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.

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A060320 Index of smallest Fibonacci number with exactly n distinct prime factors.

Original entry on oeis.org

1, 3, 8, 15, 20, 30, 40, 70, 60, 80, 90, 140, 176, 120, 168, 180, 324, 252, 240, 378, 450, 432, 552, 360, 420, 690, 504, 880, 630, 600, 756, 720, 900, 792, 840, 1296, 1050, 1350, 1140, 1080, 1200, 1824, 1260, 1512, 1320, 1560, 1680
Offset: 0

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Author

Labos Elemer, Mar 28 2001

Keywords

Comments

From Jon E. Schoenfield, Dec 28 2016: (Start)
Note that the presence of incompletely factored Fibonacci numbers with indices as low as 1301 does not prevent the drawing of conclusions such as "a(44) = 1320" with certainly. Using F(1301) as an example, the compact table of Fibonacci results at the Kelly site indicates that F(1301) = p*q*r*c where p=6400921, q=14225131397, r=100794731109596201, and c is a 238-digit unfactored composite number. The complete factorization of every Fibonacci number up to F(1000) is explicitly given elsewhere on the site, and those results allow quick verification that a(n) <= 900 for all n in [0..34], so 1301 cannot be a term unless F(1301) has at least 35 distinct prime factors, which would require c to have at least 32 distinct prime factors, at least one of which would have to be less than ceiling(c^(1/32)) = 26570323, but trial division of c by every prime less than 26570323 shows that c has no prime factors that small. Thus, while A022307(1301) is unknown, it is certain that 1301 is not a term in this sequence. Similarly, making use of known factors, it can be proved that F(n) cannot have 44 or more distinct prime factors for any n < 1320, so since F(1320) has exactly 44 distinct prime factors, it is established that a(44) = 1320. (End)
a(47) >= 2835, a(48..68) = (2040, 1800, 2736, 2730, 1890, 1980, 2520, 2280, 2100, 2160, 2640, 3300, 3060, 3150, 2520, 3120, 3696, 3240, 3990, 3360, 3420), a(69) >= 4400, a(75) = 4320, a(77) = 4200, a(79) = 3780. - Max Alekseyev, Feb 03 2025

Examples

			n=9: F(80) = 23416728348467685 = 3 * 5 * 7 * 11 * 41 * 47 * 1601 * 2161 * 3041.
n=25: F(690) = 2^3 * 5 * 11 * 31 * 61 * 137 * 139 * 461 * 691 * 829 * 1151 * 1381 * 4831 * 5981 * 18077 * 28657 * 186301 * 324301 * 686551 * 1485571 * 4641631 * 117169733521 * 2441738887963981 * 3490125311294161 * 25013864044961447973152814604981 is the smallest Fibonacci number with exactly 25 distinct prime factors.
		

Crossrefs

Programs

  • Mathematica
    First /@ SortBy[#, Last] &@ Map[First@ # &, Values@ GroupBy[#, Last]] &@ Table[{n - Boole[n == 2], #, PrimeNu@ #} &@ Fibonacci@ n, {n, 2, 300}] (* Michael De Vlieger, Feb 18 2017, Version 10 *)
    Module[{ff=Table[{n,PrimeNu[Fibonacci[n]]},{n,1400}]},Table[ SelectFirst[ ff,#[[2]]==k&],{k,0,40}]][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 28 2018 *)
  • PARI
    my(o=[],s); print1(1); for(n=1,20, s=0; until( o[s]==n, #o
    				

Formula

a(n) = min (k : A022307(k) = n).

