login
Maximum number of copies of unique prime divisors of n required to express n as a sum; a(1) = 0 by convention.
2

%I #2 Mar 30 2012 17:35:23

%S 0,1,1,2,1,2,1,4,3,2,1,3,1,2,3,8,1,4,1,4,3,2,1,6,5,2,9,4,1,3,1,16,3,2,

%T 5,8,1,2,3,6,1,4,1,4,6,2,1,10,7,8,3,4,1,12,5,7,3,2,1,6,1,2,7,32,5,5,1,

%U 4,3,5,1,15,1,2,10,4,7,5,1,12,27,2,1,7,5,2,3,8,1,9,7,4,3,2,5,20,1,12,9,15,1

%N Maximum number of copies of unique prime divisors of n required to express n as a sum; a(1) = 0 by convention.

%C For p prime, a(p^k) = p^(k-1). For p and q distinct primes, a(pq) = min(p,q).

%C a(n) >= n/sopf(n), where sopf is A008472; when the right hand side is an integer, this is an equality.

%e For n = 12, the sum 2+2+2+3+3 uses each prime factor at most 3 times, so a(12) = 3.

%o (PARI) a(n)=local(fm,p);if(n<=1,return(0));fm=factor(n);p=prod(i=1,matsize(fm)[1],1+x^fm[i,1]);for(k=0,n,if(polcoeff(p^k,n)!=0,return(k)))

%Y Cf. A164878, A164880, A027748.

%K nonn

%O 1,4

%A _Franklin T. Adams-Watters_, Aug 29 2009