A001602 -id:A001602 - cp's OEIS frontend

A114419 a(n) is the smallest number k such that Fibonacci(k) is a multiple of primorial(n).

3, 12, 60, 120, 120, 840, 2520, 2520, 2520, 2520, 2520, 47880, 47880, 526680, 1053360, 3160080, 91642320, 91642320, 1557919440, 1557919440, 57643019280, 749359250640, 749359250640, 749359250640, 5245514754480, 26227573772400

Offset: 1

Shyam Sunder Gupta, Feb 12 2006

nonn

Because the Fibonacci numbers form a divisibility sequence, each term of this sequence is a multiple of the previous term. The multiple can be computed using A001602. - T. D. Noe, May 04 2009

			a(2)=12 because 12th Fibonacci number (144) is the smallest Fibonacci number which is a multiple of primorial(2), i.e., 6.

Zak Seidov, Table of n, a(n) for n = 1..100
Index to divisibility sequences

Cf. A002110, A000045.

Cf. A267095.

a(n) = {prn = prod(k=1, n, prime(k)); k = 1; while(fibonacci(k) % prn, k++); k;} \\ Michel Marcus, Jan 13 2016

a(n) = {min j: A002110(n) | A000045(j)}. - R. J. Mathar, Jan 31 2008

a(n) = lcm(A001602(1),...,A001602(n)). - T. D. Noe, May 04 2009

a(1) corrected and a(14) added by R. J. Mathar, Jan 31 2008
a(14)-a(18) from Donovan Johnson, Sep 03 2008
Extended by T. D. Noe, May 04 2009

A116514 a(1) = 1; thereafter a(n) = (p - (5|p)) divided by the smallest m such that p divides Fibonacci(m), where p is the n-th prime and (5|p) is the Legendre symbol.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 3, 2, 1, 4, 1, 1, 2, 1, 1, 8, 2, 2, 1, 3, 4, 6, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 1, 2, 1, 2, 2, 9, 5, 1, 1, 2, 18, 1, 2, 1, 2, 3, 4, 1, 2, 10, 1, 2, 7, 1, 2, 2, 3, 2, 3, 2, 6, 1, 1, 2, 1, 1, 4, 2, 4, 2, 1, 20, 1, 2, 1, 1, 2, 2, 10, 1, 1, 1, 1, 1, 1, 1, 2, 20, 1, 6, 1, 18, 3

Offset: 1

Nick Krempel, Mar 24 2006

Lucas showed that A001602 divides p-1 or p+1, according as (5|p) = 1 or -1 respectively. This is the quotient.

			a(6) = 2, as 13 is the 6th prime, 5 is not a quadratic residue mod 13, 13 first occurs as a prime factor of Fibonacci(7) and (13 - (-1)) / 7 = 2.

Patrick McKinley, Table of n, a(n) for n = 1..10000

Cf. A001602.

a(n) = (prime(n) - (5|prime(n))) / A001602(n).

a(1)=1 added by N. J. A. Sloane, Dec 07 2020

A262708 a(n) = p-(p/5) where p = prime(n) and (p/5) is a Legendre symbol.

Original entry on oeis.org

8, 10, 14, 18, 18, 24, 28, 30, 38, 40, 44, 48, 54, 58, 60, 68, 70, 74, 78, 84, 88, 98, 100, 104, 108, 108, 114, 128, 130, 138, 138, 148, 150, 158, 164, 168, 174, 178, 180, 190, 194, 198, 198, 210, 224, 228, 228, 234, 238, 240, 250, 258, 264, 268, 270, 278, 280

Offset: 4

Shane Findley, Sep 27 2015

nonn

The sequence lists Fibonacci indices q that are conjectured to produce Fibonacci numbers divisible by p^2, where p is a Fibonacci-Wieferich prime.

			For n=4, prime(4)=7, and a(4)=8.

