A254571 Least multiplier k such that k*n is abundant or perfect (A023196).

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, 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, 2, 1, 6, 1, 6, 3, 2, 3, 4, 1, 6, 1, 2, 3, 6, 1, 4, 3, 2

Offset: 1

Author

Jeppe Stig Nielsen, Feb 01 2015

Keywords

• nonn
• easy

Comments

See A254572(n)=a(n)*n for the actual non-deficient numbers.

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.

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A023196, A254572.

Programs

Maple

f:= proc(n) local k; uses numtheory;
      for k from 1 to 4 do if sigma(k*n)>=2*k*n then return k fi od:
      6
end proc:
map(f, [$1..100]); # Robert Israel, Feb 10 2019

Mathematica

a[n_] := Do[If[DivisorSigma[1, k*n] >= 2*k*n, Return[k]], {k, {1, 2, 3, 4, 6}}];
Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Feb 09 2023 *)

PARI

a(n) = for(k=1,6,if(sigma(k*n)>=2*k*n,return(k)))

Formula

a(A023196(n)) = 1. - Michel Marcus, Feb 02 2015