Extensions

Corrected by Shyam Sunder Gupta, Jul 20 2002
Edited by M. F. Hasler, Nov 01 2012
a(35)-a(40), a(42), a(44) computed based on Kelly's data in A022307 by Jon E. Schoenfield, Dec 28 2016
a(41), a(43), a(45)-a(46) from Max Alekseyev, Feb 03 2025

A090206 Nonprime Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 8, 21, 34, 55, 144, 377, 610, 987, 2584, 4181, 6765, 10946, 17711, 46368, 75025, 121393, 196418, 317811, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169
Offset: 1

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Author

Felix Tubiana, Jan 22 2004

Keywords

Comments

It is possible to find a run of at least length n (not necessarily exactly n) such that n consecutive terms in this sequence are also consecutive in the sequence of Fibonacci numbers. However, it is not possible for such a run to be of exactly length n if n is even. - Alonso del Arte, Nov 23 2010
Some terms of this sequence have prime indices in the sequence of Fibonacci numbers (A000045), see A050937. - Alonso del Arte, Aug 16 2013

Examples

			34 is in the sequence as it is a Fibonacci number and it is composite, the product of 2 and 17.
55 is in the sequence as it is a Fibonacci number and it is composite, the product of 5 and 11.
89 is not in the sequence because, although it is a Fibonacci number, it is prime.
		

Crossrefs

Programs

A119984 Numbers k such that Fibonacci(prime(k)) is prime.

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 9, 10, 14, 15, 23, 32, 33, 72, 83, 84, 87, 97, 104, 105, 429, 637, 710, 1152, 1194, 1692, 2814, 3316, 3824, 3971, 5206, 8002, 10016, 12161, 13681, 18069, 33653, 36467, 48355, 48629, 49455, 73574, 82128, 99005, 123685, 135276, 146779, 210404, 233207, 239581
Offset: 1

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Author

Alexander Adamchuk, Aug 04 2006

Keywords

Comments

All prime Fibonacci numbers have prime indices, except prime F(4) = 3; a(n) is such that Fibonacci(prime(a(n))) is prime. - Robert G. Wilson v, Aug 05 2006

Crossrefs

Programs

  • Mathematica
    Select[ Range@3000, PrimeQ@ Fibonacci@ Prime@ # &] (* Robert G. Wilson v, Aug 05 2006 *)

Formula

a(n) = pi(A001605(n+1)). This holds for all n including n=1, since pi(4) = pi(3) = 2. - Jens Kruse Andersen, Jul 24 2014

Extensions

a(21)-a(27) from Robert G. Wilson v, Aug 05 2006
More terms (from A001605) from T. D. Noe, Aug 18 2006
a(42)-a(48) (from A001605, found by Henri Lifchitz) from Jens Kruse Andersen, Jul 24 2014
a(49)-a(50) (from A001605, found by Henri Lifchitz) from Amiram Eldar, Sep 01 2019

A338762 Greatest prime Fibonacci divisor of F(n)^2 + 1 where F(n) is the n-th Fibonacci number, or 0 if no such prime factor exists.

Original entry on oeis.org

2, 2, 5, 5, 13, 13, 5, 13, 89, 89, 233, 233, 89, 233, 1597, 1597, 5, 1597, 1597, 13, 28657, 28657, 13, 28657, 28657, 5, 514229, 514229, 2, 514229, 514229, 89, 13, 89, 89, 13, 233, 233, 0, 233, 433494437, 433494437, 5, 433494437, 2971215073, 2971215073, 13, 2971215073
Offset: 1

Views

Author

Michel Lagneau, Nov 07 2020

Keywords

Comments

a(n) = 0 for n = 39, 60, 69, 72, ... .
a(5385) has 1126 decimal digits. - Chai Wah Wu, Nov 19 2020

Examples

			a(6) = 13 because F(6)^2 + 1 = 8^2 + 1 = 65 = 5*13 and 13 is the greatest prime Fibonacci divisor.
		

Crossrefs

Programs

  • Maple
    a:= proc(n) local F, m, t; F, m, t:=
          [1, 2], 0, (<<0|1>, <1|1>>^n)[2, 1]^2+1;
          while F[2]<=t do if isprime(F[2]) and irem(t, F[2])=0
            then m:=F[2] fi; F:= [F[2], F[1]+F[2]]
          od; m
        end:
    seq(a(n), n=1..50);  # Alois P. Heinz, Nov 07 2020
  • Mathematica
    a[n_] := Module[{F, m, t}, F = {1, 2}; m = 0; t = MatrixPower[{{0, 1}, {1, 1}}, n][[2, 1]]^2 + 1; While[F[[2]] <= t, If[PrimeQ[F[[2]]] && Mod[t, F[[2]]] == 0, m = F[[2]]]; F = {F[[2]], F[[1]] + F[[2]]}]; m];
    Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Feb 09 2025, after Alois P. Heinz *)
  • PARI
    isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
    a(n) = my(f=factor(fibonacci(n)^2+1)[,1]~, v=select(x->isfib(x), f)); if (#v, vecmax(v), 0); \\ Michel Marcus, Nov 07 2020

A245236 Numbers n such that the Fibonacci number F(n) satisfies F(n)^2+1 = f1*f2 where f1, f2 are prime Fibonacci numbers.

