A000057 -id:A000057 - cp's OEIS frontend

A001177 Fibonacci entry points: a(n) = least k >= 1 such that n divides Fibonacci number F_k (=A000045(k)).

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

Offset: 1

N. J. A. Sloane

nonn

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
Conjecture: Sum_{d|n} phi(d)*a(d) = A232656(n). - Logan J. Kleinwaks, Oct 28 2017
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

			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.
From _Wolfdieter Lang_, Jan 19 2015: (Start)
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)

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.

T. D. Noe, Table of n, a(n) for n = 1..10000
A. Allard and P. Lecomte, Periods and entry points in Fibonacci sequence, Fib. Quart. 17 (1) (1979) 51-57.
R. C. Archibald (?), Review of B. H. Hannon and W. L. Morris, Tables of arithmetical functions related to the Fibonacci numbers, Math. Comp., 23 (1969), 459-460.
B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 25.
J. D. Fulton and W. L. Morris, On arithmetical functions related to the Fibonacci numbers, Acta Arithmetica, 16 (1969), 105-110.
Molly FitzGibbons, Steven J. Miller, and Amanda Verga, Dynamics of the Fibonacci Order of Appearance Map, arXiv:2309.14501 [math.NT], 2023.
Ramon Glez-Regueral, An entry-point algorithm for high-speed factorization, Thirteenth Internat. Conf. Fibonacci Numbers Applications, Patras, Greece, 2008.
B. H. Hannon and W. L. Morris, Tables of Arithmetical Functions Related to the Fibonacci Numbers [Annotated and scanned copy]
Ron Knott, The first 300 Fibonacci numbers factorized
Paolo Leonetti and Carlo Sanna, On the greatest common divisor of n and the nth Fibonacci number, arXiv:1704.00151 [math.NT], 2017. See z(n).
Diego Marques, Fixed points of the order of appearance in the Fibonacci sequence, Fibonacci Quart. 50:4 (2012), pp. 346-352.
Diego Marques, The order of appearance of the product of consecutive Lucas numbers, Fibonacci Quarterly, 51 (2013), 38-43.
Diego Marques, Sharper upper bounds for the order of appearance in the fibonacci sequence, Fibonacci Quarterly, 51 (2013), pp. 233-238.
Zuzana Masáková and Edita Pelantová, Midy's Theorem in non-integer bases and divisibility of Fibonacci numbers, arXiv:2401.03874 [math.NT], 2024. See page 9.
Marc Renault, The Fibonacci sequence under various moduli, Masters Thesis, Wake Forest University, 1996.
Samin Riasat, Z[phi] and the Fibonacci Sequence Modulo n, Mathematical Reflections 1 (2011): 1 - 7.
H. J. A. Salle, A Maximum Value for the Rank of Apparition of Integers in Recursive Sequences , Fibonacci Quart. 13.2 (1975) 159-161.
U. Stroinski, Connection between Euler's totient function and Fibonacci numbers, Mathematics Stack Exchange, Feb 17 2015.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly 67 (6) (1960) 525-532.
Eric Weisstein, MathWorld: Wall-Sun-Sun Prime

Cf. A000045, A001175, A001176, A060383, A001602. First occurrence of k is given in A131401. A233281 gives such k that a(k) is a prime.
From Antti Karttunen, Dec 21 2013: (Start)
Various derived sequences:
A047930(n) = A000045(a(n)).
A037943(n) = A000045(a(n))/n.
A217036(n) = A000045(a(n)-1) mod n.
A132632(n) = a(n^2).
A132633(n) = a(n^3).
A214528(n) = a(n!).
A215011(n) = a(A000217(n)).
A215453(n) = a(n^n).
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)

a001177 n = head [k | k <- [1..], a000045 k `mod` n == 0]
-- Reinhard Zumkeller, Jan 15 2014

