A199583 a(n) is the smallest number such that the sum of the n-th powers of its distinct prime divisors is divisible by n.
2, 2, 3, 2, 5, 70, 7, 2, 3, 33, 11, 1155, 13, 78, 26, 2, 17, 2156564410, 19, 6006, 26, 114, 23, 2156564410, 5, 33, 3, 1365, 29, 110, 31, 2, 62, 15, 201, 2156564410, 37, 30, 14, 961380175077106319535, 41, 1385670, 43, 2805, 26, 266, 47, 961380175077106319535
Offset: 1
Keywords
Examples
a(6) = 70 = 2*5*7; 2^6 + 5^6 + 7^6 = 133338 = 22223*6. a(18)= 2*5*7*11*13*17*19*23*29 = 2156564410 because: 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.
Programs
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Maple
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]: 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
Comments