A349483 Length of cycle reached when iterating the mapping x-> n*A035116(x) on 1.

1, 2, 2, 2, 4, 2, 5, 2, 2, 7, 2, 1, 2, 5, 6, 1, 2, 2, 2, 3, 2, 2, 2, 1, 1, 2, 4, 1, 2, 1, 2, 3, 1, 2, 1, 1, 2, 2, 1, 3, 2, 4, 2, 1, 3, 2, 2, 4, 3, 6, 1, 1, 2, 2, 3, 3, 1, 2, 2, 4, 2, 2, 1, 3, 3, 3, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 3, 2, 8, 1, 2, 2, 3, 3, 2, 1, 3, 2, 3, 1, 1, 1, 2, 3, 1, 2, 4, 1, 2

Offset

1,2

Comments

The terms 1-25 all appear below 10^8; the last of these are a(12545280) = 21, a(12684672) = 24, and a(96940800) = 25. - Charles R Greathouse IV, Nov 23 2021

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Example

For n = 2, 1 --> 2 --> 8 --> 32 --> 72 --> 288 --> 648 --> 800 --> 648. The cycle reached has just two terms: 648 and 800. Therefore, a(2) = 2.

Mathematica

a[n_] := Module[{s = NestWhileList[n*DivisorSigma[0, #]^2 &, 1, UnsameQ, All]}, Differences[Position[s, s[[-1]]]][[1, 1]]]; Array[a, 100] (* Amiram Eldar, Nov 19 2021 *)

Prog

(PARI) brent(f, x)=my(pow=1, lam=1, tortoise=x, hare=f(x)); while(tortoise!=hare, if(pow==lam, tortoise=hare; pow<<=1; lam=0); hare=f(hare); lam++); lam
a(n)=brent(k->n*numdiv(k)^2, 1) \\ Charles R Greathouse IV, Nov 19 2021

Crossrefs

Cf. A035116.

Similar sequences: A349410.

Sequence in context: A173738 A135838 A279966 * A114349 A186749 A133265

Adjacent sequences: A349480 A349481 A349482 * A349484 A349485 A349486

Keyword

nonn

Author

Tejo Vrush, Nov 19 2021

Status

approved