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A182174 a(n) = prime(n)^2 - n. 5

%I

%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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Last modified February 19 16:02 EST 2019. Contains 320311 sequences. (Running on oeis4.)