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

1,1

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

Formula

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

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 *)

Prog

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

Crossrefs

Cf. A023196, A254572.

Sequence in context: A362871 A068924 A106224 * A360562 A244815 A226579

Adjacent sequences: A254568 A254569 A254570 * A254572 A254573 A254574

Keyword

nonn,easy

Author

Jeppe Stig Nielsen, Feb 01 2015

Status

approved