perplexus.info :: Sequences : Sequence Addition

Sequence Addition (Posted on 2011-08-03)

A, B and C are polynomial increasing sequences of positive integers.

1. A(k) + B(k) = C(k)
2. A(8) is a square
3. C(8)= 128.
Both A and B series have only one prime number as a member.
Each member of C is evenly divisible by at least one square (other than 1) number.

How are the sequences A, B and C defined?
What's so special about them?

No Solution Yet Submitted by brianjn

Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

Solution

Comment 2 of 2 |

Going off of Ady's hint I set C(k)=2*k^2 so that 8th square (64) corresponds to the 8th term (128).

Trying A(k) = k^2-k and B(k) = k^2+k is almost a solution. The only issue is A(1) = 0, which is nonpositive. But since B(k) is always an even positive integer, its terms can be cut in half and still be integers.

With a new B(k) = (k^2+k)/2 then A(k) = (3k^2-k)/2.

A(k) = 1, 3, 6, 10, 15, 21, 28, 36 (Triangular numbers)

B(k) = 1, 5, 12, 22, 35, 51, 70, 92 (Pentagonal numbers)

C(k) = 2, 8, 18, 32, 50, 72, 98, 128 (Doubled square numbers)

Posted by Brian Smith on 2018-12-16 19:56:14

Please log in:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info