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%I #32 Jan 17 2019 13:44:08
%S 1,4,6,7,384,666,675,3165
%N Numbers n such that (10^(2n+1)+63*10^n-1)/9 is prime.
%C 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.
%C a(9) > 10^5. - _Robert Price_, Oct 30 2017
%D C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.
%D 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.
%H Patrick De Geest, World!Of Numbers, <a href="http://www.worldofnumbers.com/wing.htm#pwp181">Palindromic Wing Primes (PWP's)</a>
%H Makoto Kamada, <a href="https://stdkmd.net/nrr/1/11811.htm#prime">Prime numbers of the form 11...11811...11</a>
%H <a href="/index/Pri#Pri_rep">Index entries for primes involving repunits</a>.
%F a(n) = (A077791(n)-1)/2.
%e 7 is in the sequence because (10^15+63*10^7-1)/9=1(7).8.1(7)=111111181111111 is prime.
%e 666 is in the sequence because (10^(2*666+1)+63*10^666-1)/9=1(666).8.1(666) is prime.
%t Do[If[PrimeQ[(10^(2n + 1) + 63*10^n - 1)/9], Print[n]], {n, 4000}]
%o (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
%Y Cf. A004023, A077775-A077798, A107123-A107127, A107648, A107649, A115073, A183174-A183187.
%K nonn,more,base
%O 1,2
%A _Farideh Firoozbakht_, May 19 2005
%E Edited by _Ray Chandler_, Dec 28 2010
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