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A244424
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Least number k > 0 such that concatenating k consecutive natural numbers beginning with n is prime, or 0 if no such number exists.
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3
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0, 1, 1, 4, 1, 2, 1, 2, 179, 0, 1, 2, 1, 4, 5, 28, 1, 3590, 1, 4, 0, 0, 1, 0, 25, 122, 0, 46, 1, 0, 1, 0, 71, 4, 569, 2, 1, 20, 5, 0, 1, 2, 1, 8, 0, 0, 1, 0, 193, 2, 0, 0, 1, 0, 0, 2, 5, 4, 1, 0, 1, 2, 0, 4, 5, 938, 1, 2, 119, 58, 1, 116, 1, 0, 125, 346, 5, 2, 1, 2, 0, 0, 1, 0, 0, 32
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OFFSET
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1,4
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COMMENTS
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The current zero values are only conjectural: a(n) > 5000 - n for all a(n) = 0 shown. [Edited by M. F. Hasler, Apr 27 2017]
Probably a(n) > 0 for all n. Appending k integers gives a number of size ~10^(k log_10 k) and so the expected number of primes with k < x is about the integral of 1/(k log k) up to x which is log log x. This diverges, so by the Borel-Cantelli lemma we expect that there will be a prime eventually. (Corrections for the particular base at hand affect the expected number but not its order of growth.) On the other hand, log log x grows slowly so finding the values of a(1), a(10), a(21), etc. may be hard. - Charles R Greathouse IV, Jul 10 2014 [Corrected by Pontus von Brömssen, Oct 12 2021]
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LINKS
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EXAMPLE
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14 is not prime. 1415 is not prime. 141516 is not prime. 14151617 is prime. Thus a(14) = 4 since 4 consecutive numbers were concatenated.
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PROG
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(PARI) a(n) = {p=""; tot=0; for(i=n, 5000, p=concat(p, Str(i)); tot++; if(ispseudoprime(eval(p)), return(tot)))}
n=1; while(n<100, print1(a(n), ", "); n++)
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CROSSREFS
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KEYWORD
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nonn,base,more,hard
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AUTHOR
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STATUS
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approved
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