OFFSET
0,1
COMMENTS
For n > 1, the last digit cannot be 2, so a(n) has at least n+1 digits. The probability is big that none of [10^n/9]*20 + {1,3,7,9} is prime, in which case a(n) must have at least n+2 digits. This is the most frequent case. We can even conjecture that for all n > 1, a(n) equals [10^(n+1)/9]*20 + b with 1 <= b <= 9 and one of the (first) digits 2 replaced by 0 or 1. - M. F. Hasler, Feb 22 2016
LINKS
M. F. Hasler, Table of n, a(n) for n = 0..200
MATHEMATICA
f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 2], {n, 1, 18}]
PROG
(PARI) A037057(n)={my(p, t=10^(n+1)\9*20); n>1 && forvec(v=[[-1, n], [-2, -1]], nextprime(p=t+10^(n-v[1])*v[2])-p<10 && return(nextprime(p))); 3-n} \\ M. F. Hasler, Feb 22 2016
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Jan 04 1999
EXTENSIONS
More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 23 2003
More terms and a(0) = 3 from M. F. Hasler, Feb 22 2016
STATUS
approved