allocate mem(2^30)

\\ A055932 Numbers all of whose prime divisors are consecutive primes starting at 2.
list(lim, p=2)=my(v=[1], q=nextprime(p+1), t=1); while((t*=p)<=lim, v=concat(v, t*list(lim\t, q))); vecsort(v) \\ Charles R Greathouse IV, Oct 02 2012

toruns(n) = { my (r=[]); while (n, my (v=valuation(n+n%2, 2)); n\=2^v; r=concat(v, r)); r }
fromruns(r) = { my (v=0); for (k=1, #r, v=(v+k%2)*2^r[k]-k%2); v }

{
	b = vector(10 001, n, -1);
	vv = list(11257259750506770);
	for (n = 1, #vv,
		v = vv[n];
		r = [];
		forprime (p = 2, oo,
			if (v==1,
					a = fromruns(r);
					if (a+1 <= #b,
						b[a+1] = n;
					);
					break,

					e = valuation(v, p);
					r = concat(e, r);
					v /= p^e;
			);
		);
	);

	for (n = 1, #b,
		if (b[n] >= 0,
				print (n-1 " " b[n]),
				break
		);
	);
}

quit