This problem is much like A PowerSum Puzzle, so I will use the technique from

my solution there on this problem.

Given is S(1) = 3, S(2) = n, S(3) = 15, S(4) = 35, S(5) = m.

Then from the Newton-Girard Formulas P(1) = 3, P(2) = (-n+9)/2, and P(3) = (-3n+19)/2.

Then a, b, c are the roots of z^3 - 3z^2 + (-n+9)*z/2 + (3n-19)/2 = 0. And S(n) = 3*S(n-1) + (n-9)*S(n-2)/2 - (3n-19)*S(n-3)/2.

Then with S(1)-S(4): 35 = 3*15 + (n-9)*n/2 - (3n-19)*3/2, which simplifies to n^2 - 18n + 77 = 0, or n={7, 11}.

Given is n<10, then n=7, which simplifies the cubic and the S(n) relations to z^3 - 3z^2 + z + 1 = 0. And S(n) = 3*S(n-1) - S(n-2) - S(n-3).

Finally, m = S(5) = 3*35 - 15 - 7 = **83**.