A257484 Numbers k such that lambda(sum of divisors of k that are congruent to 0 mod 3) = lambda(sum of divisors of k that are congruent to 1 mod 3) = lambda(sum of divisors of k that are congruent to 2 mod 3) where lambda is the Carmichael function (A002322).

78, 222, 234, 294, 312, 366, 375, 438, 582, 618, 666, 834, 876, 882, 888, 936, 942, 1086, 1095, 1098, 1125, 1158, 1176, 1236, 1314, 1464, 1482, 1536, 1545, 1662, 1746, 1752, 1842, 1878, 2013, 2022, 2028, 2094, 2166, 2274, 2316, 2328, 2382, 2472, 2502, 2526

Offset

1,1

Comments

A majority of numbers having three distinct prime divisors are in the sequence, but the number 1482 contains four distinct prime divisors {2,3,13,19}.

If a(n) is a squarefree number (subsequence 78, 222, 366, 438, 582, 618, 834, 942, 1086, 1095, 1158, 1482, 1545, 1662, 1842, 1878, 2013, 2022, 2094, 2274, 2382, 2526,...), the number 3*a(n) is also in the sequence because the sum of divisors of a(n) that are congruent to 0 mod 3 is equal to k, the sum of divisors of 3*a(n) that are congruent to 0 mod 3 is equal to 4k, and lambda(4*k) = lcm(lambda(4),lambda(k)) = lcm(2,lambda(k)) = lambda(k).

The sequence of the corresponding values lambda is {6, 18, 6, 18, 12, 30, 12, 36, 42, 12, 18, 12, 36, 18, 36, 12, 78, 12, 36, 30, 12, 96, 36, 12, 36, 60, 12, 30, 12, 138, 42, 36, 30, 156, 30, 156, 60, 60, 126, 36, 96, 84, 198, 12, 12, 210, 30,...}.

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Example

78 is in the sequence because the divisors of 78 are {1,2,3,6,13,26,39,78} and the divisors congruent to 0 mod 3 are {3,6,39,78} => sum=126, the divisors congruent to 1 mod 3 are {1,13} => sum=14, the divisors congruent to 2 mod 3 are {2,26} => sum=28, and lambda(126)=lambda(14)=lambda(28) = 6.

Maple

with(numtheory):nn:=2600:
  for n from 1 to nn do:
   s0:=0:s1:=0:s2:=0:
   x:=divisors(n):n0:=nops(x):
    for i from 1 to n0 do:
     q:=x[i]:
      if irem(q, 3)=0 then s0:=s0+q:
       else
        if irem(q, 3)=1 then s1:=s1+q:
       else
        s2:=s2+q:
        fi:fi:
       od:
      if lambda(s0)=lambda (s1) and lambda(s1)=lambda(s2)
      then
      printf(`%d, `, n):
      else
      fi:
      od:

Mathematica

lst={}; f[x_] := Plus @@ Select[Divisors[x], Mod[#, 3]==0 &]; g[x_] := Plus @@ Select[Divisors[x], Mod[#, 3]==1 &]; h[x_] := Plus @@ Select[Divisors[x], Mod[#, 3]==2 &]; Do[If[CarmichaelLambda[f[n]]== CarmichaelLambda[g[n]]&& CarmichaelLambda[f[n]]== CarmichaelLambda[h[n]], AppendTo[lst, n]], {n, 1, 2600}]; lst

Prog

(PARI) lambda(n)=lcm(znstar(n)[2]);
isok(n) = {my(sd0=sumdiv(n, d, d*((d % 3)==0))); my(sd1=sumdiv(n, d, d*((d % 3)==1))); my(sd2=sumdiv(n, d, d*((d % 3)==2))); sd0 && sd1 && sd2 && (lambda(sd0) == lambda(sd1)) && (lambda(sd0)==lambda(sd2)); }
lista(nn) = for (n=1, nn, if (isok(n), print1(n, ", "))); \\ Michel Marcus, May 02 2015

Crossrefs

Cf. A002322, A248881.

Sequence in context: A389203 A389179 A232386 * A068130 A118938 A362546

Adjacent sequences: A257481 A257482 A257483 * A257485 A257486 A257487

Keyword

nonn

Author

Michel Lagneau, Apr 26 2015

Status

approved