Theorem Let Σ = {0, 1, 8} ⊂ ℕ₀.

For n ≥ 2, let



Then

Proof
Define the seed map by
σ : 𝒫ₙ → Σ^⌈n/2⌉ σ(d) = (d_n−1, d_n−2, …, d_⌈n/2⌉)
The palindrome constraint d_i = d_n−1−i
makes σ bijective onto


First coordinate has 2 admissible values
each of the remaining⌈n/2⌉ − 1 coordinates has 3

Hence


The result of the bijection ∎

Corollary (Parity Split)
For k ≥ 1:

|𝒫₂ₖ| = 2 · 3^k−1,     |𝒫₂ₖ₊₁| = 2 · 3^k

Proof

Immediate from the theorem by substituting
n = 2k and n = 2k+1 ∎

Corollary (Ordinary Generating Function)


Proof


From the parity split:



Summing: G(x) = (2x² + 6x³)/(1−3x²) = 2x²(1+3x)/(1−3x²) ∎

Remark (Strobogrammatic Coincidence)

The map ρ : {0,1,8} → {0,1,8} defined by ρ(0)=0, ρ(1)=1, ρ(8)=8

is the 180°-rotation involution on Σ.

Since every element of 𝒫ₙ satisfies

d_i = d_n−1−i and ρ is the identity on Σ,
each member of 𝒫ₙ is simultaneously palindromic and strobogrammatic

∴  The theorem enumerates the n-digit tetradic integers

Verification

n	Formula	Palindromes
2	2·3⁰ = 2	11, 88
3	2·3¹ = 6	101, 111, 181, 808, 818, 888
4	2·3¹ = 6	1001, 1111, 1881, 8008, 8118, 8888
5	2·3² = 18	—