sqrt(2) ≃ mean(2 1 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 2 1 1 2 …

sequence.c

#include <stdio.h>

//See http://oeis.org/A001030
unsigned int maxdepth, nbrtermes;

unsigned int recur(char num, unsigned int depth) {
	if (depth>=maxdepth) {
		nbrtermes++;
		return (unsigned int)num;
	}
	depth++;
	if (num==2){
		return recur(2, depth) + recur(1, depth) + recur(1, depth);
	} else {//num==1
		return recur(2, depth) + recur(1, depth);
	}
}

int main(int argc, char *argv[]){
	nbrtermes=0;
	printf("Profondeur de récursion?");
	scanf("%d", &maxdepth);
	printf("sqrt(2) ~ %d/%d\n", recur(2,1), nbrtermes);
}

sequence.py

#!/usr/bin/env python3

def seq(n, repl):
	a=[2]
	while len(a)<n: a=[l for i in a for l in (repl[0] if i==1 else repl[1])]
	return a[:n]

def mean(iterbl):
	ssum=0
	for i in iterbl: ssum+=i
	return ssum/len(iterbl)

seqlen = int(1e6) #the larger seqlen, the better the approximation, the slower this program

print("""
	φ  ≃ %f
	√2 ≃ %f
""" % (mean(seq(seqlen, ((2,), (2,1)))), mean(seq(seqlen, ((2,1), (2,1,1))))))

sqrt2.c

#include <stdio.h>

/* Ophir LOJKINE
Compute a sequence which mean converges to sqrt(2)
The computation is done using only integer arithmetics
*/

int abs (int x) {
	return (x>0)?x:-x;
}

int main (void){
	long int sum=0, iter=0;
	long int step=0;
	long int invepsilon=1e5; //The precision of the approximation will be at least 1/invepsilon

	//Thanks to Ulysse LOJKINE for the formula:
	//5*abs(sum^2-2*iter^2)*1/epsilon-14*iter^2 > 0 ==> | sqrt(2) - sum/iter | < epsilon
	while (5*abs(sum*sum-2*iter*iter)*invepsilon-14*iter*iter>=0){
		if(sum*sum > 2 * iter*iter) step=1;
		else step=2;
		printf("%ld ", step);
		sum+=step;
		iter++;
	}
	printf("\nsqrt(2) ~ %ld/%ld\n", sum, iter);
	return 0;
}