[THOUGHT] Map x in [2026,2027] to t=x-2026 in [0,1]. Use the recurrence f(x)=2 f(x-2) to express f(2026+t)=2^{1013} f(t)=2^{1013} t sin(πt). The line is y=2^{1012}t. The vertical scale factor 2^{1012} will factor out, leaving an integral of t*|2 sin(πt)−1| over t∈[0,1]. Determine sign changes where sin(πt)=1/2, i.e., t=1/6 and 5/6, and integrate piecewise. Then match the result to p+q/π+r√3/π and compute 2028·p·q·r. [/THOUGHT] [PYTHON] import sympy as sp

t = sp.symbols('t', real=True) expr = t*(2sp.sin(sp.pit) - 1)

I = sp.integrate(t*(1-2sp.sin(sp.pit)), (t, 0, sp.Rational(1,6))) \

• sp.integrate(t*(2sp.sin(sp.pit)-1), (t, sp.Rational(1,6), sp.Rational(5,6))) \
• sp.integrate(t*(1-2sp.sin(sp.pit)), (t, sp.Rational(5,6), 1))

I_simplified = sp.simplify(I) print("I =", I_simplified)

Express I as p + q/pi + r*sqrt(3)/pi

p, q, r = sp.symbols('p q r') target = p + q/sp.pi + r*sp.sqrt(3)/sp.pi

Solve for p,q,r by matching coefficients in basis {1, 1/pi, sqrt(3)/pi}

Rewrite I in that basis

I_rewritten = sp.nsimplify(I_simplified, [sp.pi, sp.sqrt(3)], rational=True) print("I nsimplify =", I_rewritten)

Extract p,q,r by solving linear system after multiplying by pi

Let I = p + (q + rsqrt3)/pi. So piI = ppi + q + rsqrt3.

eq = sp.Eq(sp.piI_simplified, psp.pi + q + r*sp.sqrt(3)) sol = sp.solve(eq, [p, q, r], dict=True) print("sol =", sol)

Since single equation for three unknowns, use coefficient comparison:

Represent pi*I in basis {pi, 1, sqrt3}

expr_piI = sp.simplify(sp.piI_simplified) print("piI =", expr_piI)

Collect in terms of pi and sqrt(3)

expr_collected = sp.collect(expr_piI, [sp.pi, sp.sqrt(3)], evaluate=False) print("collected dict:", expr_collected)

Determine coefficients:

expr_piI should be Api + B + Csqrt3

A = sp.simplify(sp.expand(expr_piI).coeff(sp.pi, 1)) # coefficient of pi rest = sp.simplify(expr_piI - Asp.pi) C = sp.simplify(rest.coeff(sp.sqrt(3), 1)) B = sp.simplify(rest - Csp.sqrt(3)) print("A,B,C =", A,B,C)

p_val = A q_val = B r_val = C

print("p,q,r =", p_val, q_val, r_val)

value = sp.simplify(2028 * p_val * q_val * r_val) print("2028pqr =", value) print("as integer?", sp.nsimplify(value)) [/PYTHON][THOUGHT] Map x in [2026,2027] to t=x-2026 in [0,1]. Use the recurrence f(x)=2 f(x-2) to express f(2026+t)=2^{1013} f(t)=2^{1013} t sin(πt). The line is y=2^{1012}t. The vertical scale factor 2^{1012} will factor out, leaving an integral of t|2 sin(πt)−1| over t∈[0,1]. Determine sign changes where sin(πt)=1/2, i.e., t=1/6 and 5/6, and integrate piecewise. Then match the result to p+q/π+r√3/π and compute 2028·p·q·r. [/THOUGHT] [PYTHON] import sympy as sp

t = sp.symbols('t', real=True) expr = t*(2sp.sin(sp.pit) - 1)

I = sp.integrate(t*(1-2sp.sin(sp.pit)), (t, 0, sp.Rational(1,6))) \

• sp.integrate(t*(2sp.sin(sp.pit)-1), (t, sp.Rational(1,6), sp.Rational(5,6))) \
• sp.integrate(t*(1-2sp.sin(sp.pit)), (t, sp.Rational(5,6), 1))

I_simplified = sp.simplify(I) print("I =", I_simplified)

Work with piI = ppi + q + r*sqrt(3)

expr_piI = sp.simplify(sp.piI_simplified) print("piI =", expr_piI)

A = sp.simplify(sp.expand(expr_piI).coeff(sp.pi, 1)) # coefficient of pi rest = sp.simplify(expr_piI - Asp.pi) C = sp.simplify(rest.coeff(sp.sqrt(3), 1)) B = sp.simplify(rest - Csp.sqrt(3))

p_val, q_val, r_val = A, B, C print("p,q,r =", p_val, q_val, r_val)

value = sp.simplify(2028 * p_val * q_val * r_val) print("2028pq*r =", value) print("integer check:", sp.Integer(value)) [/PYTHON]

[THOUGHT] From the computed integral, $p=-\frac16$, $q=-2$, $r=2$. Thus $2028\cdot p\cdot q\cdot r=1352$.