A242188 a(n) = Sum_{i=1..n} (-1)^(i+1) prime(i)^3.

0, 8, -19, 106, -237, 1094, -1103, 3810, -3049, 9118, -15271, 14520, -36133, 32788, -46719, 57104, -91773, 113606, -113375, 187388, -170523, 218494, -274545, 297242, -407727, 504946

Offset: 0

Author

Timothy Varghese, May 22 2014

Keywords

• sign

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

Crossrefs

Cf. A008347, A240860.

Programs

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 *)

PARI

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