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%I #33 Dec 13 2024 18:21:21
%S 1,5,19,41,109,155,271,341,505,811,929,1331,1639,1805,2161,2755,3421,
%T 3659,4421,4969,5255,6161,6805,7831,9311,10099,10505,11341,11771,
%U 12655,16001,17029,18631,19181,22051,22649,24491,26405,27721,29755,31861,32579,36289
%N a(n) = p^2 - p - 1 where p = prime(n), the n-th prime.
%C Terms are divisible by 5 iff p is of the form 10*m + 3 (A030431).
%H Robert Israel, <a href="/A306190/b306190.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A036689(n) - 1.
%F a(n) = A036690(n) - A072055(n).
%F a(n) = A060800(n) - A089241(n).
%F From _Amiram Eldar_, Nov 07 2022: (Start)
%F Product_{n>=1} (1 + 1/a(n)) = A065488.
%F Product_{n>=2} (1 - 1/a(n)) = A065479. (End)
%F a(n) = A033879(A001248(n)). [Deficiency of squares of primes] - _Antti Karttunen_, Dec 13 2024
%e a(3) = 19 because 5^2 - 5 - 1 = 19.
%p map(p -> p^2-p-1, [seq(ithprime(i),i=1..100)]); # _Robert Israel_, Mar 11 2019
%t Table[Prime[n]^2-Prime[n]-1, {n, 1, 100}] (* _Jinyuan Wang_, Feb 02 2019 *)
%o (PARI) a(n) = {p=prime(n);p^2-p-1;} \\ _Jinyuan Wang_, Feb 02 2019
%Y Supersequence of A091568.
%Y Subsequence of A028387 or A165900.
%Y Second column of A378979.
%Y Cf. A000040, A001248, A030431, A033879, A036689, A036690, A060800, A065479, A065488, A072055, A089241.
%Y A039914 is an essentially identical sequence.
%K nonn
%O 1,2
%A _Kritsada Moomuang_, Jan 28 2019