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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.

%I A213891 #56 Jun 28 2025 22:15:27
%S A213891 3,11,19,43,67,83,107,131,139,163,211,283,307,331,347,467,491,499,523,
%T A213891 547,563,571,587,619,659,691,739,787,811,859,883,907,947,971,1019,
%U A213891 1051,1123,1163,1171,1283,1291,1307
%N 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.
%C A213891 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.
%C A213891 Let [a,b,...,c] = a+1/(b+(1/...+1/c)) represent a simple continued fraction.
%C A213891 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.
%C A213891 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.
%C A213891 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.
%C A213891 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.
%C A213891 If we do the same for inserting 2s as we just described for 1s, we get this sequence here.
%C A213891 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).
%e A213891 The basic sequence h(n) (= A262212) is for n = 3,4,5,..:
%e A213891 3*[3, 2, 2, 2, 3] = [10,4,10], h(3) = 3: the first fixed point a(1) = 3.
%e A213891 4*[4, 2, 2, 2, 4] = [17, 1, 1, 1, 17], h(4) = 3;
%e A213891 5*[5, 2, 2, 5] = [27, 27], h(5) = 2;
%e A213891 6*[6, 2, 2, 2, 6] = [38, 2, 38], h(6) = 3;
%e A213891 (...)
%e A213891 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.
%p A213891 simpcf := proc(L)
%p A213891         if nops(L) = 1 then
%p A213891                 op(1,L) ;
%p A213891         else
%p A213891                 op(1,L)+1/procname([op(2..nops(L),L)]) ;
%p A213891         end if;
%p A213891 end proc:
%p A213891 A213891aux := proc(n)
%p A213891         local h,ins,c ;
%p A213891         for ins from 1 do
%p A213891                 c := [n,seq(2,i=1..ins),n] ;
%p A213891                 h := numtheory[cfrac](n*simpcf(c),quotients) ;
%p A213891                 if op(1,h) = op(-1,h) then
%p A213891                         return ins;
%p A213891                 end if;
%p A213891         end do:
%p A213891 end proc:
%p A213891 A213891 := proc(n)
%p A213891         if n = 1 then
%p A213891                 3;
%p A213891         else
%p A213891                 for a from procname(n-1)+1 do
%p A213891                         if A213891aux(a) = a then
%p A213891                                 return a;
%p A213891                         end if;
%p A213891                 end do:
%p A213891         end if;
%p A213891 end proc: # _R. J. Mathar_, Jul 06 2012
%t A213891 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 *)
%o A213891 (PARI) {a(n) = local(t, m=1); if( n<2, 0, while( 1,
%o A213891    t = contfracpnqn( concat([n, vector(m,i,2), n]));
%o A213891    t = contfrac(n*t[1,1]/t[2,1]);
%o A213891    if(t[1]<n^2 || t[#t]<n^2, m++, break));
%o A213891 m)};
%o A213891 for(k=1,1500,if(k==a(k),print1(a(k), ", "))); \\ based on code from Michael Somos
%o A213891 (PARI) forprime(p=2,999, A262212(p)==p&&print1(p",")) \\ _M. F. Hasler_, Sep 30 2015
%Y A213891 Cf. A000057, A213892 - A213899, A261311; A213358.
%Y A213891 Cf. A213648, A262212 - A262220, A213900, A262211.
%K A213891 nonn
%O A213891 1,1
%A A213891 _Art DuPre_, Jun 23 2012
%E A213891 Edited by _R. J. Mathar_ and _T. D. Noe_, Jul 06 2012
%E A213891 Edited by _M. F. Hasler_, Sep 30 2015