OFFSET
1,1
COMMENTS
a(n) = prime(n) for almost all n. Probably a(n) = prime(n) for all n > N for some N, but N must be very large. - Charles R Greathouse IV, Jul 20 2011
FORMULA
a(n) ~ n log n. - Charles R Greathouse IV, Jul 20 2011
MATHEMATICA
p=Prime[3]; b={p}; d=p; Do[Do[r=Prime[c]; If[FreeQ[b, r]&&Intersection@@IntegerDigits/@{d, r}=!={}, b=Append[b, r]; d=r; Break[]], {c, 1000}], {k, 60}]; b
PROG
(PARI) common(a, b)=a=vecsort(eval(Vec(Str(a))), , 8); b=vecsort(eval(Vec(Str(b))), , 8); #a+#b>#vecsort(concat(a, b), , 8)
in(v, x)=for(i=1, #v, if(v[i]==x, return(1))); 0
lista(nn) = {my(v=[5]); for(n=2, nn, forprime(p=2, default(primelimit), if(!in(v, p)&&common(v[#v], p), v=concat(v, p); break))); v; }
\\ Charles R Greathouse IV, Jul 20 2011
CROSSREFS
Cf. A107353.
KEYWORD
nonn,base
AUTHOR
Zak Seidov & Eric Angelini, May 24 2005
STATUS
approved