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A213320
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Numbers such that the number of nonprime substrings equals the number of digits (substrings with leading zeros are considered to be nonprime).
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2
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1, 4, 6, 8, 9, 11, 12, 15, 19, 20, 21, 24, 26, 28, 30, 34, 36, 38, 39, 41, 42, 45, 50, 51, 54, 56, 58, 61, 62, 63, 65, 70, 74, 76, 78, 82, 85, 87, 89, 92, 93, 95, 117, 123, 127, 132, 133, 135, 139, 153, 157, 167, 171, 172, 175
(list;
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refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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Also numbers such that the number of prime substrings is A000217(m-1) = m(m-1)/2, where m is the number of digits.
The sequence is finite. Proof: Let p be a number >= 10^17 and let m = 9k+j be the number of digits of p, where k = floor(m/9) >= 2 and j = m mod 9. Since each 9-digit number has at least 15 nonprime substrings, it follows that p has at least 15k = 9k + 6k > 9k + j = m nonprime substrings (since 6k >= 12> j for k >= 2). Consequently, no number >= 10^17 can be a term of the sequence.
The last term is a(858)=3733739. Proof: Each 9-digit number has at least 15 nonprime substrings, thus, the numbers 10^8 <= p < 10^14 also have at least 15 nonprime substrings and therefore cannot be terms of the sequence. Same is true for numbers 10^14 <= p < 10^17 since each 6-digit number has at least 4 nonprime substrings, and thus each number with >= 15 digits has at least 15+4 = 19 nonprime substrings. Since each 8-digit number has at least 10 nonprime substrings, it follows that the last term of the sequence must be less than 10^7. By direct search we find a(858) = 3733739.
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LINKS
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EXAMPLE
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a(1) = 1, since 1 has 1 nonprime substrings.
a(43) = 117, since 117 has 3 digits and also 3 nonprime substrings (1, 1, 117).
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CROSSREFS
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KEYWORD
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nonn,fini,full,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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