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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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%I #46 Sep 17 2015 04:40:04

%S 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,

%T 23,13,59,29,23,19,8,39,11,18,17,27,29,19,23,43,29,59,23,15,11,55,74,

%U 35,41,26,35,9,23,35,41,57,59,14,29,23,47,34,59,67

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

%C 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.

%C Conjecture: the fixed points of this sequence are in A000057.

%C 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

%C 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

%D A. Hurwitz, Über die Kettenbrüche, deren Teilnenner arithmetische Reihen bilden, Vierteljahrsschrift der Naturforschenden Gesellschaft in Zürich, Jahrg XLI, 1896, Jubelband II, S. 34-64.

%H Bill Gosper, <a href="http://perl.plover.com/yak/cftalk/INFO/gosper.txt">Appendix 2 Continued Fraction Arithmetic</a>

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

%e 3* [3,1,1,1,3] = [10,1,10],so a(3)=3

%e 4* [4,1,1,1,1,1,4] = [18,2,18],so a(4)=5

%e 5* [5,1,1,1,1,5] = [28,28],so a(5)=4

%e 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

%e 7* [7,1,1,1,1,1,1,1,7] = [53,3,53], so a(7)=7

%p A213648 := proc(n)

%p local h,ins,c ;

%p for ins from 1 do

%p c := [n,seq(1,i=1..ins),n] ;

%p h := numtheory[cfrac](n*simpcf(c),quotients) ;

%p if op(1,h) = op(-1,h) then

%p return ins;

%p end if;

%p end do:

%p end proc: # _R. J. Mathar_, Jul 06 2012

%t 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 *)

%o (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 */

%o (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 */

%Y Cf. A000057, A262212 - A262220.

%K nonn

%O 2,1

%A _Art DuPre_, Jun 17 2012