ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Iterating a recursive specification to determine coefficients through degree $6$

Example

For the plane-tree equation

$T = \frac{x}{1 - T},$

start with $T_{0} : = 0$ and define $T_{j + 1} : = x / (1 - T_{j})$ , truncating modulo $x^{7}$ . The iterates are

$T_{1} = x,$

$T_{2} = x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6},$

$T_{3} = x + x^{2} + 2 x^{3} + 4 x^{4} + 8 x^{5} + 16 x^{6},$

$T_{4} = x + x^{2} + 2 x^{3} + 5 x^{4} + 13 x^{5} + 34 x^{6},$

$T_{5} = x + x^{2} + 2 x^{3} + 5 x^{4} + 14 x^{5} + 41 x^{6},$

$T_{6} = x + x^{2} + 2 x^{3} + 5 x^{4} + 14 x^{5} + 42 x^{6} .$

So the coefficients through degree $6$ are $1, 1, 2, 5, 14, 42$ .

Facts & Assumptions

Given: The plane-tree series is the unique fixed point of $Y \mapsto x / (1 - Y)$ (Rooted plane trees satisfy $T (x) = x / (1 - T (x))$ ), and order-raising recursion converges coefficientwise by successive truncation (An order-raising recursive specification has a unique solution).

Verification

technique · direct

1.1

Substituting each displayed iterate into $x / (1 - Y)$ and truncating modulo $x^{7}$ gives the next one in the list.

givenalgebra

2.1

By the sixth iterate, every coefficient through degree $6$ has stabilized, so the unique fixed point begins $T (x) = x + x^{2} + 2 x^{3} + 5 x^{4} + 14 x^{5} + 42 x^{6} + \dots$ .

step 1.1given∎

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Iterating a recursive specification to determine coefficients through degree 66