

A342731


a(1)=0; for n >= 2, a(n) is the number of iterations needed for the map (x/y) > (A000203(x)*A000005(y)) / (A000005(x)*A000203(y)) to reache a cycle, when starting from x = n. If no cycle is reachable, a(n) = 1.


0



0, 2, 3, 4, 4, 4, 5, 8, 10, 14, 5, 5, 6, 5, 5, 16, 11, 6, 15, 6, 9, 11, 6, 6, 10, 16, 15, 6, 6, 11, 17, 16, 6, 11, 6, 10, 16, 6, 6, 11, 10, 6, 12, 6, 7, 7, 7, 10, 16, 11, 7, 12, 16, 6, 7, 6, 7, 11, 12, 6, 18, 7, 8, 18, 10, 7, 12, 10, 7, 7, 11, 12, 17, 9, 7, 12
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OFFSET

1,2


COMMENTS

Extension of A342718 to rational numbers. I assume the map reaches only 3 possible cycles: [1], [(4/3) <> (7/6)], [(3/4) <> (6/7)], but have no proof for this.


LINKS

Table of n, a(n) for n=1..76.


EXAMPLE

n = 11; 11 > s(11)/d(11) = 6 > s(6)/d(6) = 3 > s(3)/d(3) = 2 > s(2)/d(2) = 3/2 > (s(3)*d(2))/(d(3)*s(2)) = 4/3 > (s(4)*d(3))/(d(4)*s(3)) = 7/6 > (s(7)*d(6))/(d(7)*s(6)) = 4/3 ..., so a(11) = 5 steps needed to reach a cycle [4/3 <> 7/6]; s(x) = A000203(x), d(x) = A000005(x).
n = 16; 16 > 31/5 > 16/3 > 31/10 > 32/9 > 63/26 > 104/63 > 315/208 > 260/217 > 49/64 > 133/127 > 5/8 > 4/5 > 7/9 > 12/13 > 2/3 > 3/4 > 6/7 > 3/4 >..., so a(16) = 16 steps needed to reach a cycle [3/4 <> 6/7].


MATHEMATICA

f[n_] := Divide @@ DivisorSigma[{1, 0}, n]; g[x_] := Divide @@ (f /@ {Numerator[x], Denominator[x]}); a[1] = 0; a[n_] := Length @ NestWhileList[g, n, UnsameQ, All]  3; Array[a, 100] (* Amiram Eldar, Mar 21 2021 *)


PROG

(PARI) a(n) = my(v=List([n]), s, t, x, y); for(k=1, oo, x=numerator(v[k]); y=denominator(v[k]); s=1; listput(v, t=sigma(x)*numdiv(y)/sigma(y)/numdiv(x)); while(v[s]!=t, s++); if(s<=k, return(s1))); \\ Jinyuan Wang, Mar 21 2021


CROSSREFS

Cf. A000005, A000203, A003601, A049642, A342718.
Sequence in context: A029085 A087875 A195848 * A099777 A221917 A131798
Adjacent sequences: A342728 A342729 A342730 * A342732 A342733 A342734


KEYWORD

nonn


AUTHOR

Ctibor O. Zizka, Mar 20 2021


STATUS

approved