A001177 := proc(n)
        for k from 1 do
                if combinat[fibonacci](k) mod n = 0 then
                        return k;
                end if;
        end do:
end proc: # R. J. Mathar, Jul 09 2012
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:
seq(A[n],n=1..N); # Robert Israel, Oct 14 2015

fibEntry[n_] := Block[{k = 1}, While[ Mod[ Fibonacci@k, n] != 0, k++ ]; k]; Array[fibEntry, 74] (* Robert G. Wilson v, Jul 04 2007 *)

a(n)=if(n<0,0,s=1;while(fibonacci(s)%n>0,s++);s) \\ Benoit Cloitre, Feb 10 2007

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<Charles R Greathouse IV, May 08 2017

(define (A001177 n) (let loop ((k 1)) (cond ((zero? (modulo (A000045 k) n)) k) (else (loop (+ k 1)))))) ;; Antti Karttunen, Dec 21 2013

A001175(n) = A001176(n) * a(n) for n >= 1.
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
Apparently a(n) = A213648(n) + 1 for n >= 2. - Art DuPre, Jul 01 2012
a(n) < n^2. [Vorob'ev]. - Zak Seidov, Jan 07 2016
a(n) < n^2 - 3n + 6. - Jinyuan Wang, Oct 13 2018
a(n) <= 2n [Salle]. - Jon Maiga, Apr 25 2019

Definition corrected by Wolfdieter Lang, Jan 19 2015

A213891 Fixed points of the sequence A262212 defined by the minimum number of 2's in the relation n*[n,2,2,...,2,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

3, 11, 19, 43, 67, 83, 107, 131, 139, 163, 211, 283, 307, 331, 347, 467, 491, 499, 523, 547, 563, 571, 587, 619, 659, 691, 739, 787, 811, 859, 883, 907, 947, 971, 1019, 1051, 1123, 1163, 1171, 1283, 1291, 1307

Offset: 1

Art DuPre, Jun 23 2012

nonn

It has long been a problem to find "natural" functions which will produce only primes. The sequence here apparently does just that, and it may well be the most natural function yet doing just that. There is apparently no reason why these sequences should produce only primes.
Let [a,b,...,c] = a+1/(b+(1/...+1/c)) represent a simple continued fraction.
Consider for n=2 the continued fraction [2,1,2] = 8/3. If we multiply 8/3 by 2, we get 16/3. If we write 16/3 as a continued fraction, we get [5,3]. Since the first entry 5 of this sequence is not equal to the last, 3, we insert another 1 in [2,1,2] between n and n to get [n,1,1,n] = 13/5. If we multiply 13/5 by 2, we get 26/5. If we write 26/5 as a continued fraction, we get [5,5] and now the first entry 5, of [5,5] is the same as the last entry 5 of [5,5]. Therefore 2 is the first number of 1s that we had to insert between the 1s in order for twice the resulting continued fraction to have equal first and last entries. Therefore, we define g(2)=2.
If we do the same for n=3, [3,1,3], we see that 3 is the minimum number of 1s that we have to insert between the 3s in order that when we multiply the continued fraction [3,1,1,1,3] by 3, we get [10,1,10], so the first and last entries are the same, namely 10. Therefore we define g(3)=3.
If we do this for n=4, [4,1,4] we see that 5 is the minimum number of 1s we have to insert before the first and last entries of 4*[4,1,1,1,1,1,4] are the same, namely, we get [18,2,18]. If we had multiplied [4,1,4], [4,1,1,4], [4,1,1,1,4],[4,1,1,1,1,4] by 4 we get, respectively [19,5],[18,4,2],[18,1,1,3],[18,2,2,3], none of which has its first and last entries equal. Therefore we define g(4)=5.
It turns out, proceeding as we just have, we get g(5)=4, g(6)=11, g(7)=7, which is A213648. If we define a sequence b(n) to contain the fixed points for which g(n)=n, considering that the sequence A213648 starts with 2 as its second term, then we get A000057 connected with the prime divisors of all the Fibonacci sequences.
If we do the same for inserting 2s as we just described for 1s, we get this sequence here.
These primes arise by first looking at the sequence h(n), whose n-th term is the minimum number of twos in [n,2,2,....,2,n], so that the continued fraction of n times the fraction corresponding to the above quotients has its first and last term equal. Next we construct the sequence of fixed points where h(n)=n. This sequence consists of prime numbers (conjecture). We conjecture that this sequence of prime numbers is analogous to A000057, in the sense that, instead of referring to the Fibonacci sequences it refers to the generalized Fibonacci sequences satisfying f(n)=2*f(n-1)+f(n-2). This would mean that a prime is in this sequence here if and only if it divides some term in each of the sequences satisfying f(n)=2*f(n-1)+f(n-2).

