A000057 -id:A000057 - cp's OEIS frontend

A262214 Minimum number of 4's such that n*[n; 4, ..., 4, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

1, 3, 1, 4, 3, 7, 3, 3, 9, 9, 3, 6, 7, 19, 7, 2, 3, 5, 9, 7, 9, 7, 3, 24, 13, 11, 7, 13, 19, 9, 15, 19, 5, 39, 3, 18, 5, 27, 19, 19, 7, 43, 9, 19, 7, 15, 7, 55, 49, 11, 13, 8, 11, 9, 7, 11, 13, 57, 19, 4, 9, 7, 31, 34, 19, 67, 5, 7, 39, 69, 3, 36, 37, 99, 5, 39, 27, 25, 39, 35, 19, 27, 7, 14, 43, 27, 19, 10, 19, 55, 7, 19, 15, 29, 15, 48, 55, 19

Offset: 2

M. F. Hasler, Sep 15 2015

nonn

Sequence A213893 lists fixed points of this sequence.

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

Cf. A000057, A213891 - A213899, A261311.

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[4, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)

cf(v)={t=v[#v];forstep(i=#v-1,1,-1,t=v[i]+1/t);t}
A262214(n,d=4)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

A262215 Minimum number of 5's such that n*[n; 5, ..., 5, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Original entry on oeis.org

2, 3, 5, 1, 11, 5, 5, 3, 5, 11, 11, 2, 5, 3, 11, 8, 11, 19, 5, 11, 11, 21, 11, 9, 2, 3, 5, 28, 11, 31, 23, 11, 8, 5, 11, 18, 59, 11, 5, 6, 11, 43, 11, 3, 65, 47, 11, 41, 29, 35, 5, 12, 11, 11, 5, 19, 86, 57, 11, 30, 95, 11, 47, 5, 11, 65, 17, 43, 5, 69, 11, 36, 56, 19, 59, 11, 11, 79, 11, 11, 20, 81, 11, 17, 131, 115, 11, 44, 11, 5, 65, 31, 47, 19, 23, 48

Offset: 2

M. F. Hasler, Sep 15 2015

nonn

Sequence A213894 lists fixed points of this sequence.

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

Cf. A000057, A213891 - A213899, A261311: fixed points of the above.

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[5, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)

cf(v)={t=v[#v];forstep(i=#v-1,1,-1,t=v[i]+1/t);t}
A262215(n,d=5)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

A262216 Minimum number of 6's such that n*[n; 6, ..., 6, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Original entry on oeis.org

1, 1, 3, 4, 1, 7, 7, 5, 9, 11, 3, 5, 7, 9, 15, 8, 5, 3, 19, 7, 11, 23, 7, 24, 5, 17, 7, 14, 9, 29, 31, 11, 17, 39, 11, 2, 3, 5, 39, 19, 7, 41, 11, 29, 23, 47, 15, 55, 49, 17, 11, 25, 17, 59, 7, 3, 29, 19, 19, 30, 29, 23, 63, 29, 11, 21, 35, 23, 39, 69, 23, 36, 5, 49, 3, 23, 5, 77

Offset: 2

M. F. Hasler, Sep 15 2015

nonn

Sequence A213895 lists fixed points of this sequence.

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

Cf. A000057, A213891 - A213899, A261311: fixed points of the above.

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[6, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)

cf(v)={t=v[#v];forstep(i=#v-1,1,-1,t=v[i]+1/t);t}
A262216(n,d=6)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

A262217 Minimum number of 7's such that n*[n; 7, ..., 7, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Original entry on oeis.org

2, 3, 5, 2, 11, 1, 5, 11, 2, 9, 11, 5, 5, 11, 11, 3, 11, 19, 5, 3, 29, 7, 11, 2, 5, 35, 5, 6, 11, 31, 23, 19, 11, 5, 11, 8, 59, 11, 5, 20, 11, 13, 29, 11, 23, 45, 11, 13, 2, 3, 5, 52, 35, 29, 5, 19, 20, 57, 11, 30, 95, 11, 47, 5, 59, 67, 11, 7, 5, 23, 11, 36, 8, 11, 59, 9, 11, 79, 11, 107, 20, 27, 11, 11, 41, 27, 29, 43, 11, 5, 23, 31, 137, 59, 23, 47, 41, 59

Offset: 2

M. F. Hasler, Sep 15 2015

nonn

Sequence A213896 lists fixed points of this sequence.

