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Source. Erdős and Taylor (1960), printed pp. 138–139, equations (2.1), (2.5)–(2.8), in the canonical PDF.

Statement. Let $(S_j)_{j\ge0}$ be the symmetric nearest-neighbor walk on $\mathbb Z^2$, starting at zero, with independent increments uniform on $(1,0),(-1,0),(0,1),(0,-1)$. Put

$$q_n=\mathbb P(S_j\ne0\text{ for }1\le j\le n),\qquad q_0=1.$$

Then, as $n\to\infty$,

$$q_n=\frac{\pi}{\log n}+O\!\left(\frac1{(\log n)^2}\right).$$

All logarithms are natural. The source writes $\gamma_2(n)$ for avoidance through time $n-1$; thus $q_n=\gamma_2(n+1)$. This indexing convention keeps the finite renewal identity below exact, including its final-time endpoint. The one-step shift does not change the stated asymptotic.

Proof. Write $u_j=\mathbb P(S_j=0)$. Odd return probabilities vanish. For even times,

$$u_{2r}=\frac{\binom{2r}{r}^2}{16^r} =\frac1{\pi r}+O(r^{-2})\qquad(r\ge1),\qquad u_0=1.$$

To see the exact expression, map coordinates $(x,y)$ to $(x+y,x-y)$. Each step becomes a uniformly chosen pair in $\{-1,1\}^2$, so the two transformed walks are independent symmetric sign walks. Both are at zero after $2r$ steps with the displayed probability. The estimate follows from Stirling's formula. Consequently

$$A(N):=\sum_{r=0}^{\lfloor N/2\rfloor}u_{2r} =\frac1\pi\log N+O(1).$$

Partition paths according to their last visit to the origin by time $N$. The increments after that visit are independent of the preceding path, giving

$$1=\sum_{r=0}^{\lfloor N/2\rfloor}u_{2r}q_{N-2r}. \tag{1}$$

Since $q_j$ decreases in $j$, (1) gives $q_NA(N)\le1$. Inverting the estimate for $A(N)$ proves

$$q_N\le\frac\pi{\log N}+O((\log N)^{-2}). \tag{2}$$

For the lower bound, take $N$ a sufficiently large multiple of four and set

$$K=N/4,\qquad L=\left\lfloor N/2-N/\log N\right\rfloor.$$

Then $0<K<L<N/2$. Split (1) at $K$ and $L$, use monotonicity on its first two parts, and use $q_j\le1$ on its last part. This yields

$$1\le q_{N/2}\sum_{r=0}^K u_{2r} +q_{N-2L}\sum_{r=K+1}^L u_{2r} +\sum_{r=L+1}^{N/2}u_{2r}. \tag{3}$$

The first sum is $(\log N)/\pi+O(1)$. The middle sum is $O(1)$, since $K$ and $L$ are both proportional to $N$. Also $N-2L=2N/\log N+O(1)$, so (2) gives $q_{N-2L}=O(1/\log N)$. The last sum has $O(N/\log N)$ terms, each $O(1/N)$, and hence is $O(1/\log N)$. Rearranging (3) now gives

$$q_{N/2}\ge \frac{1-O(1/\log N)}{(\log N)/\pi+O(1)} =\frac\pi{\log N}-O((\log N)^{-2}).$$

Changing $\log N$ to $\log(N/2)$ alters the main term by $O((\log N)^{-2})$. Thus the desired lower bound holds at every sufficiently large even time. Monotonicity between consecutive even times gives it at odd times as well. Combining it with (2) proves the result. $\square$

Depends on. The elementary return count, the last-visit renewal identity, Stirling's formula, and harmonic-sum estimates. All probabilistic deductions used here are included above; no recurrence theorem is needed as an input.

Bears on. #1165: the first estimate of Hao, Li, Okada and Zheng, Lemma 2.1 is this equation, and the corpus records that lemma among the inputs to their Theorem 1.1, which answers the problem. #1166: the equation is the input to the planar maximum local time bound used in the recorded deduction for that problem.