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A276672
Numbers k such that (19*10^k + 101) / 3 is prime.
0
1, 3, 4, 9, 10, 12, 13, 16, 20, 37, 57, 66, 106, 116, 127, 355, 396, 547, 2289, 3777, 4500, 7821, 15663, 22746, 25978, 30434, 39682, 119716, 133390, 145093, 200260
OFFSET
1,2
COMMENTS
For k > 1, numbers k such that the digit 6 followed by k-2 occurrences of the digit 3 followed by the digits 67 is prime (see Example section).
a(32) > 3*10^5. - Robert Price, Jul 13 2023
EXAMPLE
3 is in this sequence because (19*10^3 + 101) / 3 = 6367 is prime.
Initial terms and associated primes:
a(1) = 1, 97;
a(2) = 3, 6367;
a(3) = 4, 63367;
a(4) = 9, 6333333367;
a(5) = 10, 63333333367, etc.
MATHEMATICA
Select[Range[0, 100000], PrimeQ[(19*10^# + 101) / 3] &]
PROG
(Magma) [n: n in [0..400] |IsPrime((19*10^n + 101) div 3)]; // Vincenzo Librandi, Sep 13 2016
(PARI) is(n)=ispseudoprime((19*10^n + 101)/3) \\ Charles R Greathouse IV, Jun 13 2017
KEYWORD
nonn,more
AUTHOR
Robert Price, Sep 12 2016
EXTENSIONS
a(28)-a(30) from Robert Price, Sep 01 2019
a(31) from Robert Price, Jul 13 2023
STATUS
approved