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
Keywords
Examples
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.
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Programs
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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 *)
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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
Comments