%I #24 Sep 08 2022 08:46:13
%S 8,26,98,218,602,866,1538,1946,2906,4706,5402,7778,9602,10586,12698,
%T 16226,20186,21602,26138,29402,31106,36506,40346,46466,55298,60002,
%U 62426,67418,69986,75266,95258,101402,110978,114266,131426,135002,146018,157466
%N Number of unit cubes that have a side on the surface of a p X p X p cube composed of p^3 unit cubes, where p is the n-th prime.
%C Inspired by geometric interpretation of A030078.
%C Obviously, there is only one solution for a, b and c to the equation p X p X p = a X b X c where p is prime, if a > 1, b > 1 and c > 1. So there can be only one rectangular prism that is composed of p^3 unit cubes, if a > 1, b > 1 and c > 1. That rectangular prism is the p X p X p cube. Its uniqueness is a motivation for this sequence.
%C The number of unit cubes that have a side on the surface of the p X p X p cube is p^3 - (p-2)^3 = 6*p^2 - 12*p + 8.
%F a(n) = 6*prime(n)^2 - 12*prime(n) + 8 = 6*A001248(n) - 12*A000040(n) + 8.
%e For the 5 X 5 X 5 cube, composed of 125 unit cubes, there are 98 unit cubes that have a side on the surface of the cube.
%t Table[6 Prime[n]^2 - 12 Prime[n] + 8, {n, 50}] (* _Vincenzo Librandi_, Sep 08 2015 *)
%o (PARI) vector(45, n, 6*prime(n)^2-12*prime(n)+8)
%o (PARI) a(n,p=prime(n))=6*p^2-12*p+8 \\ _Charles R Greathouse IV_, Sep 08 2015
%o (Magma) [6*NthPrime(n)^2-12*NthPrime(n)+8: n in [1..40]]; // _Vincenzo Librandi_, Sep 08 2015
%Y Cf. A000040, A001248, A030078.
%K nonn,easy
%O 1,1
%A _Altug Alkan_, Sep 07 2015