In a very simplistic and step by step example, how do I get the 'w' values for SHA256? I've been to https://docs.google.com/spreadsheets/d/1mOTrqckdetCoRxY5QkVcyQ7Z0gcYIH-Dc0tu7t9f7tw/edit#gid=1194752368, http://www.righto.com/2014/09/mining-bitcoin-with-pencil-and-paper.html, and https://en.wikipedia.org/wiki/SHA-2#Pseudocode (the pseudocode is for sha256)

The use of those links, not withstanding, I cannot create the 'W' values beyond the first 16 values.  Everything I read is complex beyond my understanding.

In a very simplistic and step by step example, how do I get the 'w' values for SHA256?
In SHA-256, $W$ is a vector of 64 values $W_i$ each 32-bit, collectively forming an expanded form of the input to a compression step. Said input is a 512-bit sub-block of the message to be hashed par SHA-256 after that message was padded. The $W_i$ depend only on said input.

$W_0$ to $W_{15}$ are said input, formatted as 32-bit words. $W_{16}$ to $W_{63}$ are computed from these by applying for $i$ from 16 to 63 the recurrence defined by FIPS&nbsp;180-4 section 6.2.2 as (within a reordering of terms):
$$W_i\gets W_{i-16}\;\tilde+\;\sigma_0(W_{i-15})\;\tilde+\;W_{i-7}\;\tilde+\;\sigma_1(W_{i-2})$$
where $\;\tilde+\;$ is addition with truncation of the result to 32-bit.

In the above expression, $\sigma_0$ and $\sigma_1$ are 32-bit functions of a 32-bit argument defined by FIPS&nbsp;180-4 section 4.1.2 as:
$$\begin{align}
\sigma_0(x)&amp;\gets(x\ggg\,\,\,7)\oplus(x\ggg18)\oplus(x\gg\,\,\,3)\\
\sigma_1(x)&amp;\gets(x\ggg17)\oplus(x\ggg19)\oplus(x\gg10)
\end{align}$$
where $\oplus$ is bitwise-XOR, $\ggg$ is right-rotation of a 32-bit word (also known as right-rotate-no-carry or right-circular-shift), $\gg$ is right-shift of a 32-bit word (also known as right-logical-shift).

For example, to compute $W_{16}$, we


apply the function $\sigma_0$ to $W_1$ (right-rotating that value by 7 and 18 bits, right-shifting it by 3 bits, then combining the resulting three 32-bit values by XOR);
apply the function $\sigma_1$ to $W_{14}$ (right-rotating that value by 17 and 19 bits, right-shifting it by 10 bits, then combining the resulting three 32-bit values by XOR);
add $W_0$, $W_9$, and the two above results, truncating the result to 32-bit.




Note: the reason for using shift rather than rotate for one of the terms of $\sigma_0$ and $\sigma_1$ is discussed here.

Note: often (including in almost any careful hardware implementation) it is kept only the chronologically last obtained 16 values of $W$, evolving as the compression rounds are performed per the transformation:
$$W_i\gets W_i\;\tilde+\;\sigma_0(W_{(i+1\bmod16)})\;\tilde+\;W_{(i+9\bmod16)}\;\tilde+\;\sigma_1(W_{(i+14\bmod16)})$$