Search: a214031 -id:a214031
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A001177
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Fibonacci entry points: a(n) = least k >= 1 such that n divides Fibonacci number F_k (=A000045(k)).
(Formerly M2314 N0914)
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+10
75
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1, 3, 4, 6, 5, 12, 8, 6, 12, 15, 10, 12, 7, 24, 20, 12, 9, 12, 18, 30, 8, 30, 24, 12, 25, 21, 36, 24, 14, 60, 30, 24, 20, 9, 40, 12, 19, 18, 28, 30, 20, 24, 44, 30, 60, 24, 16, 12, 56, 75, 36, 42, 27, 36, 10, 24, 36, 42, 58, 60, 15, 30, 24, 48, 35, 60, 68, 18, 24, 120
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OFFSET
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1,2
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COMMENTS
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In the formula, the relation a(p^e) = p^(e-1)*a(p) is called Wall's conjecture, which has been verified for primes up to 10^14. See A060305. Primes for which this relation fails are called Wall-Sun-Sun primes. - T. D. Noe, Mar 03 2009
All solutions to F_m == 0 (mod n) are given by m == 0 (mod a(n)). For a proof see, e.g., Vajda, p. 73. [Old comment changed by Wolfdieter Lang, Jan 19 2015]
If p is a prime of the form 10n +- 1 then a(p) is a divisor of p-1. If q is a prime of the form 10n +- 3 then a(q) is a divisor of q+1. - Robert G. Wilson v, Jul 07 2007
Definition 1 in Riasat (2011) calls this k(n), or sometimes just k. Corollary 1 in the same paper, "every positive integer divides infinitely many Fibonacci numbers," demonstrates that this sequence is infinite. - Alonso del Arte, Jul 27 2013
If p is a prime then a(p) <= p+1. This is because if p is a prime then exactly one of the following Fibonacci numbers is a multiple of p: F(p-1), F(p) or F(p+1). - Dmitry Kamenetsky, Jul 23 2015
From Renault 1996:
1. a(lcm(n,m)) = lcm(a(n), a(m)).
2. if n|m then a(n)|a(m).
3. if m has prime factorization m=p1^e1 * p2^e2 * ... * pn^en then a(m) = lcm(a(p1^e1), a(p2^e2), ..., a(pn^en)). - Dmitry Kamenetsky, Jul 23 2015
a(n)=n if and only if n=5^k or n=12*5^k for some k >= 0 (see Marques 2012). - Dmitry Kamenetsky, Aug 08 2015
Every positive integer (except 2) eventually appears in this sequence. This is because every Fibonacci number bigger than 1 (except Fibonacci(6)=8 and Fibonacci(12)=144) has at least one prime factor that is not a factor of any earlier Fibonacci number (see Knott reference). Let f(n) be such a prime factor for Fibonacci(n); then a(f(n))=n. - Dmitry Kamenetsky, Aug 08 2015
We can reconstruct the Fibonacci numbers from this sequence using the formula Fibonacci(n+2) = 1 + Sum_{i: a(i) <= n} phi(i)*floor(n/a(i)), where phi(n) is Euler's totient function A000010 (see the Stroinski link). For example F(6) = 1 + phi(1)*floor(4/a(1)) + phi(2)*floor(4/a(2)) + phi(3)*floor(4/a(4)) = 1 + 1*4 + 1*1 + 2*1 = 8. - Peter Bala, Sep 10 2015
a(F_m) = m for all m > 1. Indeed, let (b(j)) be defined by b(1)=b(2)=1, and b(j+2) = (b(j) + b(j+1)) mod n. Then a(n) equals the index of the first occurrence of 0 in (b(j)). Example: if n=4 then b = A079343 = 1,1,2,3,1,0,1,1,..., so a(4)=6. If n is a Fibonacci number n=F_m, then obviously a(n)=m. Note that this gives a simple proof of the fact that all integers larger than 2 occur in (a(n)). - Michel Dekking, Nov 10 2017
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REFERENCES
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A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 25.
B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers. Report ORNL-4261, Oak Ridge National Laboratory, Oak Ridge, Tennessee, June 1968.
Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Afterword by Herbert A. Hauptman, Nobel Laureate, 2. 'The Minor Modulus m(n)', Prometheus Books, NY, 2007, page 329-342.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
N. N. Vorob'ev, Fibonacci numbers, Blaisdell, NY, 1961.
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LINKS
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FORMULA
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a(n) = n if and only if n is of form 5^k or 12*5^k (proved in Marques paper), a(n) = n - 1 if and only if n is in A106535, a(n) = n + 1 if and only if n is in A000057, a(n) = n + 5 if and only if n is in 5*A000057, ... - Benoit Cloitre, Feb 10 2007
a(1) = 1, a(2) = 3, a(4) = 6 and for e > 2, a(2^e) = 3*2^(e-2); a(5^e) = 5^e; and if p is an odd prime not 5, then a(p^e) = p^max(0, e-s)*a(p) where s = valuation(A000045(a(p)), p) (Wall's conjecture states that s = 1 for all p). If (m, n) = 1 then a(m*n) = lcm(a(m), a(n)). See Posamentier & Lahmann. - Robert G. Wilson v, Jul 07 2007; corrected by Max Alekseyev, Oct 19 2007, Jun 24 2011
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EXAMPLE
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a(4) = 6 because the smallest Fibonacci number that 4 divides is F(6) = 8.
a(5) = 5 because the smallest Fibonacci number that 5 divides is F(5) = 5.
a(6) = 12 because the smallest Fibonacci number that 6 divides is F(12) = 144.
a(2) = 3, hence 2 | F(m) iff m = 2*k, for k >= 0;
a(3) = 4, hence 3 | F(m) iff m = 4*k, for k >= 0;
etc. See a comment above with the Vajda reference.
