OFFSET
0,1
COMMENTS
The last digit of n cannot be 8, therefore a(n) must have at least n+1 digits. It is probable (using [] for floor) that none among [10^n/9]*80 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*80 + b with 1 <= b <= 9 and one of the (first) digits 8 replaced by a digit among {0, ..., 7}. - 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, 8], {n, 1, 18}]
PROG
(PARI) A037069(n)={my(p, t=10^(n+1)\9*80); forvec(v=[[-1, n], [-8, -1]], nextprime(p=t+10^(n-v[1])*v[2])-p<10 && return(nextprime(p)))} \\ ~
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Patrick De Geest, Jan 04 1999
EXTENSIONS
Corrected by Jud McCranie, Jan 04 2001
More terms from Erich Friedman, Jun 03 2001
More terms and a(0) = 2 from M. F. Hasler, Feb 22 2016
STATUS
approved