A213891 -id:A213891 - cp's OEIS frontend

A213898 Fixed points of a sequence h(n) defined by the minimum number of 9's in the relation n*[n,9,9,...,9,n] = [x,...,x] between simple continued fractions.

2, 11, 31, 43, 47, 67, 79, 103, 127, 199, 211, 223, 263, 307, 311, 383, 431, 439, 463, 467, 499, 523, 563, 571, 587, 691, 719, 751, 811, 839, 863, 883, 911, 967, 991, 1051, 1063, 1087, 1091, 1123, 1151, 1231, 1307, 1327, 1399, 1447, 1451, 1459, 1483, 1499

Offset: 1

Art DuPre, Jun 24 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,9,9,..,9,n] and increase the number of 9's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 9, 9, 2] = [4, 4, 1, 1, 4, 4],
3 * [3, 9, 3] = [9, 3, 9],
4 * [4, 9, 9, 9, 9, 9, 4] = [16, 2, 3, 1, 1, 1, 1, 8, 1, 1, 1, 1, 3, 2, 16] ,
5 * [5, 9, 9, 9, 9, 5] = [25, 1, 1, 4, 1, 1, 1, 1, 1, 1, 4, 1, 1, 25],
6 * [6, 9, 9, 9, 9, 9, 6] = [36, 1, 1, 1, 13, 6, 13, 1, 1, 1, 36],
7 * [7, 9, 9, 9, 9, 9, 7] = [49, 1, 3, 3, 6, 1, 6, 3, 3, 1, 49].
The number of 9's needed defines the sequence h(n) = 2, 1,5, 4, 5, 5, 5, 1, 14,...  (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 sequence (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)=9*f(n-1)+f(n-2) like A099371, A015455 etc. This would mean that a prime is in the sequence A213898 if and only if it divides some term in each of the sequences satisfying f(n)=9*f(n-1)+f(n-2).

Cf. A213358; A000057, A213891 - A213897, A213899, 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[9, #] == # &] (* 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,9), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A214030 The multiplication continued fraction sequence for 2,3,2,3...

Original entry on oeis.org

0, 1, 5, 3, 4, 5, 2, 3, 5, 9, 4, 11, 13, 5, 29, 3, 15, 5, 19, 19, 5, 9, 23, 11, 13, 17, 11, 14, 29, 7, 7, 29, 15, 14, 11, 37, 19, 41, 19, 41, 5, 13, 19, 29, 23, 47, 11, 20, 49, 47, 27, 25, 17, 4, 11, 59, 29, 28, 59, 11, 7, 5, 15, 69, 29, 65, 15, 23, 29, 9, 11, 36, 37, 149

Offset: 1

Art DuPre, Jul 12 2012

nonn

This is the first sequence which comes from continued fractions which cannot, so far, be looked at as an apparition sequence for a linearly recursive sequence.
The first term a(1), is always 0. We will always refer to the continued fraction [a,b,c] for short, when we really mean the continued fraction with quotients [a,b,c].
To calculate a(2), we consider [2,2], and take a(2) to be the least number of terms of the sequence 2,3,2,3,... which will have the property that 2[2,2,3,2,3,...,2] = [2^2+k,...,2^2+l], where k,l>=0. We find that 2[2,2,2]=[4,1,4], and both the first term 4 and the last term, 4 are >= 2^2, so 1 is the correct number of terms from 2,3,2,3,.... Thus we have shown that a(2)=1.
Next, to calculate a(3), we consider the sequence [3,2,3,...,3] and notice that if we put only one term from the sequence 2,3,2,3,... between the 3s at the ends, we get 3[3,2,3]=[3^2+1,3,2] and the last term 2 is not >= 3^2, so we consider 3[3,2,3,3]=[3^2+1,3,3,2] and 2 is not >= 3^2, so we look at 3[3,2,3,2,3]=[3^2+1,3,4,4], and 4, the last term, is still not >= 3^2. Next, we consider 3[3,2,3,2,3,3] = [3^2+1,3,4,3,4] and we see that the last term 4 is still not >= 3^2. Finally when we put 5 terms of the sequence 2,3,2,3,.. between, the 3s, we get 3[3,2,3,2,3,2,3] = [10,3,4,3,10], and now the first element 10 is >=3^2, and the last number 10 is also >=3^2, and this proves that a(3)=5. By using the PARI program below, we get the sequence a(n) above, shown in the data portion.

Cf. A213891, A213892.

{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]

A213893 Fixed points of a sequence h(n) defined by the minimum number of 4's in the relation n*[n,4,4,...,4,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

3, 7, 43, 67, 103, 127, 163, 223, 283, 367, 463, 487, 523, 547, 607, 643, 727, 787, 823, 883, 907, 1063, 1123, 1303, 1327, 1423, 1447, 1543, 1567, 1627, 1663, 1723, 1747, 1783, 1867, 1987, 2083, 2143, 2203, 2287, 2347, 2383, 2467, 2683, 2707, 2767, 2803, 2887

Offset: 1

Art DuPre, Jun 23 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,4,4,...,4,n] and increase the number of 4's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2*[2,4,2] = [4,2,4],
3*[3,4,4,4,3] = [9,1,2,2,2,1,9],
4*[4,4,4] = [16,1,16],
5*[5,4,4,4,4,5] = [26,5,1,1,5,26].
The number of 4's needed defines the sequence h(n) = 1, 3, 1, 4, 3, 7, 3, 3, 9, ... (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) = 4*f(n-1) + f(n-2), A001076, A001077, A015448, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 4*f(n-1) + f(n-2).
The above sequence h() is recorded as A262214. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213891, A213892, A213894 - A213899, A261311; A213358.