Paulo Ribenboim, My Numbers, My Friends, Springer-Verlag, 2000.
Steven Vajda, Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications, Dover. (See p. 73.)

U. Alfred, On the form of primitive factors of Fibonacci numbers, Volume 1, Fibonacci Quarterly, vol 1 (1963), page 1.
Andreas-Stephan Elsenhans, The Fibonacci sequence modulo p^2., pages 1-6.
Shane Findley, Discussion of Sequence A262708.
Richard J. McIntosh and Eric L. Roettger, A search for Fibonacci-Wieferich and Wolstenholme primes, Mathematics of Computation, vol 76 (260), Oct 2007.
John Vinson, The Relation of the Period Modulo m to the Rank of Apparition of m in the Fibonacci Sequence, Fibonacci Quarterly, vol 1 (1963), pages 37-45.

Cf. A001602, A051694.

Table[Prime@ n - JacobiSymbol[Prime@ n, 5], {n, 4, 60}] (* Michael De Vlieger, Oct 04 2015 *)

lista(nn)=forprime(p=3, nn, print1(p-kronecker(p, 5), ", ");); \\ Michel Marcus, Sep 29 2015

Edited by N. J. A. Sloane, Sep 29 2015

Edited by Jon E. Schoenfield, Oct 09 2015

A266587 Smallest index of a Lucas number (A000032) that is divisible by prime(n), if it exists, or 0 if it does not exist (for n > 1).

Original entry on oeis.org

0, 2, 0, 4, 5, 0, 0, 9, 12, 7, 15, 0, 10, 22, 8, 0, 29, 0, 34, 35, 0, 39, 42, 0, 0, 25, 52, 18, 0, 0, 64, 65, 0, 23, 0, 25, 0, 82, 84, 0, 89, 45, 95, 0, 0, 11, 21, 112, 114, 57, 0, 119, 60, 125, 0, 44, 0, 135, 0, 14, 142, 0, 22, 155, 0, 0, 55, 0, 58, 87, 0, 179, 184, 0, 189, 192, 0, 0, 50, 102, 209, 0, 215, 0, 219, 222, 112, 0, 23, 232, 234, 239, 244, 245

Offset: 1

Richard R. Forberg, Jan 01 2016

nonn

These a(n) values may be called Lucas "entry points".
a(1)=0, corresponding to prime(1)= 2, must be viewed as a special case here, because 0, in this case only, is the correct smallest index.
For all other zeros in a(n) (e.g., for a(3), a(6), a(7) corresponding to prime(3)=5, prime(6)=13, prime(7)=17), there are no Lucas numbers divisible by these primes.  The full set of primes not divisible into any Lucas number are given by A053028.
For a(n) > 0, a(n) is always equal to either (prime(n)-1)/k or (prime(n)+1)/k, for some k >= 1. This is similar to the Fibonacci entry points given by A001602.
For each value of a(n) > 0 there is an infinite, periodic subsequence within the Lucas numbers divisible by prime(n), given by Lucas(a(n) + 2*i*a(n)), for all i >= 0. Again this is analogous to the A001602 for Fibonacci. The periodicity of the subsequence is 2*a(n), twice the entry point vs. equal to the entry point for Fibonacci.
Conjecture: Infinite Lucas subsequences divisible by the powers of odd primes, p(n), for which a(n) > 0, are given by:
Lucas(a(n) + a(n)*(p(n)-1)*(Sum_{j=1..m-1} p(n)^(j-1)) + 2*i*a(n)*p(n)^(m-1)) is divisible by p(n)^m, where p(n) > 2, i >= 0, m > 1.
Note: the formula above also works for m=1 if the "Sum" is assumed to be zero when the upper summation index limit is less than initial summation index. See second Mathematica example below which works in this way for all m >= 1, to demonstrate the rule for p = p(n) with corresponding a = a(n).
The divisibility of Lucas numbers by powers of 2 is limited to 2^1 and 2^2, as follows: Lucas(3*i) is divisible by 2 and Lucas(3+6i) is divisible by 4, for all i >= 0.