			The basic sequence h(n) (= A262212) is for n = 3,4,5,..:
3*[3, 2, 2, 2, 3] = [10,4,10], h(3) = 3: the first fixed point a(1) = 3.
4*[4, 2, 2, 2, 4] = [17, 1, 1, 1, 17], h(4) = 3;
5*[5, 2, 2, 5] = [27, 27], h(5) = 2;
6*[6, 2, 2, 2, 6] = [38, 2, 38], h(6) = 3;
(...)
11*[11, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 11] = [125, 1, 1, 3, 1, 14, 1, 3, 1, 1, 125] , h(11) = 11: this is the next fixed point after 3, so a(2)=11.

Cf. A000057, A213892 - A213899, A261311; A213358.

Cf. A213648, A262212 - A262220, A213900, A262211.

simpcf := proc(L)
        if nops(L) = 1 then
                op(1,L) ;
        else
                op(1,L)+1/procname([op(2..nops(L),L)]) ;
        end if;
end proc:
A213891aux := proc(n)
        local h,ins,c ;
        for ins from 1 do
                c := [n,seq(2,i=1..ins),n] ;
                h := numtheory[cfrac](n*simpcf(c),quotients) ;
                if op(1,h) = op(-1,h) then
                        return ins;
                end if;
        end do:
end proc:
A213891 := proc(n)
        if n = 1 then
                3;
        else
                for a from procname(n-1)+1 do
                        if A213891aux(a) = a then
                                return a;
                        end if;
                end do:
        end if;
end proc: # R. J. Mathar, Jul 06 2012

f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; Select[Range[2, 1000], f[2, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)

{a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,2), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

forprime(p=2,999, A262212(p)==p&&print1(p",")) \\ M. F. Hasler, Sep 30 2015

Edited by R. J. Mathar and T. D. Noe, Jul 06 2012

Edited by M. F. Hasler, Sep 30 2015

A213648 The minimum number of 1's in the relation n*[n,1,1,...,1,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

2, 3, 5, 4, 11, 7, 5, 11, 14, 9, 11, 6, 23, 19, 11, 8, 11, 17, 29, 7, 29, 23, 11, 24, 20, 35, 23, 13, 59, 29, 23, 19, 8, 39, 11, 18, 17, 27, 29, 19, 23, 43, 29, 59, 23, 15, 11, 55, 74, 35, 41, 26, 35, 9, 23, 35, 41, 57, 59, 14, 29, 23, 47, 34, 59, 67

Offset: 2

Art DuPre, Jun 17 2012

nonn

Multiplying n by a simple continued fraction with an increasing number of 1's sandwiched between n generates fractions that have a leading term x in their continued fraction, where x is obviously > n^2. We increase the number of 1's until the first and the last term in the simple terminating continued fraction of n*[n,1,...,1,n] =[x,...,x] is the same, x, and set a(n) to the count of these 1's.
Conjecture: the fixed points of this sequence are in A000057.
We have [n,1,1,...,1,n] = n + (n*Fib(m)+Fib(m-1))/(n*Fib(m+1)+Fib(m)) and n*[n,1,1,...,1,n] = n^2 + 1 + (n^2-n-1)*Fib(m)/(n*Fib(m+1)+Fib(m)), where m is the number of 1's. - Max Alekseyev, Aug 09 2012
The analog sequence with 11 instead of 1, A213900, seems to have the same fixed points, while other variants (A262212 - A262220, A262211) have other fixed points (A213891 - A213899,  A261311). - M. F. Hasler, Sep 15 2015

			3* [3,1,1,1,3] = [10,1,10],so a(3)=3
4* [4,1,1,1,1,1,4] = [18,2,18],so a(4)=5
5* [5,1,1,1,1,5] = [28,28],so a(5)=4
6* [6,1,1,1,1,1,1,1,1,1,1,1,6] = [39,1,2,2,2,1,39], so a(6)=11
7* [7,1,1,1,1,1,1,1,7] = [53,3,53], so a(7)=7

A. Hurwitz, Über die Kettenbrüche, deren Teilnenner arithmetische Reihen bilden, Vierteljahrsschrift der Naturforschenden Gesellschaft in Zürich, Jahrg XLI, 1896, Jubelband II, S. 34-64.

