Pete's Blog

The following are the first twenty terms of a sequence:

4, 6, 9, 10, 14, 15, 21, 25, 35, 49, 121, 143, 169, 187, 209, 221, 247, 253, 289, 299, . . .

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.

If it hasn't been solved by Thursday, I'll give a hint. 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.

Solutions to the sequences of the previous post:

1. The Tribonacci sequence. Each term is the sum of the previous three terms. E.g., a(3)=1, a(4)=1, a(5)=2, so a(6)=4. Jer solved it.
   
2. The sequence of primes who were the sum of two primes. E.g., 2 and 17 are prime so 19=2+17 is in the sequence, i.e., a(4)=19. It was not solved.
   
3. The sequence of non-square free numbers. That is, numbers who have at least one prime factor with multiplicity. E.g., 21=(3)(7) has no prime factors with multiplicity so it is not in the sequence. E.g., 18=(2)(3²) has the prime factor 3 with a multiplicity of 2. So 18 is in the sequence, i.e., a(6)=18. It was not solved.
   
4. I'll repeat that this was not an easy one. The term a(n) was the largest exponent such that 2^a(n) divided 2n. E.g., take the index n=5. Then 2n=10. 2¹ divides 10, but 2² does not. So, a(5)=1. It was not solved.
   
5. First, define a(1)=1. For the rest of the sequence a(n) was the smallest positive integer that hadn't been used yet which shared a prime factor with a(n-1). E.g., consider n=7. Since a(6)=9, a(7) must be a multiple of 3, since the only prime factor of 9 is 3. Is a(7)=3? No, because 3 has already been used by a(5). Is a(7)=6? No, because 6 has already been used by a(3). Obviously a(7)≠9 since that's what a(6) is. Is a(7)=12? Since 12 hasn't appeared in the sequence yet, yes, a(7)=12. It was not solved.
   
6. Of the seven sequences, this was by far the most interesting. Define a(1)=1. The hint gave away most of the solution, that is, "i is in the sequence if and only if a(i) is odd." The only things left to say were to make clear the definition of the first term, and to say that the sequence is strictly increasing, that is to say, a(i+1)>a(i). E.g., a(2) is unquestionably the most fun to figure out why it is 4, so I'll leave that for the reader's pleasure. 3 is not in the sequence (since only 1 and 4 are, thus excluding 2 and 3), so a(3) must be even. The next lowest even number is 6. So a(3)=6. E.g., 4 is in the sequence so a(4) is odd. So we take the lowest odd number that greater than 6. Thus, a(4)=7.
   
7. The sequence of balanced primes. The sequence is formed by taking the mean average of each pair of primes (i.e., for all distinct pairs of primes p and q we find (p+q)/2) and if the value is prime it is kept in the sequence. Obviously the sequence is ordered least to greatest. E.g., the mean value of the primes 3 and 7 is 5 (i.e., (3+7)/2=5), and since 5 is prime, 5 is in the sequence, i.e., a(1)=5. As you can see, there aren't a lot of primes that are the mean of two other primes.
   

Proof of the exercise given for the sequence of Monday, April 14 (1, 2, 4, 6, 16, 12, 64, 24, 36, 48). I asked you to prove that for each positive integer i there existed a number with exactly i factors. It's fairly simple once you get the first step. The number 2^i-1 has exactly i distinct factors. So, there might be a number less than 2^i-1 with i factors (e.g., 48 is significantly less than 512, but they each have exactly 10 distinct factors), but there always is a number with i factors.