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

Cf. A000057, A213891 - A213899, A261311: fixed points of the above.

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[7, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)

cf(v)={t=v[#v];forstep(i=#v-1,1,-1,t=v[i]+1/t);t}
A262217(n,d=7)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

A262218 Minimum number of 8's such that n*[n; 8, ..., 8, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Original entry on oeis.org

1, 3, 1, 2, 3, 7, 1, 11, 5, 3, 3, 2, 7, 11, 3, 16, 11, 17, 5, 7, 3, 23, 3, 14, 5, 35, 7, 14, 11, 31, 7, 3, 33, 23, 11, 18, 17, 11, 5, 20, 7, 41, 3, 11, 23, 45, 3, 55, 29, 67, 5, 25, 35, 11, 7, 35, 29, 57, 11, 30, 31, 23, 15, 2, 3, 5, 33, 23, 23, 71, 11, 36, 37, 59, 17, 7, 11, 15, 11, 107, 41, 81, 7, 50, 41, 59, 3, 43, 11, 23, 23, 31, 45, 17, 7, 48, 55, 11

Offset: 2

M. F. Hasler, Sep 15 2015

nonn

Sequence A213897 lists fixed points of this sequence.

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

Cf. A000057, A213891 - A213899, A261311: fixed points of the above.

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[8, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)

cf(v)={t=v[#v];forstep(i=#v-1,1,-1,t=v[i]+1/t);t}
A262218(n,d=8)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

A262219 Minimum number of 9's such that n*[n; 9, ..., 9, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Original entry on oeis.org

2, 1, 5, 4, 5, 5, 5, 1, 14, 11, 5, 6, 5, 9, 11, 16, 5, 17, 29, 5, 11, 21, 5, 24, 20, 5, 5, 14, 29, 31, 23, 11, 50, 29, 5, 17, 17, 13, 29, 2, 5, 43, 11, 9, 65, 47, 11, 41, 74, 33, 41, 26, 5, 59, 5, 17, 14, 57, 29, 30, 95, 5, 47, 34, 11, 67, 101, 21, 29, 7, 5, 35, 17, 49, 17, 11, 41, 79, 59, 17, 2, 3, 5, 84, 131, 29, 11, 43, 29, 41, 65, 31, 47, 89, 23, 7, 41

Offset: 2

M. F. Hasler, Sep 15 2015

nonn

Sequence A213898 lists fixed points of this sequence.

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

Cf. A000057, A213891 - A213899, A261311: fixed points of the above.

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[9, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)

cf(v)={t=v[#v];forstep(i=#v-1,1,-1,t=v[i]+1/t);t}
A262219(n,d=9)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

A270532 Integers k such that A003266(p) is not divisible by p*(p+1)/2 where p is the k-th prime.

Original entry on oeis.org

1, 2, 4, 9, 14, 19, 23, 27, 31, 38, 39, 48, 49, 61, 73, 76, 86, 90, 91, 93, 96, 99, 101, 107, 111, 117, 118, 124, 129, 138, 143, 144, 150, 153, 154, 155, 161, 166, 179, 188, 192, 195, 208, 213, 217, 219, 224, 229, 236, 243, 247, 250, 253, 258, 261, 262, 269, 272, 276, 277, 283, 285, 292, 300

Offset: 1

Altug Alkan, Mar 18 2016

nonn

See A000057 to corresponding prime numbers. In other words, this sequence is an indirect way to generate primes dividing all Fibonacci sequences.