Bill Gosper, Appendix 2 Continued Fraction Arithmetic

Cf. A000057, A262212 - A262220.

A213648 := proc(n)
        local h,ins,c ;
        for ins from 1 do
                c := [n,seq(1,i=1..ins),n] ;
                h := numtheory[cfrac](n*simpcf(c),quotients) ;
                if op(1,h) = op(-1,h) then
                        return ins;
                end if;
        end do:
end proc: # R. J. Mathar, Jul 06 2012

f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; f[1, #] & /@ Range[2, 67] (* Michael De Vlieger, Sep 16 2015 *)

{a(n) = local(t, m=1); if( n<2, 0, while( t = contfracpnqn( concat( [n, vector(m, i, 1 ), n])), t = contfrac( n * t[1, 1] / t[2, 1]); if( t[1] < n^2 || t[#t] < n^2, m++, break)); m)} /* Michael Somos, Jun 17 2012 */

{a(n) = local(t, m=0); if( n<2, 0, until(t[1]==t[#t], m++; t = contfrac(n^2 + 1 + (n^2-n-1)*fibonacci(m)/(n*fibonacci(m+1)+fibonacci(m))); ); m )} /* Max Alekseyev, Aug 09 2012 */

Conjecture: a(n)=A001177(n)-1.

A213900 The minimum number of 11's in the relation n*[n,11,11,...,11,n] = [x,...,x] between simple terminating continued fractions.

Original entry on oeis.org

2, 3, 5, 4, 11, 7, 5, 11, 14, 1, 11, 6, 23, 19, 11, 8, 11, 17, 29, 7, 5, 23, 11, 24, 20, 35, 23, 13, 59, 5, 23, 3, 8, 39, 11, 18, 17, 27, 29, 3, 23, 43, 5, 59, 23, 15, 11, 55, 74, 35, 41, 26, 35, 9, 23, 35, 41, 57, 59, 2, 5, 23, 47, 34, 11, 67, 17, 23, 119, 13

Offset: 2

Art DuPre, Jun 24 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,11,11,..,11,n] and increase the number of 11's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 11, 11, 2] = [4, 5, 1, 1, 5, 4],
3 * [3, 11, 11, 11, 3] = [9, 3, 1, 2, 3, 2, 1, 3, 9],
4 * [4, 11, 11, 11, 11, 11, 4] = [16, 2, 1, 3, 2, 1, 1, 10, 1, 1, 2, 3, 1, 2, 16],
5 * [5, 11, 11, 11, 11, 5] = [25, 2, 4, 1, 1, 2, 2, 1, 1, 4, 2, 25] ,
6 * [6, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 6] = [36, 1, 1, 5, 1, 1, 2, 7, 16, 1, 1, 1, 2, 1, 6, 1, 2, 1, 1, 1, 16, 7, 2, 1, 1, 5, 1, 1, 36].
The number of 11's needed defines the sequence a(n).
If we consider the fixed points such that a(n)=n, we conjecture to obtain the sequence A000057. This sequence consists of prime numbers. We conjecture that this sequence of prime numbers, in addition to its well-known relation to the collection of Fibonacci sequences (sequences satisfying f(n)=f(n-1)+f(n-2) with arbitrary positive integer values for f(1) and f(2)) it also refers to the sequences satisfying f(n)=11*f(n-1)+f(n-2), A049666, A015457, etc. This would mean that a prime is in the sequence A000057 if and only if it divides some term in each of the sequences satisfying f(n)=11*f(n-1)+f(n-2).
It is surprising that the fixed points of this sequence seem to be the same as for the variant A213648 where 11 is replaced by 1, while for the other variants A262212 - A262220 (where the repeated term is 2, ..., 10) the fixed points are different, see A213891 - A213899. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213358, A213891 - A213899.

