%I #14 Jun 12 2017 17:32:26

%S 2,2,3,2,5,70,7,2,3,33,11,1155,13,78,26,2,17,2156564410,19,6006,26,

%T 114,23,2156564410,5,33,3,1365,29,110,31,2,62,15,201,2156564410,37,30,

%U 14,961380175077106319535,41,1385670,43,2805,26,266,47,961380175077106319535

%N a(n) is the smallest number such that the sum of the n-th powers of its distinct prime divisors is divisible by n.

%C a(n) > 1 and a(n) = n if n prime. All terms are squarefree.

%e a(6) = 70 = 2*5*7; 2^6 + 5^6 + 7^6 = 133338 = 22223*6.

%e a(18)= 2*5*7*11*13*17*19*23*29 = 2156564410 because:

%e p^18 == 10, 9 (mod 18) for p = 2,3 respectively, and p^18 == 1 (mod 18) for p prime > 3. The minimum sum divisible by 18 is s = 2^18 + Sum_{k=3..10} prime(k)^18 whose residues sum to 10 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 18. Hence a(18) = 2156564410.

%p with(numtheory): T:=array(1..50):for n from 1 to 50 do:q:=0:for k from 2 to 7000 while(q=0)do:x:=factorset(k):s:=sum(x[j]^n ,j=1..nops(x)) :if irem(s,n)=0 then printf ( "%d %d \n",n,k):q:=1:else fi:od:if q=0 then for i from 1 to n do: T[i]:=irem(ithprime(i)^n,n):od:W:=convert(T,set):n1:=nops(W):n2:=W[n1]:n3:=W[n1-1]:

%p s:=0:p:=1:for a from 1 to n while(s<>n) do: if T[a]= 1 or T[a]=n2 or (T[a] = n3 and n2+n3<n) then s:=s+T[a]:p:=p*ithprime(a):else fi:if s= n then printf ( "%d %d \n",n,p):

%p else fi:od:else fi:od:

%Y Cf. A005063, A005064, A005065, A199167.

%K nonn

%O 1,1

%A _Michel Lagneau_, Nov 08 2011