Why is the first value in digital first-order IIR filter impulse response not the largest?
A causal first-order IIR filter is characterized by the following difference equation:

$$y[n]=b_0x[n]+b_1x[n-1]-a_1y[n-1]\tag{1}$$

with $x[n]$ the input signal, and $y[n]$ the output signal. The impulse response of that system can be computed via the $\mathcal{Z}$-transform or otherwise, and it turns out to be

$$h[n]=b_0\delta[n]+(-a_1)^{n-1}(b_1-b_0a_1)u[n-1]\tag{2}$$

where $\delta[n]$ is the unit impulse, and $u[n]$ is the unit step sequence. Assuming that the filter is stable, we can conclude that $|a_1|&lt;1$ is satisfied. Consequently, $|h[0]|&gt;|h[n]|$ for $n&gt;0$ is satisfied if and only if

$$|b_0|&gt;|b_1-b_0a_1|\tag{3}$$

holds.

With $b_0=b_1$ and $a_1&lt;0$ as in your example, condition $(3)$ can never be satisfied, and, consequently, the maximum magnitude of the impulse response is not assumed at index $n=0$.

EDIT:

I add more details referring to your filter design problem. A Butterworth low pass filter's frequency response satisfies $H(1)=1$ (DC) and $H(-1)=0$ (Nyquist). With

$$H(z)=\frac{b_0+b_1z^{-1}}{1+a_1z^{-1}}\tag{4}$$

we have the following requirements on the coefficients:

$$H(1)=\frac{b_0+b_1}{1+a_1}=1\\
H(-1)=\frac{b_0-b_1}{1-a_1}=0\tag{5}$$

which gives

$$b_0=b_1\quad\textrm{ and }\quad a_1=2b_0-1\tag{6}$$

From the stability condition $|a_1|&lt;1$ we get

$$|2b_0-1|&lt;1\Longrightarrow 0&lt;b_0&lt;1\tag{7}$$

Combining the constraints $(6)$ and $(7)$ with the inequality $(3)$ yields

$$1 &gt;b_0&gt;\frac12\tag{8}$$

If $(8)$ is satisfied, $h[0]$ will have the largest magnitude, otherwise it won't. Now it turns out that the exact value of $b_0$ depends on the desired cut-off frequency of the low pass filter. For cut-off frequencies less than $\frac12$ (i.e., less than half the Nyquist frequency), you get $b_0&lt;\frac12$, and hence $|h[0]|$ is smaller than the maximum magnitude of the impulse response. Only for cut-off frequencies greater than half the Nyquist frequency do you get $|h[0]|&gt;|h[n]|$, $n&gt;0$.

So to directly answer the question in your title: the first impulse response value is not the largest (in magnitude) because the filter coefficients do not satisfy inequality $(3)$, and in your example of a (first-order) Butterworth low pass filter this happens because the cut-off frequency is smaller than half the Nyquist frequency.