

A217306


Minimal natural number (in decimal representation) with n prime substrings in base6 representation (substrings with leading zeros are considered to be nonprime).


2



1, 2, 11, 17, 47, 83, 269, 263, 479, 839, 1559, 1579, 2999, 5039, 9355, 9479, 14759, 56131, 56135, 61343, 56879, 336791, 341351, 336815, 341279, 341275, 2020727, 2020895, 2047651, 2020891, 4055159, 12098587, 12125347, 12285907, 15737755, 19128523, 39190247
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OFFSET

0,2


COMMENTS

The sequence is welldefined in that for each n the set of numbers with n prime substrings is not empty. Proof: Define m(0):=1, m(1):=2 and m(n+1):=6*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n1} 6^j = 2*(6^n  1)/5 or m(n)=1, 2, 22, 222, 2222, 22222, …, (in base6) for n=0,1,2,3,…. Evidently, for n>0 m(n) has n 2’s and these are the only prime substrings in base6 representation. This is why every substring of m(n) with more than one digit is a product of two integers > 1 (by definition) and can therefore not be prime number.
No term is divisible by 6.


LINKS



FORMULA

a(n) > 6^floor(sqrt(8*n7)1)/2), for n>0.
a(n) <= 2*(6^n  1)/5, n>0.
a(n+1) <= 6*a(n)+2.


EXAMPLE

a(1) = 2 = 2_6, since 2 is the least number with 1 prime substring in base6 representation.
a(2) = 11 = 15_6, since 11 is the least number with 2 prime substrings in base6 representation (5_6=5 and 15_6=11).
a(3) = 17 = 25_6, since 17 is the least number with 3 prime substrings in base6 representation (2_6, 5_6, and 25_6).
a(4) = 47 = 115_6, since 47 is the least number with 4 prime substrings in base6 representation (5_6, 11_6=7, 15_6=11, and 115_6=47).
a(8) = 479 = 2115_6, since 479 is the least number with 8 prime substrings in base6 representation (2_6, 5_6, 11_6=7, 15_6=11, 21_6=13, 115_6=47, 211_6=79, and 2115_6=479).


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



