[THOUGHT] I will use Newton's identities to obtain the elementary symmetric sums e1..e7 from the given power sums p1..p7 of a..g. These will define the monic polynomial with roots a..g. Then I'll use the induced linear recurrence on power sums s_n = a^n+...+g^n to compute s_8 through s_12. Finally, I'll add 10^12 to s_12 to get P_12. [/THOUGHT] [PYTHON] from sympy import symbols, Eq, solve

Given power sums for a..g

p = { 1: 58, 2: 666, 3: 8944, 4: 129846, 5: 1972408, 6: 30870006, 7: 493478344 }

Unknown elementary symmetric sums e1..e7

e_symbols = symbols('e1:8') # e1,e2,...,e7 e1,e2,e3,e4,e5,e6,e7 = e_symbols

eqs = []

Newton's identities for k=1..7

p1 - e1 = 0

eqs.append(Eq(p[1] - e1, 0))

p2 - e1p1 + 2e2 = 0

eqs.append(Eq(p[2] - e1p[1] + 2e2, 0))

p3 - e1p2 + e2p1 - 3*e3 = 0

eqs.append(Eq(p[3] - e1p[2] + e2p[1] - 3*e3, 0))

p4 - e1p3 + e2p2 - e3p1 + 4e4 = 0

eqs.append(Eq(p[4] - e1p[3] + e2p[2] - e3p[1] + 4e4, 0))

p5 - e1p4 + e2p3 - e3p2 + e4p1 - 5*e5 = 0

eqs.append(Eq(p[5] - e1p[4] + e2p[3] - e3p[2] + e4p[1] - 5*e5, 0))

p6 - e1p5 + e2p4 - e3p3 + e4p2 - e5p1 + 6e6 = 0

eqs.append(Eq(p[6] - e1p[5] + e2p[4] - e3p[3] + e4p[2] - e5p[1] + 6e6, 0))

p7 - e1p6 + e2p5 - e3p4 + e4p3 - e5p2 + e6p1 - 7*e7 = 0

eqs.append(Eq(p[7] - e1p[6] + e2p[5] - e3p[4] + e4p[3] - e5p[2] + e6p[1] - 7*e7, 0))

solution = solve(eqs, e_symbols, dict=True) solution print(solution)

Extract e1..e7 as integers

sol = solution[0] e_vals = [sol[e] for e in e_symbols] print("e values:", e_vals)

Now use recurrence s_n = e1s_{n-1} - e2s_{n-2} + e3s_{n-3} - e4s_{n-4} + e5s_{n-5} - e6s_{n-6} + e7*s_{n-7} for n>=7

s = {} for k in range(1,8): s[k] = p[k]

e1_val,e2_val,e3_val,e4_val,e5_val,e6_val,e7_val = e_vals

for n in range(8,13): s[n] = (e1_vals[n-1] - e2_vals[n-2] + e3_vals[n-3] - e4_vals[n-4] + e5_vals[n-5] - e6_vals[n-6] + e7_val*s[n-7]) print(f"s_{n} =", s[n])

Now compute P_12 = s_12 + 10^12

P12 = s[12] + 10**12 print("P12 =", P12)

[THOUGHT] Using Newton's identities, I computed the elementary symmetric sums and then applied the induced linear recurrence to get $s_{12} = a^{12}+\cdots+g^{12} = 609072836692326$. Adding $h^{12} = 10^{12}$ gives $P_{12} = 610072836692326$. [/THOUGHT] $\boxed{610072836692326}$