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[4, #] == # &] (* 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,4), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A213894 Fixed points of a sequence h(n) defined by the minimum number of 5's in the relation n*[n,5,5,...,5,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

2, 3, 11, 19, 31, 43, 47, 79, 127, 131, 163, 211, 251, 271, 307, 311, 331, 367, 379, 443, 503, 563, 599, 607, 659, 743, 751, 823, 839, 859, 887, 907, 911, 947, 967, 1063, 1087, 1091, 1123, 1163, 1171, 1187, 1259, 1279, 1291, 1303, 1307, 1319, 1423, 1447, 1471, 1487

Offset: 1

Art DuPre, Jun 23 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,5,5,...,5,n] and increase the number of 5's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 5, 5, 2] = [4, 2, 1, 1, 2, 4],
3 * [3, 5, 5, 5, 3] = [9, 1, 1, 2, 1, 2, 1, 1, 9],
4 * [4, 5, 5, 5, 5, 5, 4] = [16, 1, 3, 2, 1, 4, 1, 2, 3, 1, 16] ,
5 * [5, 5, 5] = [25, 1, 25].
The number of 5's needed defines the sequence h(n) = 2, 3, 5, 1, 11, 5, 5, 3, 5, 11, 11, ... (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 generalized Fibonacci sequences satisfying f(n) = 5*f(n-1) + f(n-2), A052918, A015449, A164581, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 5*f(n-1) + f(n-2).
The above sequence h() is recorded as A262215. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213891 - A213893, A213895 - A213899, A261311; A213358.

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[5, #] == # &] (* 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,5), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A213895 Fixed points of a sequence h(n) defined by the minimum number of 6's in the relation n*[n,6,6,...,6,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

7, 11, 23, 47, 127, 139, 211, 223, 251, 331, 367, 379, 383, 463, 487, 499, 607, 619, 691, 727, 739, 743, 811, 823, 863, 887, 967, 971, 983, 1051, 1063, 1087, 1171, 1291, 1303, 1327, 1367, 1423, 1447, 1451, 1459

Offset: 1

Art DuPre, Jun 23 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,6,6,...,6,n] and increase the number of 6's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 6, 2] = [4, 3, 4],
3 * [3, 6, 3] = [9, 2, 9],
4 * [4, 6, 6, 6, 4] = [16, 1, 1, 1, 5, 1, 1, 1, 16],
5 * [5, 6, 6, 6, 6, 5] = [25, 1, 4, 3, 3, 4, 1, 25],
6 * [6, 6, 6] = [36, 1, 36],
7 * [7, 6, 6, 6, 6, 6, 6, 6, 7] = [50, 7, 2, 1, 4, 4, 4, 1, 2, 7, 50].
The number of 6's needed defines the sequence h(n) = 1, 1, 3, 4, 1, 7, 7, 5, 9, ... (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 numbers is 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 generalized Fibonacci sequences satisfying f(n) = 6*f(n-1) + f(n-2), A005668, A015451, A179237, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 6*f(n-1) + f(n-2).
The above sequence h() is recorded as A262216. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213891 - A213894, A213896 - A213899, A261311; A213358.

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[6, #] == # &] (* 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,6), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A262213 Minimum number of 3's such that n*[n; 3, ..., 3, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Original entry on oeis.org

2, 1, 5, 2, 5, 7, 5, 5, 2, 3, 5, 12, 23, 5, 11, 7, 5, 19, 5, 7, 11, 21, 5, 14, 38, 17, 23, 6, 5, 31, 23, 3, 23, 23, 5, 18, 59, 25, 5, 6, 23, 41, 11, 5, 65, 47, 11, 55, 14, 7, 77, 25, 17, 11, 23, 19, 20, 11, 5, 29, 95, 23, 47, 38, 11, 67, 23, 21, 23, 71, 5, 36, 56, 29, 59, 7, 77, 25, 11, 53

Offset: 2

M. F. Hasler, Sep 15 2015

nonn

Sequence A213892 lists fixed points of this sequence.

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

cf(v)={t=v[#v];forstep(i=#v-1,1,-1,t=v[i]+1/t);t}
A262213(n,d=3)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

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

Original entry on oeis.org

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

cp's OEIS Frontend

A213898 Fixed points of a sequence h(n) defined by the minimum number of 9's in the relation n*[n,9,9,...,9,n] = [x,...,x] between simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A214030 The multiplication continued fraction sequence for 2,3,2,3...

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

A213893 Fixed points of a sequence h(n) defined by the minimum number of 4's in the relation n*[n,4,4,...,4,n] = [x,...,x] between simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A213894 Fixed points of a sequence h(n) defined by the minimum number of 5's in the relation n*[n,5,5,...,5,n] = [x,...,x] between simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A213895 Fixed points of a sequence h(n) defined by the minimum number of 6's in the relation n*[n,6,6,...,6,n] = [x,...,x] between simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A262213 Minimum number of 3's such that n*[n; 3, ..., 3, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

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