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 1-7 of 7 results.

A159052 Odd terms in A159051.

Original entry on oeis.org

1037, 1157, 20737, 250973, 854813, 1055617, 4042469, 8588477
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Mathematica
    lst={};Do[If[Mod[(Fibonacci[n-2]),n]==0,If[OddQ[n],AppendTo[lst,n]]],{n,3*8!}];lst
    a = c = 0; b = 1; lst = {}; Do[c = a + b; If[ Mod[a, n] == 0 && OddQ@n, Print@n]; a = b; b = c, {n, 2, 10^7}] (* Robert G. Wilson v, Apr 05 2009 *)

Extensions

a(4) - a(8) from Robert G. Wilson v, Apr 05 2009
Edited by N. J. A. Sloane, Apr 06 2009
Erroneous term 1 deleted by N. J. A. Sloane, Dec 20 2014

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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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

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

A069104 Numbers m such that m divides Fibonacci(m+1).

Original entry on oeis.org

1, 2, 3, 7, 13, 17, 23, 37, 43, 47, 53, 67, 73, 83, 97, 103, 107, 113, 127, 137, 157, 163, 167, 173, 193, 197, 223, 227, 233, 257, 263, 277, 283, 293, 307, 313, 317, 323, 337, 347, 353, 367, 373, 377, 383, 397, 433, 443, 457, 463, 467, 487, 503, 523, 547, 557
Offset: 1

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Author

Benoit Cloitre, Apr 06 2002

Keywords

Comments

Equals A003631 union A069107.
Let u(1)=u(2)=1 and (m+2)*u(m+2) = (m+1)*u(m+1) + m*u(m); then sequence gives values of k such that u(k) is an integer.

Crossrefs

Programs

  • Haskell
    import Data.List (elemIndices)
    a069104 n = a069104_list !! (n-1)
    a069104_list =
       map (+ 1) $ elemIndices 0 $ zipWith mod (drop 2 a000045_list) [1..]
    -- Reinhard Zumkeller, Oct 13 2011
    
  • Mathematica
    Select[Range[6! ],IntegerQ[Fibonacci[ #+1]/# ]&] (* Vladimir Joseph Stephan Orlovsky, Apr 03 2009 *)
    Select[Range[600],Mod[Fibonacci[#+1],#]==0&] (* Harvey P. Dale, Feb 24 2025 *)
  • PARI
    is(n)=((Mod([1,1;1,0],n))^n)[1,1]==0 \\ Charles R Greathouse IV, Feb 03 2014

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

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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=', ')
    

A159053 Indices n such that Fibonacci(n-3) is a multiple of n.

Original entry on oeis.org

1, 3, 87, 123, 143, 183, 267, 303, 327, 447, 483, 543, 687, 723, 807, 843, 1023, 1047, 1167, 1203, 1227, 1263, 1347, 1383, 1527, 1563, 1623, 1707, 1763, 1803, 1923, 1983, 2103, 2127, 2283, 2307, 2427, 2463, 2487, 2643, 2691, 2703, 2787, 2823, 3027, 3063
Offset: 1

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Examples

			Fibonacci(84)/87=160500643816367088/87=1844834986395024, so 87 is in the sequence. Fibonacci(120)/123=5358359254990966640871840/123 = 43563896382040379194080 so 123 is in the sequence.
		

Crossrefs

Programs

  • Mathematica
    lst={};Do[If[Mod[(Fibonacci[n-3]),n]==0,AppendTo[lst,n]],{n,3*7!}];lst
    Select[Range[3100],Mod[Fibonacci[#-3],#]==0&] (* Harvey P. Dale, Jul 18 2024 *)

Extensions

Comments edited by R. J. Mathar, Apr 05 2009

A159054 Members of A159053 which are not multiples of 3.

Original entry on oeis.org

1, 143, 1763, 5183, 10153, 10403, 15109, 36863, 40183, 79523, 97343, 130273
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Mathematica
    lst={};Do[If[Mod[(Fibonacci[n-3]),n]==0,If[Mod[n,3]!=0,AppendTo[lst,n]]], {n,5*8!}];lst

Formula

A159053 INTERSECT A001651. - R. J. Mathar, Apr 05 2009

Extensions

Definition reworded by R. J. Mathar, Apr 05 2009
Showing 1-7 of 7 results.