#a(n) = sum(exp(2*Pi*i*A038566(k)/A038567(k-1)), k=1..A002088(n), where i is the imaginary unit. - Eric Desbiaux, Jul 31 2014

v038566 = [1, 1, 1, 2, 1, 3, 1, 2, 3, 4, 1, 5, 1, 2, 3, 4, 5, 6, 1, 3, 5, 7, 1, 2, 4, 5, 7, 8, 1, 3, 7, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 5, 7, 11, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 3, 5, 9, 11, 13, 1, 2, 4, 7, 8, 11, 13, 14, 1, 3, 5, 7, 9, 11, 13, 15, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

v038567 = [1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16]

v002088 = [0, 1, 2, 4, 6, 10, 12, 18, 22, 28, 32, 42, 46, 58, 64, 72, 80, 96, 102, 120, 128, 140, 150, 172, 180, 200, 212, 230, 242, 270, 278, 308, 324, 344, 360, 384, 396, 432, 450, 474, 490, 530, 542, 584, 604, 628, 650, 696, 712, 754, 774, 806, 830, 882, 900, 940, 964]

n=14
sum([ exp(  2*pi*I*v038566[k+1]/v038567[k]  ) for k in range(v002088[n]) ]).numerical_approx();