*10^3] | Fibonacci(n) mod n eq 0 ]; // Vincenzo Librandi, Sep 17 2015
  • Maple
    fmod:= proc(n,m) local M,t; uses LinearAlgebra:-Modular;
        if m <= 1 then return 0 fi;
        if m < 2^25 then t:= float[8] else t:= integer fi;
        M:= Mod(m,<<1,1>|<1,0>>,t);
        round(MatrixPower(m,M,n)[1,2])
    end proc:
    select(n -> fmod(n,n)=0, [$1..2000]); # Robert Israel, May 10 2016
  • Mathematica
    a=0; b=1; c=1; Do[a=b; b=c; c=a+b; If[Mod[c, n]==0, Print[n]], {n, 3, 1500}]
    Select[Range[1350], Mod[Fibonacci[ # ], # ]==0&] (* Harvey P. Dale *)
  • PARI
    is(n)=((Mod([1,1;1,0],n))^n)[1,2]==0 \\ Charles R Greathouse IV, Feb 03 2014
    

Extensions

Edited by Don Reble, Sep 07 2003

A069107 Composite numbers k that divide Fibonacci(k+1).

Original entry on oeis.org

323, 377, 2834, 3827, 5777, 6479, 10877, 11663, 18407, 19043, 20999, 23407, 25877, 27323, 34943, 35207, 39203, 44099, 47519, 50183, 51983, 53663, 60377, 65471, 75077, 78089, 79547, 80189, 81719, 82983, 84279, 84419, 86063, 90287, 94667
Offset: 1

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Author

Benoit Cloitre, Apr 06 2002

Keywords

Comments

Primes p congruent to +2 or -2 (mod 5) divide Fibonacci(p+1) (cf. A003631 and [Hardy and Wright]).

References

  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers (Fifth edition), Oxford Univ. Press (Clarendon), 1979, Chap. X, p. 150.

Crossrefs

Cf. A045468, A003631, A064739, A081264 (Fibonacci pseudoprimes).

Programs

  • Haskell
    a069107 n = a069107_list !! (n-1)
    a069107_list = h 2 $ drop 3 a000045_list where
       h n (fib:fibs) = if fib `mod` n > 0 || a010051 n == 1
           then h (n+1) fibs else n : h (n+1) fibs
    -- Reinhard Zumkeller, Oct 13 2011
    
  • Mathematica
    Select[Range[2,100000],!PrimeQ[#]&&Divisible[Fibonacci[#+1],#]&] (* Harvey P. Dale, Sep 18 2011 *)
  • PARI
    is(n)=((Mod([1,1;1,0],n))^(n+1))[1,2]==0 && !isprime(n) && n>1 \\ Charles R Greathouse IV, Oct 07 2016

Formula

Fibonacci(2*a(n)) mod a(n) = a(n) - 1. - Gary Detlefs, May 26 2014

Extensions

Corrected by Ralf Stephan, Oct 17 2002

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

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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

A159051 Numbers n such that Fibonacci(n-2) is divisible by n.

Original entry on oeis.org

2, 8, 38, 62, 122, 158, 218, 278, 302, 362, 398, 422, 458, 482, 542, 662, 698, 758, 818, 842, 878, 884, 902, 998, 1037, 1082, 1142, 1157, 1202, 1238, 1262, 1322, 1382, 1418, 1478, 1502, 1538, 1622, 1658, 1718, 1838, 1982, 2018, 2042, 2078, 2102, 2138
Offset: 1

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Author

Keywords

Examples

			8=8/8=1, 38=14930352/38=392904, 62=1548008755920/62=24967883160, ...
		

Crossrefs

Programs

  • Haskell
    import Data.List (elemIndices)
    a159051 n = a159051_list !! (n-1)
    a159051_list = map (+ 2) $ elemIndices 0 $ zipWith mod a000045_list [2..]
    -- Reinhard Zumkeller, Oct 13 2011
    
  • Mathematica
    lst={};Do[If[Mod[(Fibonacci[n-2]),n]==0,AppendTo[lst,n]],{n,7!}];lst
    Select[Range[2,2200],Divisible[Fibonacci[#-2],#]&] (* Harvey P. Dale, Dec 20 2014 *)
  • PARI
    is(n)=((Mod([1,1;1,0],n))^(n-2))[1,2]==0 \\ Charles R Greathouse IV, Feb 03 2014

A100993 Indices k of Fibonacci numbers F(k) (A000045) that are divisible by k-1.

Original entry on oeis.org

2, 3, 4, 8, 14, 18, 24, 38, 44, 48, 54, 68, 74, 84, 98, 104, 108, 114, 128, 138, 158, 164, 168, 174, 194, 198, 224, 228, 234, 258, 264, 278, 284, 294, 308, 314, 318, 324, 338, 348, 354, 368, 374, 378, 384, 398, 434, 444, 458, 464, 468, 488, 504, 524, 548, 558
Offset: 1

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Author

Ron Knott, Nov 25 2004

Keywords

Comments

When k-1 is prime, it is in A003631; when k-1 is composite (such as 323), it is in A069107. - T. D. Noe, Dec 13 2004

Examples

			14 is a term because F(14) = 377 = 13*29 is divisible by 13, one less than its index number 14.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[2, 500], Mod[ Fibonacci[ # ], # - 1] == 0 &] (* Robert G. Wilson v, Nov 26 2004 *)

Formula

a(n) = A069104(n) + 1. - T. D. Noe, Dec 13 2004

A220168 Numbers k that divide Fibonacci(k+2).

Original entry on oeis.org

1, 4, 34, 46, 88, 94, 106, 166, 214, 226, 274, 334, 346, 394, 454, 466, 514, 526, 586, 634, 646, 694, 706, 754, 766, 886, 934, 1006, 1114, 1126, 1174, 1186, 1234, 1294, 1306, 1354, 1366, 1486, 1546, 1594, 1654, 1714, 1726, 1774, 1894, 1906, 1954, 1966, 2026
Offset: 1

Views

Author

Alex Ratushnyak, May 03 2013

Keywords

Crossrefs

Cf. A000045.
Cf. A023172 (numbers k that divide Fibonacci(k)).
Cf. A069104 (numbers k that divide Fibonacci(k+1)).
Cf. A123976 (numbers k that divide Fibonacci(k-1)).
Cf. A159051 (numbers k that divide Fibonacci(k-2)).

Programs

  • Mathematica
    Select[Range[2000], Mod[Fibonacci[#+2], #] == 0 &] (* T. D. Noe, Feb 05 2014 *)
  • PARI
    is(n)=((Mod([1,1;1,0],n))^(n+2))[1,2]==0 \\ Charles R Greathouse IV, Feb 03 2014
  • Python
    prpr = prev = 1
    for i in range(3, 3000):
        prpr, prev = prev, prpr+prev
        if prev % (i-2) == 0:  print(i-2, end=', ')
    
Showing 1-6 of 6 results.