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 Incoming links: Floor-half generating function

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 6A / b / Solution Created 2026-09-24 Updated 2026-09-25

The floor-half generating function is

H (x) = \sum_{n \geq 1} ⌊ \frac{n}{2} ⌋ x^{n} = \frac{x ^{2}}{( 1 - x ) ^{2} ( 1 + x )} .

Summing

u_{n} - 2 u_{n - 1} = ⌊ n /2 ⌋

for

n \geq 1

gives

G (x) - 1 - 2 x G (x) = H (x),

so

G (x) = \frac{1 + \frac{x ^{2}}{( 1 - x ) ^{2} ( 1 + x )}}{1 - 2 x} .

Since

⌊ \frac{n}{2} ⌋ = \frac{n}{2} - \frac{1}{4} + \frac{1}{4} (- 1)^{n},

a particular solution is

- \frac{n}{2} - \frac{3}{4} + \frac{1}{12} (- 1)^{n} .

Adding the homogeneous term

C 2^{n}

and imposing

u_{0} = 1

gives

C = 5/3

. Therefore

u_{n} = \frac{5}{3} 2^{n} - \frac{n}{2} - \frac{3}{4} + \frac{1}{12} (- 1)^{n} .

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