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a(n) = A008475(n) + 1.
3

%I #22 Jan 05 2017 21:01:26

%S 1,3,4,5,6,6,8,9,10,8,12,8,14,10,9,17,18,12,20,10,11,14,24,12,26,16,

%T 28,12,30,11,32,33,15,20,13,14,38,22,17,14,42,13,44,16,15,26,48,20,50,

%U 28,21,18,54,30,17,16,23,32,60,13,62,34,17,65,19,17,68,22,27,15,72,18,74

%N a(n) = A008475(n) + 1.

%C If n = Product (p_i^k_i) for i = 1, …, j then a(n) is sum of divisor d from set of divisors{1, p_1^k_1, p_2^k_2, …, p_j^k_j}.

%F a(n) = [Sum_(i=1,…, j) p_i^k_i] + 1 = A000203(n) - A178636(n).

%F a(1) = 1, a(p) = p+1, a(pq) = p+q+1, a(pq...z) = p+q+...+z+1, a(p^k) = p^k+1, for p, q = primes, k = natural numbers, pq...z = product of k (k > 2) distinct primes p, q, ..., z.

%e For n = 12, set of divisors {1, p_1^k_1, p_2^k_2, …, p_j^k_j}: {1, 3, 4}. a(12) = 1+3+4=8.

%t f[n_] := 1 + Plus @@ Power @@@ FactorInteger@ n; f[1] = 1; Array[f, 60]

%o (PARI) a(n)=local(t); if(n<1, 0, t=factor(n); 1+sum(k=1, matsize(t)[1], t[k, 1]^t[k, 2])) /* _Anton Mosunov_, Jan 05 2017 */

%Y Cf. A008475, A023888.

%K nonn

%O 1,2

%A _Jaroslav Krizek_, Apr 04 2009

%E Edited by _N. J. A. Sloane_, Apr 07 2009