A058972 For a rational number p/q let f(p/q) = sum of aliquot divisors of p+q divided by number of divisors of p+q; sequence gives numbers k such that, starting at k/1 and iterating f, an integer is eventually reached.
3, 9, 15, 24, 25, 29, 33, 35, 50, 51, 55, 57, 59, 63, 73, 79, 80, 81, 85, 87, 89, 90, 95, 99, 105, 119, 120, 121, 128, 131, 139, 143, 145, 169, 177, 179, 181, 183, 193, 195, 201, 203, 204, 211, 215, 217, 218, 219, 221, 225, 227, 233, 247, 248, 255, 273, 275, 288
Offset: 1
Examples
f(9/1) = 8/4 = 2, an integer, so 9 is in the sequence; f(10/1) = 1/2 and f(1/2)=1/2, so 10 is not in the sequence.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- P. Schogt, The Wild Number Problem: math or fiction?, arXiv preprint arXiv:1211.6583 [math.HO], 2012. - From _N. J. A. Sloane_, Jan 03 2013
Programs
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Haskell
import Data.Ratio ((%), numerator, denominator) a058972 n = a058972_list !! (n-1) a058972_list = map numerator $ filter ((f [])) [1..] where f ys q = denominator y == 1 || not (y `elem` ys) && f (y : ys) y where y = a001065 q' % a000005 q' q' = numerator q + denominator q -- Reinhard Zumkeller, Jun 15 2013 -
Mathematica
f[r_] := If[init == False && IntegerQ[r], r, init = False; p = Numerator[r]; q = Denominator[r]; d = Most[Divisors[p+q]]; Total[d]/(Length[d]+1)]; ok[n_] := IntegerQ[ init = True; FixedPoint[f, n/1]]; ok[1] = False; A058972 = Select[ Range[300], ok] (* Jean-François Alcover, Dec 21 2011 *)
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PARI
f2(p,q) = (sigma(p+q)-p-q)/numdiv(p+q); f1(r) = f2(numerator(r), denominator(r)); loop(list) = {my(v=Vecrev(list)); for (i=2, #v, if (v[i] == v[1], return(1)););} ff(n) = {my(ok=0, m=f2(n,1), list=List()); while(denominator(m) != 1, m = f1(m); listput(list, m); if (loop(list), return (0));); return(m);} isok(m) = ff(m) > 0; \\ Michel Marcus, Feb 09 2022
Extensions
Corrected and extended by Matthew Conroy, Apr 18 2001