Following up on i9Fn's suggestion, there does seem to be at least one more case along these lines: $$660F_{11} = 60n^{11} + 330n^{10} + 550n^9 - 660n^7 + 660n^5 - 330n^3 + 50n.$$ This might be related to a genus 70 Riemann surface with 11 embedded buckyballs; see Martin & Singerman From biplanes to the Klein quartic and the Buckyball. Each buckyball/ buckminsterfullerene has 60 vertices, 90 edges, and 32 faces. The number 660 arises as both the order of $PSL(2,11)$ and the count of particular triangles.

Following up on i9Fn's suggestion, there does seem to be at least one more case along these lines: $$660F_{11}(n) = 60n^{11} + 330n^{10} + 550n^9 - 660n^7 + 660n^5 - 330n^3 + 50n.$$ This might be related to a genus 70 Riemann surface with 11 embedded buckyballs; see Martin & Singerman From biplanes to the Klein quartic and the Buckyball. Each buckyball/ buckminsterfullerene has 60 vertices, 90 edges, and 32 faces. The number 660 arises as both the order of $PSL(2,11)$ and the count of particular triangles.

Following up on i9Fn's suggestion, there does seem to be at least one more case along these lines: $$660F_{11} = 60n^{11} + 330n^{10} + 550n^9 - 600n^7 + 660n^5 - 330n^3 + 50n.$$ This might be related to a genus 70 Riemann surface with 11 embedded buckyballs; see Martin & Singerman From biplanes to the Klein quartic and the Buckyball. Each buckyball/ buckminsterfullerene has 60 vertices, 90 edges, and 32 faces. The number 660 arises as both the order of $PSL(2,11)$ and the count of particular triangles.

Following up on i9Fn's suggestion, there does seem to be at least one more case along these lines: $$660F_{11} = 60n^{11} + 330n^{10} + 550n^9 - 660n^7 + 660n^5 - 330n^3 + 50n.$$ This might be related to a genus 70 Riemann surface with 11 embedded buckyballs; see Martin & Singerman From biplanes to the Klein quartic and the Buckyball. Each buckyball/ buckminsterfullerene has 60 vertices, 90 edges, and 32 faces. The number 660 arises as both the order of $PSL(2,11)$ and the count of particular triangles.