A094759 - OEIS

%I #35 Jul 10 2019 21:25:36

%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,

%T 27,6,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,

%U 50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74

%N Least k <= n such that n*sigma(k) = k*sigma(n), where sigma(n) is the sum of divisors of n (A000203).

%C Conjecture: There are infinitely many terms such that a(n)<n. A050973 has those n, A050972 has the a(n).

%C See A095301 for a version of A050973 that do not duplicate every n that has several smaller k of the same abundancy. - _Jeppe Stig Nielsen_, Jul 09 2015

%C That conjecture is an easy fact: Since, e.g., (6,28) is a friendly pair, then so is (6k,28k) for any multiplier k with gcd(42,k)=1. So any n=28k, where gcd(42,k)=1, satisfies a(n)<n. This even shows that A095301 does not have asymptotic density zero. - _Jeppe Stig Nielsen_, Jul 09 2015

%C This sequence is related to Theorem 1 on p. 173 of the Erdős link in the following way. For a given x, let us consider the set of integers such that a(n) <= x, which is equivalent to removing duplicates from the current sequence. This set would begin with: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 29, ... So this set has the same number of elements as the number of distinct terms numbers of the form sigma(n)/n with 1 <= n <=x. Then by Erdős, it is c1*x + o(x), with 6/Pi^2 < c1 < 1. With x = 10^7, we find c1 ~= 0.98208... - _Michel Marcus_, Jul 21 2015

%C a(n) is the least k which has the same abundancy index as n, that is, minimal k for which sigma(k)/k = sigma(n)/n. - _Antti Karttunen_, Jul 10 2019

%H Robert Israel, <a href="/A094759/b094759.txt">Table of n, a(n) for n = 1..10000</a>

%H P. Erdős, <a href="https://www.renyi.hu/~p_erdos/1959-21.pdf">Remarks on number theory II: Some problems on the sigma function</a>, Acta Arith., 5 (1959), 171-177.

%p N:= 100: # to get a(1) to a(N)

%p for n from 1 to N do

%p    v:= numtheory:-sigma(n)/n;

%p    if not assigned(R[v]) then R[v]:= n fi;

%p    A[n]:= R[v];

%p od:

%p seq(A[n],n=1..N); # _Robert Israel_, Jul 21 2015

%o (PARI) for(n=1,74,s=sigma(n);k=1;while(n*sigma(k)!=k*s,k++);print1(k,","));

%Y Cf. A000203, A050972, A050973, A094757, A094758, A326200.

%Y Cf. A095301 for n such that a(n) < n.

%Y Cf. A000396 (positions of 6's), A005820 (positions of 120's).

%K nonn

%O 1,2

%A _Amarnath Murthy_, May 30 2004

%E Edited and extended by _Don Reble_ and _Klaus Brockhaus_, May 31 2004