Cf. A213648, A262212 - A262220, A213900.

f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; f[11, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)

\\ This PARI program will generate sequence A000057
{a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,11), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A213899 Fixed points of a sequence h(n) defined by the minimum number of 10's in the relation n*[n,10,10,...,10,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

3, 7, 31, 43, 47, 71, 107, 151, 167, 179, 211, 223, 239, 251, 271, 283, 419, 431, 463, 467, 487, 491, 523, 547, 563, 571, 631, 839, 859, 883, 907, 967, 971, 1087, 1103, 1171, 1187, 1279, 1283, 1291, 1367, 1399, 1423, 1459, 1471, 1483, 1487, 1499

Offset: 1

Art DuPre, Jun 24 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,10,10,...,10,n] and increase the number of 10's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 10, 2] = [4, 5, 4],
3 * [3, 10, 10, 10, 3] = [9, 3, 2, 1, 2, 1, 2, 3, 9],
4 * [4, 10, 10, 10, 4] = [16, 2, 1, 1, 9, 1, 1, 2, 16],
5 * [5, 10, 5] = [25, 2, 25],
6 * [6, 10, 10, 10, 6] = [36, 1, 1, 2, 6, 2, 1, 1, 36],
7 * [7, 10, 10, 10, 10, 10, 10, 10, 7] = [49, 1, 2, 3, 1, 6, 2, 1, 2, 2, 2, 1, 2, 6, 1, 3, 2, 1, 49].
The number of 10's needed defines the sequence h(n) = 1, 3, 3, 1, 3, 7, 7, 11, 1, ... (n>=2).
The current sequence contains the fixed points of h, i.e., those n where h(n)=n.
We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (sequences satisfying f(n) = f(n-1) + f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the sequences satisfying f(n) = 10*f(n-1) + f(n-2), A041041, A015456, etc. This would mean that a prime is in the sequence A213899 if and only if it divides some term in each of the sequences satisfying f(n) = 10*f(n-1) + f(n-2).
The sequence h() is given in A262220. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213891 - A213898, A261311.

Cf. A213648, A262212 - A262220, A213900, A262211.

f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; Select[Range[2, 1000], f[10, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)

{a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,10), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

Just as A000057 can be generated by looking at the subscripts of the sequence A001177 which are one less than their values, A106535 can be generated by looking at the subscripts of the sequence A001177 which are one greater than their values.
It is a surprising fact that these two sequences A000057 and A106535 are disjoint. The also have approximately the same density, if these densities exist.
It would be interesting to be able to interpret the relation of this prime sequence to the entire set of Fibonacci sequences, i.e., those sequences satisfying f(n+2) = f(n+1) + f(n) with various initial conditions.

Sequence A261311 lists fixed points of this sequence.

It is surprising that the variant A213900 with 11 instead of 12 has the same fixed points A000057 as the variant A213648 with 1 instead of 12, but other variants (A262212 - A262220 and this one) have different sets of fixed points (A213891 - A213899 and A261311).

Surprisingly, the variant A213900 with 11 instead of 12 has the same fixed points A000057 as the variant with 1 instead of 12, but other variants (A262212 - A262220 and A262211) have different sets of fixed points (A213891 - A213899 and this).

cp's OEIS Frontend

A214029 The union of the disjoint prime sequences A000057 and A106535.

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A373774 a(n) is the number of terms of A000057 in [10^n].

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A001177 Fibonacci entry points: a(n) = least k >= 1 such that n divides Fibonacci number F_k (=A000045(k)).

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A213891 Fixed points of the sequence A262212 defined by the minimum number of 2's in the relation n*[n,2,2,...,2,n] = [x,...,x] between simple continued fractions.

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A213648 The minimum number of 1's in the relation n*[n,1,1,...,1,n] = [x,...,x] between simple continued fractions.

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A213900 The minimum number of 11's in the relation n*[n,11,11,...,11,n] = [x,...,x] between simple terminating continued fractions.

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A213899 Fixed points of a sequence h(n) defined by the minimum number of 10's in the relation n*[n,10,10,...,10,n] = [x,...,x] between simple continued fractions.

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