(End)
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MAPLE
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for k from 1 do
if combinat[fibonacci](k) mod n = 0 then
return k;
end if;
end do:
N:= 1000: # to get a(1) to a(N)
L:= ilcm($1..N):
count:= 0:
for n from 1 while count < N do
fn:= igcd(L, combinat:-fibonacci(n));
divs:= select(`<=`, numtheory:-divisors(fn), N);
for d in divs do if not assigned(A[d]) then count:= count+1; A[d]:= n fi od:
od:
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MATHEMATICA
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fibEntry[n_] := Block[{k = 1}, While[ Mod[ Fibonacci@k, n] != 0, k++ ]; k]; Array[fibEntry, 74] (* Robert G. Wilson v, Jul 04 2007 *)
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PROG
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(PARI) a(n)=if(n<0, 0, s=1; while(fibonacci(s)%n>0, s++); s) \\ Benoit Cloitre, Feb 10 2007
(PARI) ap(p)=my(k=p+[0, -1, 1, 1, -1][p%5+1], f=factor(k)); for(i=1, #f[, 1], for(j=1, f[i, 2], if((Mod([1, 1; 1, 0], p)^(k/f[i, 1]))[1, 2], break); k/=f[i, 1])); k
a(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e14, ap(f[i, 1]^f[i, 2]), ap(f[i, 1])*f[i, 1]^(f[i, 2]-1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<<max(f[1, 2]-2, 1)); lcm(v) \\ Charles R Greathouse IV, May 08 2017
(Haskell)
a001177 n = head [k | k <- [1..], a000045 k `mod` n == 0]
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CROSSREFS
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Various derived sequences:
Analogous sequence for the tribonacci numbers: A046737, for Lucas numbers: A223486, for Pell numbers: A214028.
Cf. also A000057, A106535, A120255, A120256, A175026, A213648, A214031, A214781, A214783, A230359, A233283, A233285, A233287. (End)
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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17, 67, 109, 137, 181, 191, 229, 233, 239, 257, 283, 307, 311, 349, 353, 359, 479, 523, 547, 593, 599, 617, 643, 709, 719, 829, 839, 657, 883, 907, 911, 953, 977, 1021, 1031, 1069, 1097, 1123, 1151, 1193, 1217, 1319, 1433, 1439, 1483
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OFFSET
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1,1
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COMMENTS
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This sequence is to A214030 as A000057 is to A001177. It would be nice to have an interpretation of this sequence akin to the interpretation of A000057 as the set of primes which divide all Fibonacci sequences, having arbitrary initial values for a(1),a(2). The linearly recursive sequence which seems to be associated to this is 3*f(n)=6*f(n-1)+2*f(n-2), but this does not have integral values.
If we use the sequence 3,2,3,2,3,2.. instead of 2,3,2,3,... we end up with the same sequence a(n).
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LINKS
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PROG
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(PARI)
{b23(n)=local(t, m=1, s=[n]); if (n<2, 0, while(1,
if(m%2, s=concat(s, 2), s=concat(s, 3));
t=contfracpnqn(concat(s, n));
t=contfrac(n*t[1, 1]/t[2, 1]);
if(t[1]<n^2||t[#t]<n^2, m++, break)); m)};
/* To print the sequence A214032(n) to the screen, */
for(i=1, 1500, if(b23(i)==i-2, print1(i, ", ")));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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13, 17, 19, 23, 37, 41, 47, 67, 89, 109, 137, 139, 157, 181, 191, 211, 229, 233, 239, 257, 277, 281, 283, 307, 311, 331, 349, 353, 359, 373, 379, 397, 479, 499, 503, 521, 523, 547, 571, 593, 599, 613, 617, 619, 641
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OFFSET
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1,1
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COMMENTS
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It always has been one of the great mysteries of mathematics, that the superdiagonal sequence of A001177 consists of prime numbers A000057.
Here, regarding A214031 and A214032,there is the further conjecture that these two disjoint sequences are primes and roughly comparable in density. It isn't clear that these two sequences have a density, without appealing to the Riemann Hypothesis, but they are certainly close to one another in growing size.
Since these two sequences are disjoint, it is natural to take their union.
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LINKS
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PROG
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(PARI)
{b23(n)=local(t, m=1, s=[n]); if (n<2, 0, while(1,
if(m%2, s=concat(s, 2), s=concat(s, 3));
t=contfracpnqn(concat(s, n));
t=contfrac(n*t[1, 1]/t[2, 1]);
if(t[1]<n^2||t[#t]<n^2, m++, break)); m)};
To print the sequence a(n) to the screen,
for(i=1, 500, if(b23(i)==i||b23(i)==i-2,
print1(i, ", ")));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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