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A257115
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Smallest k such that none of k + 1, k + 3,... k + 2n - 1 are squarefree and all of k + 2, k + 4,... k + 2n are squarefree.
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2
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OFFSET
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0,2
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COMMENTS
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For any x, one of x+2, x+4, ..., x+18 is divisible by 9 and thus not squarefree, so a(n) does not exist for n >= 9. - Robert Israel, Apr 27 2015
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LINKS
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EXAMPLE
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a(0) = 1 because 1 + 0 = 1 is squarefree.
a(1) = 3 because 3 + 1 = 4 is not squarefree and 3 + 2 = 5 is squarefree.
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PROG
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(PARI) a(n)=k=1; while(k, c=0; for(i=1, n, if(!issquarefree(k+2*i-1)&&issquarefree(k+2*i), c++); if(issquarefree(k+2*i-1)||!issquarefree(k+2*i), c=0; break)); if(c==n, return(k)); k++)
vector(9, n, n--; a(n)) \\ Derek Orr, Apr 27 2015
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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EXTENSIONS
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Corrected and extended by Derek Orr, Apr 27 2015
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STATUS
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approved
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