diff --git a/generated/chudnovsky_formula_for_pi_inv/README.md b/generated/chudnovsky_formula_for_pi_inv/README.md
index 712fdee..962cf63 100644
--- a/generated/chudnovsky_formula_for_pi_inv/README.md
+++ b/generated/chudnovsky_formula_for_pi_inv/README.md
@@ -5,9 +5,9 @@ Chudnovsky formula for pi inverse
 - Problem ID: `chudnovsky_formula_for_pi_inv`
 - Test Problem: no
 - Submitter: Kim Morrison
-- Notes: Uses the existing Mathlib definition of the Chudnovsky series and asks for the missing identity with pi inverse.
-- Source: https://link.springer.com/article/10.1007/s40316-018-0102-6
-- Informal solution: Evaluate the Chudnovsky series and prove that it sums to 1 / pi.
+- Notes: Uses the existing Mathlib definition of the Chudnovsky series and asks for the missing identity with pi inverse. The Chudnovsky formula states 1/π = (12 / 640320^(3/2)) · ∑_{n=0}^∞ (-1)^n · (6n)! · (545140134·n + 13591409) / ((3n)! · (n!)^3 · 640320^(3n)).
+- Source: https://arxiv.org/abs/1809.00533
+- Informal solution: Prove that the Chudnovsky series 1/π = (12 / 640320^(3/2)) · ∑_{n=0}^∞ (-1)^n · (6n)! · (545140134·n + 13591409) / ((3n)! · (n!)^3 · 640320^(3n)) holds, e.g. following Milla's detailed complex-analytic proof.
 
 Do not modify `Challenge.lean` or `Solution.lean`. Those files are part of the
 trusted benchmark and fixed by the repository.
diff --git a/manifests/problems.toml b/manifests/problems.toml
index 0bc4664..d0d66c4 100644
--- a/manifests/problems.toml
+++ b/manifests/problems.toml
@@ -40,9 +40,9 @@ test = false
 module = "LeanEval.Analysis.Chudnovsky"
 holes = ["chudnovsky_formula_for_pi_inv"]
 submitter = "Kim Morrison"
-notes = "Uses the existing Mathlib definition of the Chudnovsky series and asks for the missing identity with pi inverse."
-source = "https://link.springer.com/article/10.1007/s40316-018-0102-6"
-informal_solution = "Evaluate the Chudnovsky series and prove that it sums to 1 / pi."
+notes = "Uses the existing Mathlib definition of the Chudnovsky series and asks for the missing identity with pi inverse. The Chudnovsky formula states 1/π = (12 / 640320^(3/2)) · ∑_{n=0}^∞ (-1)^n · (6n)! · (545140134·n + 13591409) / ((3n)! · (n!)^3 · 640320^(3n))."
+source = "https://arxiv.org/abs/1809.00533"
+informal_solution = "Prove that the Chudnovsky series 1/π = (12 / 640320^(3/2)) · ∑_{n=0}^∞ (-1)^n · (6n)! · (545140134·n + 13591409) / ((3n)! · (n!)^3 · 640320^(3n)) holds, e.g. following Milla's detailed complex-analytic proof."
 
 [[problem]]
 id = "pi1_circle_mulEquiv_int"