Original entry on oeis.org

4, 5, 6, 9, 12, 15, 45, 432, 570
Offset: 1

Views

Author

Michel Lagneau, Jul 14 2014

Keywords

Comments

Or index i of any Fibonacci number F(i) such that F(i-1) and F(i+1) are primes if i is even or F(i-2) and F(i+2) are primes if i is odd where F(i) is the i-th Fibonacci number.
In the general case, F(i+1)*F(i-1) = F(i)^2 + 1 if i even or F(i+2)*F(i-2) = F(i)^2 + 1 if i odd (Cassini’s identity).
The corresponding Fibonacci numbers are 3, 5, 8, 34, 144, 610, 1134903170,...
If a(10) exists, it is greater than 30000. - Robert Israel, Jul 14 2014

Examples

			4 is a term because F(4)^2+1 = F(3)*F(5)=> 3^2+1 = 2*5;
5 is a term because F(5)^2+1 = F(3)*F(7)=> 5^2+1 = 2*13;
6 is a term because F(6)^2+1 = F(5)*F(7)=> 8^2+1 = 5*13;
9 is a term because F(9)^2+1 = F(7)*F(11)=> 34^2+1 = 13*89;
12 is a term because F(12)^2+1 = F(11)*F(13)=> 144^2+1 = 89*233;
15 is a term because F(13)*F(17)=> 610^2+1 = 233* 1597.
		

Crossrefs

Programs

  • Maple
    with(combinat,fibonacci):with(numtheory):nn:=1000:for n from 1 to nn do:if (type(fibonacci(n+1),prime) and type(fibonacci(n-1),prime) and irem(n,2)=0) or (type(fibonacci(n+2),prime) and type(fibonacci(n-2),prime) and irem(n,2)=1) then print(n):else fi:od:
    # Alternative:
    filter:= proc(n) uses combinat;
        if n::even then isprime(n-1) and isprime(n+1) and isprime(fibonacci(n-1)) and isprime(fibonacci(n+1))
      else isprime(n-2) and isprime(n+2) and isprime(fibonacci(n-2)) and isprime(fibonacci(n+2))
    fi end proc:
    select(filter, [$1..10^4]); # Robert Israel, Jul 14 2014

A080327 Numbers k for which Lucas(k) and Fibonacci(k) are both prime.

Original entry on oeis.org

4, 5, 7, 11, 13, 17, 47, 148091
Offset: 1

Views

Author

T. D. Noe, Feb 15 2003

Keywords

Comments

The intersection of A001605 and A001606. Fibonacci(148091) and Lucas(148091) are probable primes.
Corresponding Fibonacci-Lucas prime twins are listed in A121533. Corresponding Lucas-Fibonacci prime twins are listed in A121534. Fibonacci(148091) and Lucas(148091) are probable Fibonacci-Lucas and Lucas-Fibonacci prime twins. They have 30949 and 30950 digits. - Alexander Adamchuk, Aug 05 2006
Heuristically, this sequence is finite. It is quite probable, but presently unprovable, that it is now complete. - David Broadhurst, Jun 25 2008
Western Number Theory problem 007:13 by Gary Walsh asks to prove that a(8) = 148091 is in this sequence. - Charles R Greathouse IV, May 21 2014

References

  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 246.

