A213648 The minimum number of 1's in the relation n*[n,1,1,...,1,n] = [x,...,x] between simple continued fractions.

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

Author

Art DuPre, Jun 17 2012

Keywords

• nonn

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

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

Examples

			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
		

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.

Links

• Bill Gosper, Appendix 2 Continued Fraction Arithmetic

Crossrefs

Cf. A000057, A262212 - A262220.

Programs

Maple

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

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 *)

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 */

Formula

Conjecture: a(n)=A001177(n)-1.