In the infinite series: 1 + 3 + 3 + 3 + 5 + 5 + 5 + 5 + 5 + 7 + ... each positive odd integer k appears k times in consecutive order.

1) Write a general formula that can be used to find the n_th term of the series.

2) Prove that if n is not 3, then the n_th partial sum of the series is never prime.

Part of part 1 can be solved where n is a perfect square by using the sum of squares formula. Let n=k², and combine like terms of the sequence to get 1², 3², 5², ... so term k of the sequence is (2k-1)² and that the sum of the two sequences to k² and n are the same.

Expanding and rearranging the terms gives:
 (1+1+...+1)+(-4)(1+2+...k)+4(1²+2²+...k²)
=k-4(k)(k+1)/2+4(k)(k+1)(2k+1)/6
=k(1)-k(2k+2)+2(k)(k+1)(2k+1)/3
=-(k)(2k+1)+k(2k+2)(2k+1)/3
=k(-3)(2k+1)/3+k(2k+2)(2k+1)/3
=k(2k-1)(2k+1)/3

It is easy to prove that these numbers must be composite if k is greater than 1 since the terms are all different, and one must be divisible by 3. The odd numbers in the series after term k^2 are 2(k+1)-1=2k+1, and it stays composite adding any number of 2k+1.

If 2k+1 is divisible by 3, it must have another prime factor, and thus it is a factor of k(2k-1)(2k+1)/3 and so no matter how many more 2k+1 are added, it will retain that factor.

If 2k+1 is not divisible by 3, then k or 2k-1 must be, and so (2k-1)(k)/3 is an integer, so no matter how many 2k+1 are added to ((2k-1)(k)/3)(2k+1)

The remaining cases when k=1 or k=0, can be tested manually, resulting in only n=3 being prime.

Posted by Gamer on 2007-01-08 19:14:20