Crossrefs

Programs

  • Mathematica
    Select[Range[0, 100], PrimeQ[Fibonacci[#]] && PrimeQ[LucasL[#]] & ] (* Robert Price, May 27 2019 *)
  • PARI
    is(n)=isprime(n) && ispseudoprime(fibonacci(n)) && ispseudoprime(fibonacci(n-1)+fibonacci(n+1)) \\ Charles R Greathouse IV, May 21 2014

A121534 Lucas-Fibonacci prime twins: Prime Lucas numbers Lucas(k) such that Fibonacci numbers Fibonacci(k) are also prime.

Original entry on oeis.org

7, 11, 29, 199, 521, 3571, 6643838879
Offset: 1

Views

Author

Alexander Adamchuk, Aug 05 2006

Keywords

Comments

Indices for Lucas-Fibonacci prime twins are A080327(n). Corresponding Fibonacci-Lucas prime twins are A121533(n). Probable primes Fibonacci(148091) and Lucas(148091) are the next probable Fibonacci-Lucas and Lucas-Fibonacci prime twins. They have 30949 and 30950 digits.

Examples

			a(1) = 7 because Lucas(4) = 7 is prime and Fibonacci(4) = 3 is prime too.
		

Crossrefs

Programs

  • Mathematica
    Do[f=Fibonacci[n]; l=Fibonacci[n-1]+Fibonacci[n+1]; If[PrimeQ[f]&&PrimeQ[l], Print[{f,l}]], {n,10000}]
    nn=1000;Transpose[Select[Thread[{Fibonacci[Range[nn]], LucasL[ Range[nn]]}],And@@PrimeQ[#]&]][[2]] (* Harvey P. Dale, Jul 08 2011 *)

Extensions

a(1) and example corrected by Harvey P. Dale, Jul 08 2011

A303216 A(n,k) is the n-th Fibonacci number with exactly k prime factors (counted with multiplicity); square array A(n,k), n>=1, k>=1, read by antidiagonals.

Original entry on oeis.org

2, 21, 3, 8, 34, 5, 6765, 610, 55, 13, 2584, 196418, 987, 377, 89, 144, 701408733, 317811, 10946, 4181, 233, 832040, 102334155, 1134903170, 2178309, 75025, 17711, 1597, 86267571272, 267914296, 12586269025, 365435296162, 32951280099, 3524578, 121393, 28657
Offset: 1

Views

Author

Alois P. Heinz, Apr 19 2018

Keywords

Examples

			Square array A(n,k) begins:
    2,    21,       8,         6765,           2584,                 144, ...
    3,    34,     610,       196418,      701408733,           102334155, ...
    5,    55,     987,       317811,     1134903170,         12586269025, ...
   13,   377,   10946,      2178309,   365435296162,      10610209857723, ...
   89,  4181,   75025,  32951280099,  6557470319842,    2111485077978050, ...
  233, 17711, 3524578, 139583862445, 72723460248141, 7540113804746346429, ...
		

Crossrefs

Columns k=1-2 give: A005478, A053409.
Row n=1 gives A072397.

Programs

  • Maple
    F:= combinat[fibonacci]: with(numtheory):
    A:= proc() local h, p, q; p, q:= proc() [] end, 2;
          proc(n, k)
            while nops(p(k))
    				
  • Mathematica
    A[n_, k_] := Module[{F = Fibonacci, h, p, q = 2}, p[_] = {}; While[ Length[p[k]] < n, q = q+1; h = PrimeOmega[F[q]]; p[h] = Append[p[h], F[q]]]; p[k][[n]]];
    Table[Table[A[n, 1+d-n], {n, 1, d}], {d, 1, 10}] // Flatten (* Jean-François Alcover, Feb 05 2021, after Alois P. Heinz *)

Formula

A(n,k) = A000045(A303215(n,k)).
A001222(A(n,k)) = k.

A339461 Number of Fibonacci divisors of n^2 + 1.

