A058971 For a rational number p/q let f(p/q) = sum of divisors of p+q divided by number of divisors of p+q; a(n) is obtained by iterating f, starting at n/1, until an integer is reached, or if no integer is ever reached then a(n) = 0.

3, 2, 6, 3, 3, 4, 10, 87, 6, 6, 9, 7, 6, 6, 87, 9, 6, 10, 7, 8, 9, 12, 9, 15, 12, 10, 16, 15, 9, 16, 12, 12, 15, 12, 87, 19, 15, 14, 19, 21, 12, 22, 14, 13, 18, 24, 34, 19, 12, 18, 0, 27, 15, 18, 15, 20, 24, 30, 14, 31, 24, 18, 51, 21, 18, 34, 21, 24, 18, 36, 24, 37, 30, 21, 37

Offset

1,1

Comments

a(p-1) = (p+1)/2 for all odd primes p. Thus there are infinitely many distinct terms. - Ely Golden, Mar 03 2018

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

Example

1 -> (1+2)/2 = 3/2 -> (1+5)/2 = 3, so a(1) = 3.
51 -> 49/3 -> 49/3 -> ..., so a(51) = 0.

Maple

with(numtheory); f := proc(n) if whattype(n) = integer then sigma(n+1)/sigma[0](n+1) else sigma(numer(n)+denom(n))/sigma[0](numer(n)+denom(n)); fi; end;

Mathematica

f[x_] := With[{p = Numerator[x], q = Denominator[x]}, DivisorSigma[1, p+q]/DivisorSigma[0, p+q]]; a[n_] := If[ IntegerQ[ r = FixedPoint[f, n, SameTest -> (#1 == #2 || IntegerQ[#2] &)]], r, 0]; Table[a[n], {n, 1, 75}] (* Jean-François Alcover, Jul 18 2012 *)

Prog

(Haskell)
import Data.Ratio ((%), numerator, denominator)
a058971 n = f [n % 1] where
   f xs@(x:_) | denominator y == 1 = numerator y
              | y `elem` xs        = 0
              | otherwise          = f (y : xs)
              where y = (a000203 x') % (a000005 x')
                    x' = numerator x + denominator x
-- Reinhard Zumkeller, Aug 02 2012

Crossrefs

Cf. A058972, A058977.

Sequence in context: A263353 A248945 A131969 * A186204 A137324 A011209

Adjacent sequences: A058968 A058969 A058970 * A058972 A058973 A058974

Keyword

nonn,easy,nice

Author

N. J. A. Sloane, Jan 14 2001

Extensions

More terms from Matthew Conroy, Apr 18 2001, who remarks that a(51) = a(655) = a(1039) = 0 are all the zeros of a(n) for n < 10^5

No more zero terms <= 10^6 found by Reinhard Zumkeller, Aug 02 2012

Status

approved