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%I #27 Sep 08 2022 08:45:55
%S 3,7,22,45,116,163,282,353,520,831,950,1357,1668,1835,2194,2793,3464,
%T 3703,4470,5021,5308,6219,6866,7897,9384,10175,10582,11421,11852,
%U 12739,16098,17129,18736,19287,22166,22765,24612,26531,27850,29889,32000,32719,36438,37205,38764,39555,44474,49681
%N a(n) = prime(n)^2 - n.
%C One way to find a run of n consecutive nonsquarefree numbers such that the first n primes appear in order as factors of numbers in the run is to use the Chinese remainder theorem (though this run is most likely not the earliest of length n).
%C The moduli are then of course the squares of the first n primes, while the remainders are then the first n terms of this sequence. (See A182433.)
%H Vincenzo Librandi, <a href="/A182174/b182174.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A000040(n)^2 - n = A001248(n) - n. - _Omar E. Pol_, Apr 16 2012
%e a(4) = 45 because the 4th prime is 7, and 7^2 - 4 = 49 - 4 = 45.
%t Table[Prime[n]^2 - n, {n, 50}]
%o (Magma) [NthPrime(n)^2-n: n in [1..50]]; // _Bruno Berselli_, Apr 16 2012
%Y Cf. A001248 squares of primes; A045882 and A078144 pertain to runs of consecutive nonsquarefree numbers.
%Y Cf. A014689. [_Bruno Berselli_, Mar 19 2013]
%K nonn,easy
%O 1,1
%A _Alonso del Arte_, Apr 16 2012
%E a(36) inserted by _Vincenzo Librandi_, Mar 19 2013
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