cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Views

Author

Art DuPre, Jun 24 2012

Keywords

Comments

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

Crossrefs

Cf. A000057, A213891 - A213898, A261311.
Cf. A213648, A262212 - A262220, A213900, A262211.

Programs

  • Mathematica
    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 *)
  • PARI
    {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]

A213897 Fixed points of a sequence h(n) defined by the minimum number of 8's in the relation n*[n,8,8,...,8,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

3, 7, 23, 31, 71, 107, 131, 139, 163, 199, 211, 227, 283, 347, 367, 379, 419, 431, 439, 487, 499, 503, 547, 571, 607, 619, 643, 691, 719, 751, 787, 811, 823, 827, 907, 911, 983, 991, 1031, 1051, 1091, 1151, 1163, 1231, 1303, 1319, 1367, 1399, 1423, 1439, 1459, 1499
Offset: 1

Views

Author

Art DuPre, Jun 24 2012

Keywords

Comments

In a variant of A213891, multiply n by a number with simple continued fraction [n,8,8,..,8,n] and increase the number of 8's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 8, 2] = [4, 4, 4],
3 * [3, 8, 8, 8, 3] = [9, 2, 1, 2, 2, 2, 1, 2, 9],
4 * [4, 8, 4] = [16, 2, 16],
5 * [5, 8, 8, 5] = [25, 1, 1, 1, 1, 1, 1, 25],
6 * [6, 8, 8, 8, 6] = [36, 1, 2, 1, 4, 1, 2, 1, 36],
7 * [7, 8, 8, 8, 8, 8, 8, 8, 7] = [49, 1, 6, 4, 3, 2, 1, 2, 1, 2, 3, 4, 6, 1, 49].
The number of 8's needed defines the sequence h(n) = 1, 3, 1, 2, 3, 7, 1, 11, 5,.. (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)=8*f(n-1)+f(n-2), A041025, A015454, etc. This would mean that a prime is in the sequence A213897 if and only if it divides some term in each of the sequences satisfying f(n)=8*f(n-1)+f(n-2).
The sequence h() is recorded as A262218. - M. F. Hasler, Sep 15 2015

Crossrefs

Cf. A000057, A213891 - A213896, A213898, A213899, A261311; A213358.
Cf. A213648, A262212 - A262220, A213900, A262211.

Programs

  • Mathematica
    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[8, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
  • PARI
    {a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,8), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

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

Views

Author

M. F. Hasler, Sep 15 2015

Keywords

Comments

Sequence A213898 lists fixed points of this sequence.

Crossrefs

Cf. A213648, A262212 - A262220, A213900, A262211.
Cf. A000057, A213891 - A213899, A261311: fixed points of the above.

Programs

  • Mathematica
    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 *)
  • PARI
    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
Showing 1-3 of 3 results.

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