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A358310
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Index in A145985 where n-th odd prime p first appears, or -1 if p never appears.
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2
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3, 2, 1, 13, -1, 12, -1, 59, 11, -1, -1, 10, -1, 9, 8, 7, -1, -1, 6, -1, -1, 5, 4, -1, 2528242167, -1
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OFFSET
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1,1
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COMMENTS
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Let p = prime(n+1). Certainly if p == 1 (mod 6) then p cannot appear in A145985 (because 10^k-p will be a multiple of 3), so a(n) = -1. In all other cases it appears that a(n) > 0.
a(27) (corresponding to the prime 107) is presently unknown.
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LINKS
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EXAMPLE
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The 8th odd prime 23 first occurs in A145985 at index 59, so a(8) = 59.
For n=25, the 25th odd prime is 101. The first time when 10^k - 101 is a prime is when k = 12, where 10^12 - 101 = 999999999899. Furthermore, when we look at the numbers t = 10^12 - q for q an increasing prime, t is a prime for q = 11, 41, 101, ..., that is, 101 is the third success. It follows that 101 is three steps back from the end of the 12th block of descending terms in A145985. The lengths of the blocks are given by A359120. Therefore a(25) = Sum_{j=1..12} A359120(j) - (3-1) = 2528242169 - 2 = 2528242167. - N. J. A. Sloane, Dec 18 2022
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MATHEMATICA
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With[{prs=Prime[Range[20000000]]}, Table[Position[Select[Table[10^IntegerLength[p]-p, {p, prs}], PrimeQ], n, 1, 1], {n, Prime[Range[24]]}]]/.({}->-1) (* Harvey P. Dale, Dec 17 2022 *)
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CROSSREFS
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KEYWORD
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sign,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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