			For prime(10) = 29, we get a(10) = 7, because Lucas(7)= 29 is the first Lucas number divisible by 29. Also note 7 = (29-1)/4.
For prime(11) = 31, we get a(11) = 15, because Lucas(15) = 1364 is the first Lucas number divisible by 31. Also note 15 = (31-1)/2.

Cf. A000032, A001602, A053028.

Essentially the same as A194363.

result={}; Do[iresult=0; Do[If[Divisible[LucasL[i], Prime[k]], iresult=i; Break[]], {i, 1, 2000}]; AppendTo[result, iresult], {k, 2, 200}]; result
p = 23; a = 12; m = 4; Table[Divisible[LucasL[a + a*(p - 1)*Sum[p^(j - 1), {j, 1, m - 1}] + 2a*i*p^(m - 1)], p^m], {i, 1, 100}]

A065107 Start of the permutation of the primes in the order in which p^2 first appears as a factor of a number in the Fibonacci sequence.

Original entry on oeis.org

2, 3, 5, 7, 13, 11, 17, 19, 29, 23, 37, 47, 41, 61, 31, 89

Offset: 1

Len Smiley, Nov 21 2001

Assuming that there are no square primitive factors in the Fibonacci sequence (an open question), then this sequence continues 53, 43, 113, 73, 109, 233, 59, 107, 199, 67, 97, 71, 101, 149, 79, 139, 83, 151, 281, 421, 211, 137, 103, 157, 307, 521. This is obtained by sorting the pairs (prime(n)*A001602(n), prime(n)) by the first position and noting the order of the primes in the second position. - T. D. Noe, Apr 15 2004

Cf. A065069, A065106.

Cf. A001602 (smallest m such that prime(n) divides Fibonacci(m)).

A116515 a(n) = the period of the Fibonacci numbers modulo p divided by the smallest m such that p divides Fibonacci(m), where p is the n-th prime.

Original entry on oeis.org

1, 2, 4, 2, 1, 4, 4, 1, 2, 1, 1, 4, 2, 2, 2, 4, 1, 4, 2, 1, 4, 1, 2, 4, 4, 1, 2, 2, 4, 4, 2, 1, 4, 1, 4, 1, 4, 2, 2, 4, 1, 1, 1, 4, 4, 1, 1, 2, 2, 1, 4, 1, 2, 1, 4, 2, 4, 1, 4, 2, 2, 4, 2, 1, 4, 4, 1, 4, 2, 1, 4, 1, 2, 4, 1, 2, 4, 4, 2, 2, 1, 4, 1, 4, 1, 2, 2, 4, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 4, 2, 2, 1

Offset: 1

Nick Krempel, Mar 24 2006

Conditions on p_n mod 4 and mod 5 restrict possible values of a(n). The unknown (?) case is p = 1 mod 4 and (5|p) = 1, equivalently, p = 1 or 9 mod 20, where {1, 2, 4} all occur.

Number of zeros in fundamental period of Fibonacci numbers mod prime(n). [From T. D. Noe, Jan 14 2009]

			a(4) = 2, as 7 is the 4th prime, the Fibonacci numbers mod 7 have period 16, the first Fibonacci number divisible by 7 is F(8) = 21 = 3*7 and 16 / 8 = 2.
One period of the Fibonacci numbers mod 7 is 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, which has two zeros. Hence a(4)=2. [From _T. D. Noe_, Jan 14 2009]

Cf. A060305, A001602.

