A044050 a(n) = "length" of the aliquot sequence for n.
1, 2, 2, 3, 2, 1, 2, 3, 4, 4, 2, 7, 2, 5, 5, 6, 2, 4, 2, 7, 3, 6, 2, 5, 1, 7, 3, 1, 2, 15, 2, 3, 6, 8, 3, 4, 2, 7, 3, 4, 2, 14, 2, 5, 7, 8, 2, 6, 4, 3, 4, 9, 2, 13, 3, 5, 3, 4, 2, 11, 2, 9, 3, 4, 3, 12, 2, 5, 4, 6, 2, 9, 2, 5, 5, 5, 3, 11, 2, 7, 5, 6, 2, 6, 3, 9, 7, 7, 2, 10, 4, 6, 4, 4, 2, 9, 2, 3, 4, 5, 2, 18
Offset: 1
Examples
a(12) = 7: 12 is divisible by 1,2,3,4 and 6 so sigma(12)=16; 16 is divisible by 1,2,4 and 8 so sigma(16)=15; 15 is divisible by 1,3 and 5 so sigma(15)=9; 9 is divisible by 1 and 3 so sigma(9)=4; 4 is divisible by 1 and 2 so sigma(4)=3; 3 is divisible only by 1 so sigma(3)=1; 1 is not divisible by anything less than 1 so sigma(1)=0. The aliquot sequence is therefore 16, 15, 9, 4, 3, 1, 0 which is 7 elements long. Therefore a(12) = 7.
Links
- T. D. Noe, Table of n, a(n) for n=1..275
- Wolfgang Creyaufmüller, Prime Families
- Eric Weisstein's World of Mathematics, Aliquot Sequence
- P. Zimmermann, Aliquot Sequences
Programs
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Mathematica
f[n_]:=Plus@@Divisors[n]-n;lst2={};Do[lst={};a=k;Do[b=a;a=f[a];AppendTo[lst,a];If[a==0||a==b,Break[]],{n,7!}];AppendTo[lst2,Length[lst]],{k,5!}];lst2 (* Vladimir Joseph Stephan Orlovsky, Apr 24 2010 *)
Comments