A240860 a(n) = Sum_{i=1..n} (-1)^{i+1} prime(i)^2, where prime(k) denotes the k-th prime: alternating sum of the squares of the first n primes.

4, -5, 20, -29, 92, -77, 212, -149, 380, -461, 500, -869, 812, -1037, 1172, -1637, 1844, -1877, 2612, -2429, 2900, -3341, 3548, -4373, 5036, -5165, 5444, -6005, 5876, -6893, 9236, -7925, 10844, -8477, 13724, -9077, 15572, -10997, 16892, -13037, 19004, -13757

Offset: 1

Author

Timothy Varghese, May 06 2014

Keywords

• sign

Comments

For n even this is the negative of the sum of (3^2 - 2^2) + (7^2 - 5^2) + ... + (prime(n)^2 - prime(n-1)^2). But this is half of the terms in the sum of (3^2 - 2^2) + (5^2 - 3^2) + (7^2 - 5^2) + ... + (prime(n)^2 - prime(n-1)^2) which has a sum that telescopes to prime(n)^2 - 4. Thus a good estimate of a(n) (half the terms) is prime(n)^2/2 (half the square of the n-th prime) which works well up to n = 10000. For odd n, add prime(n)^2 to the estimate for even n.

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• Timothy Varghese, Alternating sums, MathOverflow, May 2014.

Crossrefs

Cf. A000040, A008347, A024450.

Programs

PARI

a(n) = sum(i=1, n, (-1)^(i+1)*prime(i)^2); \\ Michel Marcus, May 09 2014