A226182 a(n) is the smallest integer k >= 2 such that the number of divisors d>1 of n + k with k|n + d equals A225867(n).

2, 2, 2, 3, 2, 2, 4, 2, 2, 3, 2, 3, 2, 2, 4, 3, 2, 2, 3, 5, 2, 3, 2, 3, 2, 2, 4, 3, 2, 5, 4, 3, 2, 3, 2, 3, 2, 2, 4, 3, 2, 2, 2, 7, 2, 3, 2, 2, 2, 5, 4, 3, 2, 4, 4, 2, 2, 3, 2, 5, 6, 3, 4, 3, 2, 4, 8, 2, 2, 5, 4, 7, 2, 2, 4, 3, 2, 2, 4, 5, 2, 3, 2, 2, 6, 7, 4

Offset: 6

Author

Vladimir Shevelev, May 30 2013

Keywords

• nonn

Examples

			Let n = 33. We begin with k = 2. Divisors>1 of 33 + 2 = 35 are d = 5,7,35. For all d, 33 + d is divisible by k = 2. But the number of such d is 3, while A225867(33)= 6. Therefore, a(33) > 2. Consider now k = 3. Divisors>1 of 33 + 3 = 36 are 2,3,4,6,9,12,18,36, but only for d = 3,6,9,12,18,36, 33 + d is divisible by k = 3. Since we have exactly A225867(33) = 6 such divisors, then k = 3 is required and a(33) = 3.
		

Links

• Peter J. C. Moses, Table of n, a(n) for n = 6..3005

Crossrefs

Cf. A225867, A225868, A188550, A188794.

Programs

Maple

A226182 := proc(n)
    local ak,k,nd,kpiv ;
    ak := 0 ;
    kpiv := 2 ;
    for k from 2 to n/2-1 do
        nd := 0 ;
        for d in numtheory[divisors](n+k) minus {1} do
            if modp(n+d,k) = 0 then
                nd := nd+1;
            end if;
        end do:
        if nd > ak then
            ak := max(ak,nd) ;
            kpiv := k ;
        end if;
    end do:
    kpiv ;
end proc: # R. J. Mathar, Jul 04 2013

Mathematica

Table[NestWhile[#+1&, 2, Max[Map[Count[(n+Rest[Divisors[n+#]])/#, Integer]&, Range[2, Floor[(n-2)/2]]]]-Count[(n+Rest[Divisors[n+#]])/#, _Integer] =!= 0&], {n,6,55}] (* _Peter J. C. Moses, Jun 03 2013 *)