Pete's Blog

The following are the first fifty terms of a sequence:

1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49, 68, 70, 79, 82, 86, 91, 94, 97, 100, 103, 109, 129, 130, 133, 139, 167, 176, 188, 190, 192, 193, 203, 208, 219, 226, 230, 236, 239, 262, 263, 280, 291, 293, 301, 302, 310, 313, 319, 320, . . .

Same idea as the other sequences. Describe the pattern of the sequence. If you use a search engine, please don't submit it in the comments.

On Wednesday, if it hasn't been solved yet, I'll give a hint (yes, earlier than usual). This one is very tough. I don't know how helpful the hint I'm planning on offering is going to be, so if it's not enough, on Friday I'll post a big hint which will be enough to get it solved.

To give you a starting point, this is a list of numbers that meet a criterion. So considering the absent numbers is as important as considering the present terms. The criterion involves the sum of the squares of the digits of the positive numbers. I don't expect this to be particularly helpful, though. Also, if you ask a polar question (i.e., a yes/no question), I will probably answer it if it relates to this sequence.

I'll post the solution on next week's sequence.

Last week's sequence (1, 4, 9, 16, 25, 36, 49, 64, 81, 1, . . .) was the sum of the squares of the digits. E.g., a(18)=1²+8²=1+64=65. KWTsui solved it.

Update: First Hint: The criterion is one of convergence. And this convergence is related to the sum of the square of the digits of the positive numbers.

Update: Second Hint: Iterate the process of finding the sum of the squares of the digits of the numbers. Do it first for the first few terms of this sequence until you think you see what happens, then do the same iteration process for the positive integers absent from this sequence and confirm that is why they are absent.