Answering just the first of your questions: “What tests would you apply to determine if this [sequence] is truly random?”

How about treating it as a time-series, and checking for auto-correlations? Here is some R code. First some test data (first 1000 digits):

digits_string="1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989"
digits=as.numeric(unlist(strsplit(digits_string,"")))

Check the counts of each digit:

> table(digits)
digits
0 1 2 3 4 5 6 7 8 9
93 116 103 102 93 97 94 95 101 106

Then turn it into a time-series, and run the Box-Pierce test:

d=as.ts( digits )
Box.test(d)

which tells me:

X-squared = 1.2449, df = 1, p-value = 0.2645

Typically you’d want the p-value to be under 0.05 to say there are auto-correlations.

Run acf(d) to see the auto-correlations. I’ve not included an image here as it is a dull chart, though it is curious that the biggest lags are at 11 and 22. Run acf(d,lag.max=40) to show that there is no peak at lag=33, and that it was just coincidence!

P.S. We could compare how well those 1000 digits of pi did, by doing the same tests on real random numbers.

probs=sapply(1:100,function(n){
digits=floor(runif(1000)*10)
bt=Box.test(ts(digits))
bt$p.value
})

This generates 1000 random digits, does the test, and repeats this 100 times.

> summary(probs)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.006725 0.226800 0.469300 0.467100 0.709900 0.969900
> sd(probs)
[1] 0.2904346

So our result was comfortably within the first standard deviation, and pi quacks like a random duck. (I used set.seed(1) if you want to reproduce those exact numbers.)

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