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%I #21 Sep 08 2022 08:45:45
%S 7,331,300030001,3000000003000000001
%N Primes of the form 3*10^(2*n) + 3*10^n + 1.
%C Primes of the form (x^3-y^3)/(x-y) with x = y+1 (which gives A002407) and also y=10^k for some k.
%C These prime numbers (differences of consecutive cubes: A002407), for k>0, have only three digits different from zero. The first is 3, the middle digit is 3 and the final digit is 1. The other 2(k-1) digits are value 0.
%C If k=6*i or k=6*i-1 the number is always divisible by 7. [_Giacomo Fecondo_, May 22 2010]
%e a(1) = 7 = (10^0+1)^3 -(10^0)^3 , 2^3-1^3.
%e a(2) = 331 =(10^1+1)^3 -(10^1)^3, 11^3-10^3.
%e a(3) = 300030001 = (10^4+1)^3 - (10^4)^3, 10001^3-10000^3.
%e a(1)= 3t(t+1)+1 with t=10^0; a(2)= 3t(t+1)+1 with t=10^1; a(3)= 3t(t+1)+1 with t=10^4.
%e For k=102 (k=6*17) the number (10^102+1)^3-(10^102)^3 is divisible by 7; for k=101 (k=6*17-1) the number (10^101+1)^3-(10^101)^3 is divisible by 7. [_Giacomo Fecondo_, May 22 2010]
%t Select[Table[3*10^(2 n) + 3*10^n + 1, {n, 0, 1000}], PrimeQ] (* _Vincenzo Librandi_, Jan 28 2013 *)
%o (Magma) [a: n in [0..30] | IsPrime(a) where a is 3*10^(2*n) + 3*10^n + 1]; // _Vincenzo Librandi_, Jan 28 2013
%o (PARI) A160432(n,print_all=0,Start=0,Limit=9e9)={for(k=Start,Limit,ispseudoprime(p=3*100^k+3*10^k+1) & !(print_all & print1(p",")) & !n-- & return(p))} \\ - _M. F. Hasler_, Jan 28 2013
%Y Cf. A002407, A003215.
%K nonn
%O 1,1
%A _Giacomo Fecondo_, May 13 2009
%E New name from _Vincenzo Librandi_, Jan 28 2013
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