OFFSET
0,2
COMMENTS
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.
LINKS
Robert Israel, Table of n, a(n) for n = 0..10000
MAPLE
ListTools:-PartialSums([0, seq((-1)^(i+1)*ithprime(i)^3, i=1..40)]); # Robert Israel, Mar 09 2020
MATHEMATICA
Table[Sum[(-1)^(i+1) Prime[i]^3, {i, n}], {n, 0, 30}] (* Harvey P. Dale, May 16 2021 *)
PROG
(PARI) a(n) = sum(i=1, n, (-1)^(i+1)*prime(i)^3);
CROSSREFS
KEYWORD
sign
AUTHOR
Timothy Varghese, May 22 2014
STATUS
approved