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%I #21 Feb 09 2023 04:42:33
%S 6,3,2,3,4,1,4,3,2,2,6,1,6,2,2,3,6,1,6,1,2,3,6,1,4,3,2,1,6,1,6,3,2,3,
%T 2,1,6,3,2,1,6,1,6,2,2,3,6,1,4,2,2,2,6,1,4,1,2,3,6,1,6,3,2,3,4,1,6,3,
%U 2,1,6,1,6,3,2,3,4,1,6,1,2,3,6,1,4,3,2
%N Least multiplier k such that k*n is abundant or perfect (A023196).
%C See A254572(n)=a(n)*n for the actual non-deficient numbers.
%C The range is {1,2,3,4,6}. Clearly a(n) <= 6 because 6*n is abundant for any n. No n can have a(n)=5. Suppose otherwise. There exists a prime p smaller than 5 which does not divide n (if not, 6|n and a(n)=1). That prime p (either 2 or 3) will boost the abundancy more than does 5. In particular (sigma(p*n))/(p*n) > (sigma(5*n))/(5*n) >= 2, but then a(n) should be p, a contradiction.
%H Robert Israel, <a href="/A254571/b254571.txt">Table of n, a(n) for n = 1..10000</a>
%F a(A023196(n)) = 1. - _Michel Marcus_, Feb 02 2015
%p f:= proc(n) local k; uses numtheory;
%p for k from 1 to 4 do if sigma(k*n)>=2*k*n then return k fi od:
%p 6
%p end proc:
%p map(f, [$1..100]); # _Robert Israel_, Feb 10 2019
%t a[n_] := Do[If[DivisorSigma[1, k*n] >= 2*k*n, Return[k]], {k, {1, 2, 3, 4, 6}}];
%t Table[a[n], {n, 1, 100}] (* _Jean-François Alcover_, Feb 09 2023 *)
%o (PARI) a(n) = for(k=1,6,if(sigma(k*n)>=2*k*n,return(k)))
%Y Cf. A023196, A254572.
%K nonn,easy
%O 1,1
%A _Jeppe Stig Nielsen_, Feb 01 2015
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