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A107648
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Numbers n such that (10^(2n+1)+63*10^n-1)/9 is prime.
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45
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OFFSET
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1,2
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COMMENTS
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n is in the sequence iff the palindromic number 1(n).8.1(n) is prime (dot between numbers means concatenation). Let f(n)=(10^(2n+1)+63*10^n-1)/9 then for all nonnegative integers m we have: I. 3 divides f(3m+2) II. 19 divides f(18m+13) III. 29 divides f(28*m+16) & 29 divides f(28*m+25) IV. 31 divides f(30*m+2) & 31 divides f(30*m+17) V. 41 divides f(5m+3), etc. So if n is in the sequence then n is not of the forms 3m+2, 18m+13, 28m+16 28m+25, 30m+2, 30m+17, 5m+3, etc.
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REFERENCES
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C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.
Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991, p. 141.
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LINKS
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FORMULA
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EXAMPLE
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7 is in the sequence because (10^15+63*10^7-1)/9=1(7).8.1(7)=111111181111111 is prime.
666 is in the sequence because (10^(2*666+1)+63*10^666-1)/9=1(666).8.1(666) is prime.
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MATHEMATICA
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Do[If[PrimeQ[(10^(2n + 1) + 63*10^n - 1)/9], Print[n]], {n, 4000}]
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PROG
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(PARI) for(n=0, 1e4, if(ispseudoprime(t=(10^(2*n+1)+63*10^n)\9), print1(t", "))) \\ Charles R Greathouse IV, Jul 15 2011
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CROSSREFS
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KEYWORD
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nonn,more,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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