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a(n) = prime(n)^5 - prime(n)^3.
10

%I #13 Sep 08 2022 08:45:33

%S 24,216,3000,16464,159720,369096,1414944,2469240,6424176,20486760,

%T 28599360,69293304,115787280,146928936,229241184,418046616,714718920,

%U 844369320,1349824344,1803871440,2072682576,3076563360,3938468856

%N a(n) = prime(n)^5 - prime(n)^3.

%H Vincenzo Librandi, <a href="/A138406/b138406.txt">Table of n, a(n) for n = 1..200</a>

%H <a href="/index/Pri#prime_powers">Index to sequences related to prime powers</a>

%t a = {}; Do[p = Prime[n]; AppendTo[a, p^5 - p^3], {n, 1, 50}]; a

%o (PARI) forprime(p=2,1e3,print1(p^5-p^3", ")) \\ _Charles R Greathouse IV_, Jun 16 2011

%o (Magma) [NthPrime((n))^5 - NthPrime((n))^3: n in [1..30] ]; // _Vincenzo Librandi_, Jun 17 2011

%K nonn,easy

%O 1,1

%A _Artur Jasinski_, Mar 19 2008