Cf. A112860, A053027, A053028 (primes producing 1, 2 and 4 zeros) [From T. D. Noe, Jan 14 2009]

a(n) = A060305(n) / A001602(n). a(n) is always one of {1, 2, 4}.

a(n) = A001176(prime(n)) [From T. D. Noe, Jan 14 2009]

A117517 Numbers k such that F(2*k + 1) is prime where F(m) is a Fibonacci number.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 11, 14, 21, 23, 41, 65, 68, 179, 215, 216, 224, 254, 284, 285, 1485, 2361, 2693, 4655, 4838, 7215, 12780, 15378, 17999, 18755, 25416, 40919, 52455, 65010, 74045, 100553, 198689, 216890, 295020, 296844, 302355, 465758, 524948, 642803, 818003, 901529, 984360, 1452176

Offset: 1

Parthasarathy Nambi, Apr 26 2006

nonn

For F(k) to be prime, with k > 4, it is necessary but not sufficient for k to be prime. Hence after F(4) = 3, every prime F(m) is of the form F(2*k+1) for some k. Every prime divides some Fibonacci number. See also comment to A093062. - Jonathan Vos Post, Apr 29 2006

			If k=68 then F(2*k + 1) = 19134702400093278081449423917, a prime, so 68 is a term.

H. Dubner and W. Keller, New Fibonacci and Lucas Primes, Math. Comp. 68 (1999) 417-427.

Cf. A000045, A001605, A117595.

Cf. A001602, A022307, A030427, A051694, A075737, A083668, A099000.

[n: n in [0..1000] | IsPrime(Fibonacci(2*n+1))]; // Vincenzo Librandi, May 24 2016

Select[Range[0, 5000], PrimeQ[Fibonacci[2 # + 1]] &] (* Vincenzo Librandi, May 24 2016 *)

a(n) = (A083668(n)-1)/2. - R. J. Mathar, Jul 08 2009

a(n) = (A001605(n+1)-1)/2, n > 1. - Vincenzo Librandi, May 24 2016

More terms from Vincenzo Librandi, May 24 2016

A175026 Fibonacci entry points: a(n) = smallest m such that prime(A075702(n)) divides Fibonacci(m).

Original entry on oeis.org

432, 127, 1426, 10488, 63221, 1328, 11136, 1291186

Offset: 1

Zak Seidov, Nov 03 2009

nonn

In all cases, a(n) is one of divisors of (A075702(n)):
{2160,3048,27094,251712,505768,936240,2182656,2582372}/
{432,127,1426,10488,63221,1328,11136,1291186} = {5,24,19,24,8,705,196,2}.
This is used in Mathematica code for faster search.

			a(1)=432 because A075702(1)=2160=5*432, prime(2160)=19009, and F(432)/19009= 45104130506533126693784341438185160821786395872599778181861900641867287643757057395776.

Cf. A001177, A001602, A075702.

s={2160,3048,27094,251712,505768,936240,2182656,2582372};
Do[sk=s[[k]]; dv=Divisors[sk]; i=2; While[Mod[Fibonacci[dvi=dv[[i]]],Prime[sk]]>0,i++ ]; Print[dvi], {k,8}]

a(n)=A001602(A075702(n)).

A264008 Index of the smallest Fibonacci number divisible by prime(n)^2.

Original entry on oeis.org

6, 12, 25, 56, 110, 91, 153, 342, 552, 406, 930, 703, 820, 1892, 752, 1431, 3422, 915, 4556, 4970, 2701, 6162, 6972, 979, 4753, 5050, 10712, 3852, 2943, 2147, 16256, 17030, 9453, 6394, 5513, 7550, 12403, 26732, 28056, 15051, 31862, 16290, 36290, 18721, 19503, 4378, 8862, 49952, 51756, 26106, 3029, 56882, 28920

Offset: 1

R. J. Mathar, Oct 31 2015

Robert Israel, Table of n, a(n) for n = 1..3584

Cf. A065069, A065106.

