%I
%S 2,5,7,10,8,15,8,18,16,18,10,29,8,19,25,28,10,33,10,35,26,20,12,50,18,
%T 20,31,36,12,51,10,42,27,23,33,62,8,22,30,60,12,53,10,40,52,22,14,78,
%U 20,41,28,38,12,63,36,63,30,24,16,95,8,23,59,60,32,54,10
%N a(n) = Sum_{i=1..A000005(n)} d(n+k(i)), where d(t) is the number of divisors of t and k(i) is the ith divisor of n.
%C Let S(n,n) be the number of solutions of the equation n/x + n/y = c where n, c, x, and y are positive integers. Then S(n,n) = Sum_{i=1..A000005(n)} d(n+k(i)), where d(t) is the number of divisors of t and k(i) is the ith divisor of n.
%C For c = 1 , S(n,n) = A000005(n).
%C Let S(n,m) be the number of solutions of the equation n/x + m/y = c where n, m, c, x, and y are positive integers, n not equal to m. Let k(i) be the ith divisor of n, and k(j) the jth divisor of m. Let d(t) be the number of divisors of t. Let R = d(k(i) + k(j)). Then S(n,m) = Sum_{i=1..A000005(n)} Sum_{j=1..A000005(m)} [R*1 if gcd(k(i),k(j)) = 1 , R*0 else].
%C For c = 1 , S(n,m) = A000005(n) * A000005(m)  P, where P is the number of divisor pairs such that gcd(k(i),k(j)) >= 2.
%H R. L. Graham, <a href="http://www.math.ucsd.edu/~ronspubs/13_03_Egyptian.pdf">Paul Erdos and Egyptian Fractions</a>, Bolyai Society Mathematical Studies 25, pp 289309, 2013.
%e For n = 4, divisors of 4 are 1, 2, 4; thus a(4) = d(4+1) + d(4+2) + d(4+4) = d(5) + d(6) + d(8) = 2 + 4 + 4 = 10.
%t a[n_] := Sum[DivisorSigma[0, d + n], {d, Divisors@n}]; Array[a, 67] (* _Giovanni Resta_, Mar 21 2017 *)
%o (PARI) for(n=1, 101, print1(sumdiv(n, d, numdiv(d + n)),", ")) \\ _Indranil Ghosh_, Mar 22 2017
%o (Python)
%o from sympy import divisor_count, divisors
%o def a(n):
%o ....D=divisors(n)
%o ....return sum([divisor_count(n + D[i]) for i in xrange(0, len(D))]) # _Indranil Ghosh_, Mar 22 2017
%Y Cf. A000005, A018892, A063647, A089233, A275387.
%K nonn
%O 1,1
%A _Ctibor O. Zizka_, Mar 21 2017
%E a(21)a(67) from _Giovanni Resta_, Mar 21 2017
