A309239 - OEIS

%I #32 Aug 19 2019 11:08:39

%S 2,4,6,8,10,12,14,16,18,22,24,26,30,32,34,36,38,40,46,48,54,56,58,60,

%T 62,64,72,74,82,84,86,90,94,96,98,100,106,108,118,120,122,126,128,132,

%U 134,142,144,146,150,158,160,162,166,168,178,180,192,194,198,202,206,210

%N 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.

%C 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.

%H Jinyuan Wang, <a href="/A309239/b309239.txt">Table of n, a(n) for n = 1..10000</a>

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

%e For m=1 the predicate would not be well defined.

%e For m=2, k=1=2-1, so 2 is a term.

%e For m=10=2*5: p=2->k=1=2-1; p=5->k=4=5-1 therefore 10 is a term.

%p filter:= proc(n) local p,k;

%p   for p in numtheory:-factorset(n) minus {2} do

%p      for k from 2 to p-3 by 2 do

%p        if (n-k) mod (p-k) = 0 then return false fi

%p   od od;

%p   true

%p end proc:

%p select(filter, [seq(i,i=2..200,2)]); # _Robert Israel_, Jul 17 2019

%t 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 *)

%o (PARI) findleast(m, p) = {for (k=1, p-1, if (!((m-k) % (p-k)), return(k)););}

%o 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

%Y Cf. A001414, A001222, A059975, A309155.

%K nonn

%O 1,1

%A _David James Sycamore_, Jul 17 2019