A242188 - OEIS

%I #27 May 16 2021 13:48:41

%S 0,8,-19,106,-237,1094,-1103,3810,-3049,9118,-15271,14520,-36133,

%T 32788,-46719,57104,-91773,113606,-113375,187388,-170523,218494,

%U -274545,297242,-407727,504946

%N a(n) = Sum_{i=1..n} (-1)^(i+1) prime(i)^3.

%C For n even this is the negative of the sum of (3^3 - 2^3) + (7^3 - 5^3) + .. (prime(n)^3 - prime(n-1)^3). But this is half of the terms in the sum of (3^3 - 2^3) + (5^3 - 3^3) + (7^3 - 5^3) + ... + (prime(n)^3 - prime(n-1)^3) which has a sum that telescopes to prime(n)^3 - 8. Thus a good estimate of a(n) (half the terms) is prime(n)^3/2 (half the square of the n-th prime) which works well. For odd n, add prime(n)^2 to the estimate for even n.

%H Robert Israel, <a href="/A242188/b242188.txt">Table of n, a(n) for n = 0..10000</a>

%p ListTools:-PartialSums([0,seq((-1)^(i+1)*ithprime(i)^3, i=1..40)]); # _Robert Israel_, Mar 09 2020

%t Table[Sum[(-1)^(i+1) Prime[i]^3,{i,n}],{n,0,30}] (* _Harvey P. Dale_, May 16 2021 *)

%o (PARI) a(n) = sum(i=1, n, (-1)^(i+1)*prime(i)^3);

%Y Cf. A008347, A240860.

%K sign

%O 0,2

%A _Timothy Varghese_, May 22 2014