f:= proc(n) local p, phi,q,k,G,Fkm,Fk,M,W,m;
  p:= ithprime(n);
  if member(p mod 5, {1,4}) then
    phi:= rhs(op(msolve(x^2-x-1,p^2)[1]));
    q:= -1-phi mod p^2;
    return numtheory:-order(q,p^2);
  fi;
  G:= GF(p,2,alpha^2-alpha-1);
  q:= G:-ConvertIn(-1-alpha);
  k:= G:-order(q);
  Fkm:= combinat:-fibonacci(k-1) mod p^2;
  Fk:= combinat:-fibonacci(k) mod p^2;
  M:= <|>;
  W:= <0,1>;
  for m from 1 do
     W:= M . W mod p^2;
     if W[1] = 0 then return(m*k) fi
  od:
end proc:
f(3):= 25:
map(f, [$1..100]); # Robert Israel, Jan 04 2018

a(n) = if(n==3, 25, my(p=prime(n)); fordiv(p^2-1, d, if(fibonacci(d)%p==0, return(d*p)))); \\ Altug Alkan, Oct 31 2015

a(n) = prime(n)*A001602(n).

a(n) = min{i: A001248(n) | A000045(i)}

A289586 Numbers k whose smallest multiple that is a Fibonacci number is Fibonacci(k).

Original entry on oeis.org

1, 5, 12, 25, 60, 125, 300, 625, 1500, 3125, 7500, 15625, 37500, 78125, 187500, 390625, 937500, 1953125, 4687500, 9765625, 23437500, 48828125, 117187500, 244140625, 585937500, 1220703125, 2929687500, 6103515625, 14648437500, 30517578125, 73242187500, 152587890625, 366210937500

Offset: 1

Jon E. Schoenfield, Aug 06 2017

Alternative names:
Numbers k such that Fibonacci(k) is the smallest positive Fibonacci number that is divisible by k.
Numbers that are their own Fibonacci entry points.
Numbers k such that k = A001177(k).
Numbers that are either a power of 5 or 12 times a power of 5. - Robert Israel, Aug 07 2017

			Fibonacci(25) = 75025 = 25*3001 is the smallest Fibonacci number that is divisible by 25, so 25 is in the sequence.
Although Fibonacci(24) = 46368 = 24*1932 is divisible by 24, it is not the smallest Fibonacci number that is divisible by 24, so 24 is not in the sequence.

Robert Israel, Table of n, a(n) for n = 1..2858
Index entries for linear recurrences with constant coefficients, signature (0,5).

Subsequence of A023172 ("Self-Fibonacci numbers").
Cf. A000045, A001177, A000351 (bisection), A216491 (bisection)
(Cf. A001602 for a different definition of "Fibonacci entry point".)

1,seq(op([5^k,12*5^(k-1)]), k=1..100); # Robert Israel, Aug 07 2017

From Robert Israel, Aug 07 2017: (Start)
a(2*k) = 5^k for k >= 1.
a(2*k-1) = 12*5^(k-2) for k >= 2.
G.f.: (1+5*x+7*x^2)/(1-5*x^2). (End)

More terms from Robert Israel, Aug 07 2017

cp's OEIS Frontend

A114419 a(n) is the smallest number k such that Fibonacci(k) is a multiple of primorial(n).

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A116514 a(1) = 1; thereafter a(n) = (p - (5|p)) divided by the smallest m such that p divides Fibonacci(m), where p is the n-th prime and (5|p) is the Legendre symbol.

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A262708 a(n) = p-(p/5) where p = prime(n) and (p/5) is a Legendre symbol.

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A266587 Smallest index of a Lucas number (A000032) that is divisible by prime(n), if it exists, or 0 if it does not exist (for n > 1).

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Mathematica

A065107 Start of the permutation of the primes in the order in which p^2 first appears as a factor of a number in the Fibonacci sequence.

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A116515 a(n) = the period of the Fibonacci numbers modulo p divided by the smallest m such that p divides Fibonacci(m), where p is the n-th prime.

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A117517 Numbers k such that F(2*k + 1) is prime where F(m) is a Fibonacci number.

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A175026 Fibonacci entry points: a(n) = smallest m such that prime(A075702(n)) divides Fibonacci(m).

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