A309239 Numbers m > 1 having the property that for any prime divisor p of m, k=p-1 is the smallest positive integer such that p-k|m-k.

2, 4, 6, 8, 10, 12, 14, 16, 18, 22, 24, 26, 30, 32, 34, 36, 38, 40, 46, 48, 54, 56, 58, 60, 62, 64, 72, 74, 82, 84, 86, 90, 94, 96, 98, 100, 106, 108, 118, 120, 122, 126, 128, 132, 134, 142, 144, 146, 150, 158, 160, 162, 166, 168, 178, 180, 192, 194, 198, 202, 206, 210

Offset: 1

Author

David James Sycamore, Jul 17 2019

Keywords

• nonn

Comments

All terms in this sequence are even. Proof: If m is odd and p a prime factor, then p is odd. Put t=p-2, then m-t is even, and since p-(p-2)=2, p-t|m-t. Therefore k <= p-2 < p-1, and p-1 is not the smallest number satisfying the definition of k. Therefore there are no odd numbers in this sequence.

Examples

			For m=1 the predicate would not be well defined.
For m=2, k=1=2-1, so 2 is a term.
For m=10=2*5: p=2->k=1=2-1; p=5->k=4=5-1 therefore 10 is a term.
		

Links

• Jinyuan Wang, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001414, A001222, A059975, A309155.

Programs

Maple

filter:= proc(n) local p,k;
  for p in numtheory:-factorset(n) minus {2} do
     for k from 2 to p-3 by 2 do
       if (n-k) mod (p-k) = 0 then return false fi
  od od;
  true
end proc:
select(filter, [seq(i,i=2..200,2)]); # Robert Israel, Jul 17 2019

Mathematica

fQ[n_, p_] := Module[{k = 1}, While[!Divisible[n - k, p - k], k++]; k == p - 1]; aQ[n_] := And @@ (fQ[n, #] & /@ FactorInteger[n][[;; , 1]]); Select[Range[2, 200], aQ] (* Amiram Eldar, Jul 17 2019 *)

PARI

findleast(m, p) = {for (k=1, p-1, if (!((m-k) % (p-k)), return(k)););}
isok(m) = {if (m == 1, return(0)); my (f = factor(m)); for (i=1, #f~, my(k = findleast(m, f[i,1])); if (k != f[i,1] - 1, return (0));); return (1);} \\ Michel Marcus, Aug 18 2019

Formula

A001414(m) - A001222(m) = A059975(m) = A309155(m), for m in this sequence.