

A217304


Minimal natural number (in decimal representation) with n prime substrings in base4 representation (substrings with leading zeros are considered to be nonprime).


2



1, 2, 7, 11, 23, 43, 93, 151, 239, 373, 479, 727, 1495, 2015, 2775, 5591, 6133, 7919, 12271, 22367, 24303, 30431, 48991, 89527, 95607, 98143, 129887, 357883, 358111, 382431, 744797, 519551, 1431007, 1432447, 1556319, 2457439
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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):=4*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n1} 4^j = 2*(4^n  1)/3 or m(n)=1, 2, 22, 222, 2222, 22222, …,for n=0,1,2,3,…. Evidently, for n>0 m(n) has n 2’s and these are the only prime substrings in base4 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 4. a(1) = 2 is the only even term.


LINKS



FORMULA

a(n) > 4^floor(sqrt(8*n7)1)/2), for n>0.
a(n) <= 2*(4^n  1)/3, n>0.
a(n+1) <= 4*a(n) + 2.


EXAMPLE

a(1) = 2 = 2_4, since 2 is the least number with 1 prime substring in base4 representation.
a(2) = 7 = 13_4, since 7 is the least number with 2 prime substrings in base4 representation (3_4=3 and 13_4=7).
a(3) = 11 = 23_4, since 11 is the least number with 3 prime substrings in base4 representation (2_4, 3_4, and 23_4).
a(5) = 43 = 223_4, since 43 is the least number with 5 prime substrings in base4 representation (2 times 2_4, 3_4, 23_4=11, and 223_4=43).
a(7) = 151 = 2113_4, since 151 is the least number with 7 prime substrings in base4 representation (2 times 2_4, 3_4, 11_4=5, 13_4=7, 113_4=23, and 2113_4=151).


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



