Displaying 1-10 of 19 results found.
Fixed points of sequence A262211 which yields the minimum number of 12's such that [n; 12, ..., 12, n] = [x; ..., x] for some x; [...] being continued fractions.
+20
19
19, 23, 31, 43, 59, 79, 103, 163, 179, 199, 227, 239, 251, 283, 331, 347, 383, 431, 439, 463, 467, 479, 487, 499, 523, 547, 587, 607, 631, 647, 683, 727, 827, 883, 907, 911, 919, 967, 991, 1019, 1031, 1051, 1087, 1123, 1171, 1303, 1327, 1423, 1499, 1511, 1523, 1531, 1567, 1571, 1667
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[12, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
PROG
(PARI) for(n=2, 9999, n== A262211(n)&&print1(n", "))
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.
+10
27
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
COMMENTS
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).
EXAMPLE
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.
MAPLE
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:
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;
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[2, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
PROG
(PARI) {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]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 1500, if(k==a(k), print1(a(k), ", "))); \\ based on code from Michael Somos
The minimum number of 1's in the relation n*[n,1,1,...,1,n] = [x,...,x] between simple continued fractions.
+10
24
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
COMMENTS
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
REFERENCES
A. Hurwitz, Über die Kettenbrüche, deren Teilnenner arithmetische Reihen bilden, Vierteljahrsschrift der Naturforschenden Gesellschaft in Zürich, Jahrg XLI, 1896, Jubelband II, S. 34-64.
EXAMPLE
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
MAPLE
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:
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[1, #] & /@ Range[2, 67] (* Michael De Vlieger, Sep 16 2015 *)
PROG
(PARI) {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 */ (PARI) {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 */
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.
+10
22
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
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).
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 *)
PROG
(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]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 1500, if(k==a(k), print1(a(k), ", ")));
Minimum number of 2's such that n*[n; 2, ..., 2, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.
+10
22
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
COMMENTS
Sequence A213891 lists fixed points of this sequence.
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[2, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)
PROG
(PARI) 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<k+2, d, n)))*n))[1]==c[#c]&&return(k))
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.
+10
5
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
COMMENTS
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).
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[3, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
PROG
(PARI)
{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]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 1500, if(k==a(k), print1(a(k), ", ")));
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.
+10
4
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
COMMENTS
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).
PROG
(PARI)
{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]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 1500, if(k==a(k), print1(a(k), ", ")));
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.
+10
4
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
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).
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 *)
PROG
(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]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 1500, if(k==a(k), print1(a(k), ", ")));
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.
+10
4
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
COMMENTS
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).
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[9, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
PROG
(PARI)
{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]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 1500, if(k==a(k), print1(a(k), ", ")));
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.
+10
3
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
COMMENTS
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).
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[4, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
PROG
(PARI)
{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]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 3000, if(k==a(k), print1(a(k), ", ")));
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