OFFSET
7,1
COMMENTS
No such p exists for n < 7.
Does p exist for every n > 6?
Answer: yes. For a given n, consider the sequence {k*n^4 + 6*n^3 + 6*n^2 + 6*n + 1}. By Dirichlet's theorem on arithmetic progressions, there exist infinitely many primes of this form, and they all end in 6661 in base n. - Jianing Song, Feb 03 2019
FORMULA
n^3 < a(n) << n^20. - Charles R Greathouse IV, May 13 2017
Probably n^3 < a(n) < n^4 for all but finitely many n. It appears the only exceptions are 21 and 52. If there are any others they are larger than 10^7; the expected number of larger exceptions is about 10^-89814. - Charles R Greathouse IV, May 13 2017
EXAMPLE
For n = 7: 2399 written in base 7 is 6665. Since 2399 is the smallest prime that contains the substring 666 in its base-7 expansion, a(7) = 2399.
MATHEMATICA
Table[k = FromDigits[#, b]; While[Nand[PrimeQ@ k, Length@ SequencePosition[IntegerDigits[k, b], #] > 0], k++]; k, {b, 7, 41}] &@ ConstantArray[6, 3] (* Michael De Vlieger, May 08 2017 *)
PROG
(PARI) a(n) = forprime(p=1, , my(subs=[6, 6, 6], dbn=digits(p, n)); for(k=1, #dbn-2, my(v=[dbn[k], dbn[k+1], dbn[k+2]]); if(v==subs, return(p))))
CROSSREFS
KEYWORD
nonn,base,less
AUTHOR
Felix Fröhlich, May 07 2017
STATUS
approved