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a(n) = 2*p(n)*p(n+2) - p(n+1)^2 where p(k) = k-th prime.
2

%I #22 Apr 25 2024 05:18:51

%S 11,17,61,61,205,205,421,573,585,1185,1173,1501,2005,2349,2737,2985,

%T 4185,4173,4741,5889,5877,7173,8181,8569,9781,11005,11005,12301,14917,

%U 13477,17637,17649,21505,19777,23985,24577,25869,28509,29857,30585,35617

%N a(n) = 2*p(n)*p(n+2) - p(n+1)^2 where p(k) = k-th prime.

%C Theorem: a(n) > 0. Proof: Use p(n+1) <= 2 p(n)^2 for n > 4. (See Sándor et al.) QED

%D József Sándor, Dragoslav S. Mitrinovic, and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, Chapter VII, p. 247, section VII.18.b.

%H Seiichi Manyama, <a href="/A324795/b324795.txt">Table of n, a(n) for n = 1..10000</a>

%t With[{p = Prime[Range[50]]}, 2 * p[[1;;-3]] * p[[3;;-1]] - p[[2;;-2]]^2] (* _Amiram Eldar_, Apr 25 2024 *)

%Y Cf. A056221 (if leading coefficient 2 is changed to 1), A327447 or A309487 (if 2 is changed to 4).

%K nonn

%O 1,1

%A _N. J. A. Sloane_, Sep 10 2019