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Let r and s be such that r + s = n; a(n) = minimum value of tau(r) + tau(s).
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%I #14 Mar 18 2023 13:00:06

%S 2,3,3,4,3,4,3,4,4,5,3,4,3,4,4,5,3,4,3,4,4,5,3,4,4,5,4,6,3,4,3,4,4,5,

%T 4,6,3,4,4,5,3,4,3,4,4,5,3,4,4,5,4,6,3,4,4,5,4,5,3,4,3,4,4,5,4,6,3,4,

%U 4,5,3,4,3,4,4,5,4,6,3,4,4,5,3,4,4,5,4,6,3,4,4,5,4,5,4,6,3,4,4,5,3,4,3,4,4

%N Let r and s be such that r + s = n; a(n) = minimum value of tau(r) + tau(s).

%C a(p+1) = 3 if p is a prime. a(n) = 4 if n is the sum of two primes. For all even numbers > 4, a(n) = 4 by Goldbach's conjecture.

%H Antti Karttunen, <a href="/A085887/b085887.txt">Table of n, a(n) for n = 2..16385</a>

%e a(8) = 3, the partitions are (1,7), (2,6), (3,5), (4,4) which give 3, 6, 4 and 6 as the sum of the number of divisors of both parts.

%t Table[Min[Total/@DivisorSigma[0,IntegerPartitions[n,{2}]]],{n,2,120}] (* _Harvey P. Dale_, Mar 18 2023 *)

%o (PARI) A085887(n) = { my(m=0,k); for(r=1,n-1,if((m > k=(numdiv(r)+numdiv(n-r)))||!m, m = k)); m; }; \\ _Antti Karttunen_, Dec 14 2017

%Y Cf. A085883.

%K nonn

%O 2,1

%A _Amarnath Murthy_, Jul 08 2003

%E More terms from _David Wasserman_, Feb 10 2005