Original entry on oeis.org

1, 2, 2, 3, 1, 3, 1, 3, 3, 2, 1, 2, 2, 4, 1, 2, 1, 3, 3, 2, 1, 4, 2, 3, 1, 2, 1, 3, 2, 2, 1, 3, 2, 3, 3, 2, 1, 3, 2, 2, 1, 2, 2, 3, 2, 2, 1, 5, 2, 2, 1, 2, 2, 3, 1, 4, 1, 4, 2, 2, 2, 2, 2, 3, 1, 2, 1, 3, 2, 2, 3, 2, 2, 4, 1, 2, 1, 3, 2, 2, 1, 3, 2, 4, 1, 2, 2
Offset: 0

Views

Author

Michel Lagneau, Dec 06 2020

Keywords

Examples

			a(13) = 4 because the divisors of 13^2 + 1 = 170 are {1, 2, 5, 10, 17, 34, 85, 170} with 4 Fibonacci divisors: 1, 2, 5 and 34.
		

Crossrefs

Programs

  • Maple
    with(numtheory):with(combinat,fibonacci):nn:=100:F:={}:
    for k from 1 to nn do:
      F:=F union {fibonacci(k)}:
    od:
       for n from 0 to 90 do:
        f:=n^2+1:d:=divisors(f):
        lst:= F intersect d: n1:=nops(lst):printf(`%d, `,n1):
       od:
  • Mathematica
    Array[DivisorSum[#^2 + 1, 1 &, Or @@ Map[IntegerQ@ Sqrt[#] &, 5 #^2 + 4 {-1, 1}] &] &, 105, 0] (* Michael De Vlieger, Dec 07 2020 *)
  • PARI
    isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || issquare(k-8);
    a(n) = sumdiv(n^2+1, d, isfib(d)); \\ Michel Marcus, Dec 06 2020

Formula

a(A005574(n)) = 1 for n > 2.
a(n) = A005086(A002522(n)). - Michel Marcus, Dec 06 2020

A338794 Indices k of Fibonacci numbers F(k) such that F(k)^2 + 1 has no Fibonacci prime factor.

Original entry on oeis.org

39, 60, 69, 72, 99, 102, 105, 108, 111, 150, 165, 180, 192, 195, 198, 225, 228, 231, 240, 270, 279, 282, 309, 312, 315, 348, 351, 381, 399, 420, 441, 459, 462, 465, 489, 501, 522, 588, 591, 600, 615, 618, 642, 645, 660, 675, 702, 741, 759, 771, 810, 822, 825, 828
Offset: 1

Views

Author

Michel Lagneau, Nov 09 2020

Keywords

Comments

Numbers k such that A338762(k) = 0.

Examples

			39 is in the sequence because F(39)^2 + 1 = 63245986^2 + 1 = 73*149*2221*2789*59369 with no Fibonacci prime factors.
38 is not in the sequence because F(38)^2 + 1 = 39088169^2 + 1 =  2*73*149*233*2221*135721. The numbers and 2, 233 are Fibonacci prime factors.
		

Crossrefs

Programs

  • Maple
    a:= proc(n) local F, m, t; F, m, t:=
          [1, 2], 0, (<<0|1>, <1|1>>^n)[2, 1]^2+1;
          while F[2]<=t do if isprime(F[2]) and irem(t, F[2])=0
            then m:=F[2] fi; F:= [F[2], F[1]+F[2]]
          od; m
        end:
    for n from 1 to 100 do :
    if a(n)=0 then printf(`%d, `,n):else fi:
    od: # program from Alois P. Heinz, adapted for the sequence. See A338762.
  • Mathematica
    A338762[n_] := Module[{F, m, t}, F = {1, 2}; m = 0; t = MatrixPower[{{0, 1}, {1, 1}}, n][[2, 1]]^2 + 1; While[F[[2]] <= t, If[PrimeQ[F[[2]]] && Mod[t, F[[2]]] == 0, m = F[[2]]]; F = {F[[2]], F[[1]] + F[[2]]}]; m];
    Reap[For[k = 1, k <= 1000, k++, If[A338762[k] == 0, Print[k]; Sow[k]]]][[2, 1]] (* Jean-François Alcover, Mar 16 2025, after Alois P. Heinz *)
  • PARI
    isok(n) = {my(i=0, f=0, x=fibonacci(n)^2+1, m=0); while(f < x, i++; f = fibonacci(i); if (ispseudoprime(f) && (x%f) == 0, return (0));); return(1);} \\ Michel Marcus, Nov 13 2020
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