Michel Marcus, Table of n, a(n) for n = 1..1000

Cf. A000057, A003266, A270475.

f(n) = prod(k=1, n, fibonacci(k)); \\ A003266
isok(n) = (lift(Mod(f(prime(n)),(prime(n)*(prime(n)+1)/2))) != 0);

isok(n) = my(p=prime(n), m=p*(p+1)/2); prod(k=1, p, Mod(fibonacci(k), m)); \\ Michel Marcus, May 14 2021

A213901 Fixed points of a sequence that gives the minimum length of a chain of 1,2,1,2,1,... so that the equation n*[n,1,2,1,...,n] = [x,...,y] between terminating continued fractions has a solution with x >= n^2 and y >= n^2.

Original entry on oeis.org

5, 7, 29, 31, 79, 103, 127, 149, 151, 173, 197, 199, 223, 269, 271, 293, 317, 367, 439, 463, 487, 557, 631, 701, 727, 751, 773, 797, 821, 823, 941, 967, 991, 1039, 1061

Offset: 1

Art DuPre, Jun 29 2012

nonn

Let [...,...,...] denote terminating continued fractions. For each n, build the value (quotient) of the continued fraction [n,1,2,1,2,...,n] by inserting the first m terms of the repeated sequence 1,2,1,2,... Note that m may be odd, so the fraction may end with [...,1,n]. Multiply this value by n and convert the product back to a continued fraction [x,...,y]. Define the sequence of minimum m(n) such that x and y in this representation are both at least n^2.
This length sequence m(n) starts 3, 2, 3, 5, 11, 7, 7, 8, 11, 4, 11, 11, 7, 5, 15, 8, 35, 9, 11, 23, 19, 21, 23, 29, 11, 26, 7, 29, 11, ... for n >= 2.
It refers to the equations
   2*[2, 1, 2, 1, 2] = [5, 2, 5],
   3*[3, 1, 2, 3] = [11, 10],
   4*[4, 1, 2, 1, 4] = [18, 1, 18],
   5*[5, 1, 2, 1, 2, 1, 5] = [28, 1, 1, 1, 28],
   6*[6, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 6] = [40, 2, 1, 1, 4, 1, 1, 2, 40],
   7*[7, 1, 2, 1, 2, 1, 2, 1, 7] = [54, 8, 54],
   8*[8, 1, 2, 1, 2, 1, 2, 1, 8] = [69, 1, 5, 1, 69],
   9*[9, 1, 2, 1, 2, 1, 2, 1, 2, 9] = [87, 1, 1, 2, 3, 84],
  10*[10, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 10] = [107, 3, 8, 3, 107],
  11*[11, 1, 2, 1, 2, 11] = [129, 125],
  12*[12, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 12] = [152, 1, 3, 1, 1, 1, 3, 1, 152],
  13*[13, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 13] = [178, 1, 1, 14, 1, 1, 178],
  14*[14, 1, 2, 1, 2, 1, 2, 1, 14] = [206, 4, 206], ...
{a(n)} contains the fixed points of the length sequence m, i.e., those n where m(n)=n.
What is surprising is that all these fixed points appear to be prime numbers. (Such a sequence of fixed points may be defined for any other pair (p,q) which defines continued fractions [n,p,q,p,q,...,n]. Here we are looking at p=1, q=2. The sequence m(n) related to the pair p=2, q=1 is 1, 2, 3, 5, 5, 7, 3, 8, 5, 4, 11, 11, 7, 5, 7, 8, ... for n >= 2.)
If we have any sequence of positive integers, we can consider the sequence of primes which divide some term of the sequence. In some cases, this sequence of primes could be finite. For the Fibonacci sequences, the sequence of primes is all primes. If we speak of Fibonacci sequences, we are referring to any sequence which satisfies the recursion relation f(n) = f(n-1) + f(n-2) with arbitrary initial conditions f(1), f(2). Each of these sequences has a set of prime divisors. None of these has the sequence of all primes as its prime sequence, but the intersection of all these sequences of primes is nonempty, and coincides with A000057.
From that perspective, the current sequence seems to relate to the prime divisors common to some family of generalized Fibonacci sequences f(n) = e*f(n-1) + g*f(n-2) for some fixed coefficients e and g.

Bill McEachen, Table of n, a(n) for n = 1..360 (terms 1..157 from Art DuPre)

Cf. A213891-A213900.

