A213891 -id:A213891 - cp's OEIS frontend

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

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

A270617 Primes p such that A256832(p) is divisible by p.

Original entry on oeis.org

2, 5, 7, 13, 17, 23, 29, 31, 37, 41, 47, 53, 59, 61, 71, 73, 79, 89, 97, 101, 103, 109, 113, 127, 137, 149, 151, 157, 167, 173, 179, 181, 191, 193, 197, 199, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 293, 311, 313, 317, 337, 349, 353, 359, 367, 373, 379, 383, 389, 397

Offset: 1

Altug Alkan, Mar 20 2016

nonn

Sequence focuses on the prime numbers because of the complement of this sequence. Primes that are listed in this sequence cannot be generated by function which is related with A213891. See comment section of A213891.

			5 is a term because A256832(5) = 3480 is divisible by 5.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A000129, A256832, A213891.

nn = 400; s = FoldList[Times, LinearRecurrence[{2, 1}, {1, 2}, nn]]; Select[Prime@ Range@ PrimePi@ nn, Divisible[s[[#]], #] &] (* Michael De Vlieger, Mar 27 2016, after Harvey P. Dale at A256832 *)

a000129(n) = ([2, 1; 1, 0]^n)[2, 1];
t(n) = prod(k=1, n, Mod(a000129(k), n));
forprime(p=2, 1e3, if(lift(t(p)) == 0, print1(p, ", ")));

is(n)=my(a=Mod(1,n),b=Mod(2,n)); for(i=2,n, if(b==0, return(isprime(n))); [a,b]=[b,2*b+a]); 0 \\ Charles R Greathouse IV, Mar 31 2016

list(lim)=my(v=List([2]), G=factorback(primes([2,lim])), a=1, b=2, t=2, p=2); forprime(q=3,lim, for(n=p+1,q, [a,b]=[b,2*b+a]; t=gcd(t*b, G)); if(t%q==0, listput(v, q)); G/=q; p=q); Vec(v) \\ Charles R Greathouse IV, Mar 31 2016

A270834 Numbers n such that A256832(n)/A000129(n-1) is not divisible by n.

Original entry on oeis.org

3, 7, 9, 11, 19, 23, 31, 43, 47, 67, 71, 83, 107, 127, 131, 139, 151, 163, 167, 191, 211, 263, 271, 283, 307, 311, 331, 347, 359, 367, 383, 431, 439, 463, 467, 479, 491, 499, 503, 523, 547, 563, 571, 587, 619, 631, 647, 659, 691, 719, 727, 739, 743, 787, 811, 823, 839, 859, 863, 883, 887

Offset: 1

Altug Alkan, Mar 23 2016

nonn

The computation of integers n such that A256832(n) is not divisible by n, leads to A213891. This sequence contains A213891 as a subsequence.
It appears that 9 is the only composite number in this sequence.
No composites below 10^7. - Charles R Greathouse IV, Apr 20 2016
No composites below 2*10^7. - Charles R Greathouse IV, May 06 2016

			7 is a term because 1*2*5*12*29*169 = 588120 is not divisible by 7.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A000129, A213891, A256832, A270617.

With[{s = Sqrt@ 2}, Select[Range[2, 90], ! Divisible[Product[Expand[((1 + s)^k - (1 - s)^k)/2 s], {k, #}]/Simplify[((1 + s)^(# - 1) - (1 - s)^(# -
1))/(2 s)], #] &]] (* Michael De Vlieger, Mar 24 2016, after Vaclav Kotesovec at A256832 and Michael Somos at A000129 *)

a000129(n) = ([2, 1; 1, 0]^n)[2, 1];
t(n) = Mod((prod(k=1, n, a000129(k)) / a000129(n-1)), n);
for(n=2, 1e3, if(lift(t(n)) != 0, print1(n, ", ")));

is(n)=my(a,b=Mod(1,n),t=b); for(k=2,n-2,[a,b]=[b,a+2*b]; t*=b; if(t==0, return(0))); t*(2*a+5*b) && n>2 \\ Charles R Greathouse IV, Mar 24 2016

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]

A270474 Integers k such that A256832(k) is not divisible by k*(k+1)/2.

Original entry on oeis.org

2, 3, 10, 11, 18, 19, 42, 43, 66, 67, 82, 83, 106, 107, 130, 131, 138, 139, 162, 163, 210, 211, 282, 283, 306, 307, 330, 331, 346, 347, 466, 467, 490, 491, 498, 499, 522, 523, 546, 547, 562, 563, 570, 571, 586, 587, 618, 619, 658, 659, 690, 691, 738, 739, 786, 787, 810, 811, 858, 859

Offset: 1

Altug Alkan, Mar 17 2016

nonn

It appears that the odd numbers in the sequence are prime.

This holds at least up to a million. - Charles R Greathouse IV, Feb 24 2022

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

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A000129, A000217, A213891, A256832.

nn = 10^3; Function[k, Select[Range@ nn, ! Divisible[k[[#]], # (# + 1)/2] &]]@ FoldList[Times, LinearRecurrence[{2, 1}, {1, 2}, nn]] (* Michael De Vlieger, Mar 19 2016, after Harvey P. Dale at A256832 *)

a000129(n) = ([2, 1; 1, 0]^n)[2, 1];
f(n) = prod(k=1, n, a000129(k)); \\ A256832
for(n=1, 1e3, if(f(n) % (n*(n+1)/2) != 0, print1(n, ", ")));

{g(n) = my(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]Bill McEachen, Feb 14 2022 (from A213891 code, faster)

is(n)=my(m=n^2+n,q=Mod([2, 1; 1, 0],m),Q=q,P=Mod(1,m)); for(k=2,n, P*=(Q*=q)[2,1]; if(P==0, return(0))); 1 \\ Charles R Greathouse IV, Feb 14 2022

cp's OEIS Frontend

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

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

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A270617 Primes p such that A256832(p) is divisible by p.

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A270834 Numbers n such that A256832(n)/A000129(n-1) is not divisible by n.

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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.

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A270474 Integers k such that A256832(k) is not divisible by k*(k+1)/2.

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