A261311 -id:A261311 - cp's OEIS frontend

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.

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.

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]

A262212 Minimum number of 2's such that n*[n; 2, ..., 2, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Original entry on oeis.org

1, 3, 3, 2, 3, 5, 7, 11, 5, 11, 3, 6, 5, 11, 15, 7, 11, 19, 11, 11, 11, 21, 7, 14, 13, 35, 11, 4, 11, 29, 31, 11, 7, 5, 11, 18, 19, 27, 23, 9, 11, 43, 11, 11, 21, 45, 15, 41, 29, 7, 27, 26, 35, 11, 23, 19, 9, 19, 11, 30, 29, 11, 63, 20, 11, 67, 7, 43, 5, 69, 23, 35, 37, 59, 19, 11, 27, 25, 47, 107, 9, 83, 11, 23, 43, 19, 23, 43, 11, 41, 43, 59, 45, 59, 31

Offset: 2

M. F. Hasler, Sep 15 2015

nonn

Sequence A213891 lists fixed points of this sequence.

Cf. A213648, A262213 - A262220, A213900, A262211; 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[2, #] & /@ 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}
A262212(n,d=2)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

A262211 Minimum number of 12's such that n*[n; 12, ..., 12, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Original entry on oeis.org

1, 1, 1, 2, 1, 5, 3, 5, 5, 9, 1, 6, 5, 5, 7, 8, 5, 19, 5, 5, 9, 23, 3, 14, 13, 17, 5, 2, 5, 31, 15, 9, 17, 5, 5, 36, 19, 13, 11, 19, 5, 43, 9, 5, 23, 45, 7, 5, 29, 17, 13, 12, 17, 29, 11, 19, 5, 59, 5, 30, 31, 5, 31, 20, 9, 65, 17, 23, 5, 13, 11, 3, 73, 29, 19, 29, 13, 79, 23, 53, 19, 81, 5, 8, 43, 5, 19, 14, 5, 41, 23, 31, 45, 59, 15, 48, 5, 29

Offset: 2

M. F. Hasler, Sep 15 2015

nonn

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).

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

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[12, #] & /@ 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}
A262211(n,d=12)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

A213892 Fixed points of a sequence h(n) defined by the minimum number of 3's in the relation n*[n,3,3,...,3,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

2, 7, 19, 31, 47, 67, 71, 83, 151, 163, 167, 223, 227, 271, 307, 331, 359, 379, 431, 463, 479, 487, 499, 631, 643, 683, 691, 743, 787, 811, 839, 863, 947, 967, 1019, 1051, 1087, 1103, 1123, 1163, 1259, 1279, 1307, 1319, 1399, 1423, 1451, 1471

Offset: 1

Art DuPre, Jun 23 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,3,3,...,3,n] and increase the number of 3's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 3, 3, 2] = [4, 1, 1, 1, 1, 4],
3 * [3, 3, 3] = [9, 1, 9],
4 * [4, 3, 3, 3, 3, 3, 4] = [17, 4, 1, 2, 1, 4, 17],
5 * [5, 3, 3, 5] = [26, 1, 1, 26],
6 * [6, 3, 3, 3, 3, 3, 6] = [37, 1, 4, 2, 4, 1, 37],
7 * [7, 3, 3, 3, 3, 3, 3, 3, 7] = [51, 8, 2, 1, 2, 8, 51].
The number of 3's needed defines the sequence h(n) = 2, 1, 5, 2, 5, 7, 5, 9, 2, ... (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) = 3*f(n-1) + f(n-2), A006190, A003688, A052924, 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) = 3*f(n-1) + f(n-2).
The above sequence h() is recorded as A262213. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213891, A213893 - 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[3, #] == # &] (* 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,3), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A213896 Fixed points of a sequence h(n) defined by the minimum number of 7's in the relation n*[n,7,7,...,7,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

2, 3, 19, 31, 67, 79, 103, 127, 139, 151, 167, 179, 191, 263, 283, 359, 383, 443, 463, 479, 491, 503, 571, 631, 691, 787, 827, 883, 919, 1019, 1087, 1171, 1291, 1303, 1307, 1327, 1399, 1423, 1451, 1487

Offset: 1

Art DuPre, Jun 23 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,7,7,..,7,n] and increase the number of 7's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 7, 7, 2] = [4, 3, 1, 1, 3, 4],
3 * [3, 7, 7, 7, 3] = [9, 2, 2, 1, 1, 1, 2, 2, 9] ,
4 * [4, 7, 7, 7, 7, 7, 4] = [16, 1, 1, 3, 1, 1, 1, 6, 1, 1, 1, 3, 1, 1, 16],
5 * [5, 7, 7, 5] = [25, 1, 2, 2, 1, 25] ,
6 * [6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6] = [36, 1, 5, 3, 1, 4, 10, 1, 2, 2, 4, 2, 2, 1, 10, 4, 1, 3, 5, 1, 36],
7 * [7, 7, 7] = [49, 1, 49] .
The number of 7's needed defines the sequence h(n) = 2, 3, 5, 2, 11, 1, 5, 11, 2,...  (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)=7*f(n-1)+f(n-2), A054413, A015453, etc. This would mean that a prime is in the sequence A213896 if and only if it divides some term in each of the sequences satisfying f(n)=7*f(n-1)+f(n-2).
The above sequence h() is recorded as A262217. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213891 - A213895, A213897 - A213899, A261311; A213358.

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

{a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,7), 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

Art DuPre, Jun 24 2012

nonn

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

Cf. A000057, A213891 - A213896, A213898, 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[8, #] == # &] (* 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,8), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

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.

Original entry on oeis.org

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]

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]

cp's OEIS Frontend

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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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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A262212 Minimum number of 2's such that n*[n; 2, ..., 2, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

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A262211 Minimum number of 12's such that n*[n; 12, ..., 12, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

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A213892 Fixed points of a sequence h(n) defined by the minimum number of 3's in the relation n*[n,3,3,...,3,n] = [x,...,x] between simple continued fractions.

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A213896 Fixed points of a sequence h(n) defined by the minimum number of 7's in the relation n*[n,7,7,...,7,n] = [x,...,x] between simple continued fractions.

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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.

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