{a(n,p,q) = local(t, m=1,s=[n]); if( n<2, 0, while( 1,
   if(component(Mod(m,2),2)==1, s=concat(s,p),s=concat(s,q));
t=contfracpnqn(concat(s,n));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A270475 Integers n such that A003266(n) is not divisible by n*(n+1)/2.

Original entry on oeis.org

2, 3, 4, 6, 7, 22, 23, 42, 43, 66, 67, 82, 83, 102, 103, 126, 127, 162, 163, 166, 167, 222, 223, 226, 227, 282, 283, 366, 367, 382, 383, 442, 443, 462, 463, 466, 467, 486, 487, 502, 503, 522, 523, 546, 547, 586, 587, 606, 607, 642, 643, 646, 647, 682, 683, 726, 727, 786, 787

Offset: 1

Altug Alkan, Mar 17 2016

nonn

This sequence contains primes dividing all Fibonacci sequences.

			6 is a term because (1*1*2*3*5*8) is not divisible by (1+2+3+4+5+6).
5 is not a term because (1*1*2*3*5) is divisible by (1+2+3+4+5).

Cf. A000040, A000057, A003266.

nn = 800; Function[k, Select[Range@ nn, ! Divisible[k[[#]], # (# + 1)/2] &]]@ FoldList[Times, Array[Fibonacci@ # &, nn]] (* Michael De Vlieger, Mar 19 2016 *)

a(n) = prod(k=1, n, fibonacci(k));
for(n=1, 1e3, if(a(n) % (n*(n+1)/2) != 0, print1(n, ", ")));

A270839 Integers k such that (A003266(k)/A000045(k-1)) is not divisible by k.

Original entry on oeis.org

2, 3, 4, 7, 9, 11, 19, 23, 31, 43, 59, 67, 71, 79, 83, 103, 127, 131, 163, 167, 179, 191, 223, 227, 239, 251, 271, 283, 311, 359, 367, 379, 383, 419, 431, 439, 443, 463, 467, 479, 487, 491, 499, 503, 523, 547, 571, 587, 599, 607, 631, 643, 647, 659, 683, 719, 727, 739, 751, 787, 823, 827, 839

Offset: 1

Altug Alkan, Mar 23 2016

nonn

A270777 is a subsequence.

It appears that this sequence generates prime numbers except 4 and 9. [Verified for the first 500 terms. - Amiram Eldar, Apr 01 2021]

			3 is a term because 1*2 is not divisible by 3.
7 is a term because 1*1*2*3*5*13 is not divisible by 7.

Amiram Eldar, Table of n, a(n) for n = 1..500

Cf. A000045, A000057, A003266, A270777.

Select[Range[2, 840], ! Divisible[Fibonorial@ #/Fibonacci[# - 1], #] &] (* Version 10, or *) Select[Range[2, 840], ! Divisible[Product[Fibonacci@ k, {k, #}]/Fibonacci[# - 1], #] &] (* Michael De Vlieger, Mar 24 2016 *)

t(n) = fibonacci(n) * prod(k=1, n-2, Mod(fibonacci(k), n));
for(n=2, 1e3, if(lift(t(n)) != 0, print1(n, ", ")));

Offset corrected by Amiram Eldar, Apr 01 2021

cp's OEIS Frontend

A262214 Minimum number of 4's such that n*[n; 4, ..., 4, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A262215 Minimum number of 5's such that n*[n; 5, ..., 5, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A262216 Minimum number of 6's such that n*[n; 6, ..., 6, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A262217 Minimum number of 7's such that n*[n; 7, ..., 7, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A262218 Minimum number of 8's such that n*[n; 8, ..., 8, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A262219 Minimum number of 9's such that n*[n; 9, ..., 9, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A270532 Integers k such that A003266(p) is not divisible by p*(p+1)/2 where p is the k-th prime.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

PARI

A213901 Fixed points of a sequence that gives the minimum length of a chain of 1,2,1,2,1,... so that the equation n*[n,1,2,1,...,n] = [x,...,y] between terminating continued fractions has a solution with x >= n^2 and y >= n^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

A270475 Integers n such that A003266(n) is not divisible by n*(n+1)/2.

Views

Author

Keywords

Comments

Examples