Pete's Blog

The following are the first forty terms of a sequence:

0, 2, 4, 6, 8, 10, 11, 12, 14, 16, 18, 22, 33, 44, 55, 66, 77, 88, 99, 101, 110, 121, 132, 141, 143, 154, 161, 165, 176, 181, 187, 198, 201, 202, 221, 222, 241, 242, 261, 262, 281, . . .

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.

Update: Hint: This is a list, so the sequence is not dependent on the index. Now, for positive integer i there exist nonnegative integers c_i and d_i (c_i greater than or equal to d_i) such that a(i)=c_i+d_i. Obviously, there are ½a(i)+1 choices for the pair (c_i, d_i) if a(i) is even, and ½(a(i)+1) choices for the pair (c_i, d_i) if a(i) is odd, but to be more specific, for a certain pair dependent on a(i) (and for full disclosure, even this pair might not be unique), there is an almost one-to-one relationship between c_i and d_i, that is to say, if you choose the correct c_i you get d_i, and most of the time if you have the correct d_i
you get c_i. An instance where knowing d_i will not give you c_i is for a(19); if you have c_i you get d_i
but if you have d_i
there are an infinite number of choices for c_i. To give some substance to this hint, think mirror for the relationship between the two elements of the pair. (This is a very strong hint, provided you don't over extend the properties of a mirror relationship.)

Update: I've added twenty more terms to the sequence. This is not to denote difficulty. Something worth noting happens at a(33) so I decided to list to the nearest multiple of ten that was greater than 33. (A similar thing happened at a(11).)

Last week's sequence (5, 13, 17, 25, 29, 37, 41, 53, 61, 65, . . .) was the set of hypotenuses of primitive Pythagorean Triples. Suppose you have integers a, b and c. They form a primitive Pythagorean Triple if 1. a²+b²=c², and 2. a and b are coprime (i.e., they share no prime factors, i.e., their greatest common denominator is 1). For the triple a, b and c that satisfies the previous equation, we say that c is the hypotenuse because it is the longest side, and it is opposite the right angle. E.g., 5 is in the sequence because 5²=3²+4², and 3 and 4 share no prime factors. E.g., 37²=12²+35², and 12 and 35 share no prime factors.

Alternatively, last week's sequence is the list of terms from the sequence two weeks ago (0, 1, 2, 4, 5, 8, 9, 10, 13, 16, . . .) with some excluded. Suppose we have a term from two weeks ago, say a(i) for some i=1, 2, 3, . . ., where a(i)= c²+d² for some integers c and d. Then a(i) is excluded from last week's sequence if and only if c and d share a prime factor or c and d are both odd or one of c and d is 0. I mentioned in the hint last week that Euclid's formula generates these terms. Suppose you have odd integer c and even integer d who share no prime factor. These generate the primitive Pythagorean Triple 2cd, |c²-d²| and c²+d². E.g., 0=0²+0², c=d=0. But 0 and 0 share an infinite number of prime factors, so 0 was not in last week's sequence.

Practically speaking, the exclusion statements in the previous paragraph exclude three categories of terms in the sequence from two weeks ago (this is not an exhaustive list):

1. The term is a square, say k² for some integer k. Then k²=k²+0². Obviously 0's in this construction, so all terms that are squares are excluded.
   
2. The term is a square doubled, say 2k² for some k=2, 3, 4, . . . . Then 2k²=k²+k². Obviously k shares at least one prime factor with itself so all the double squares are excluded.
   
3. The term is an hypotenuse of a Pythagorean Triple, but it is not of a primitive Pythagorean Triple. A Pythagorean Triple has the form (ma) ²+(mb) ²=(mc) ² where a, b and c are integers greater than 1, and a and b share no prime factors. The primitive case is for m=1. (An example of a Pythagorean Triple than is not primitive is 45. 45²=27²+36². But 27 and 36 share the prime factor 3, so 45 is a Pythagorean Triple, but is not primitive.) These are the toughest to identify for the purposes of excluding.
   

So, e.g., exclude 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, the rest of the squares, as they fit in the first category.

E.g, exclude 8, 18, 32, 50, the rest of the double squares, as they fit in the second category.

E.g., exclude 10, 20, 26, 34 the rest of the non-primitive Pythagorean Triples, as they fit in the third category.

Real examples: E.g., 2=1²+1². Obviously 1 and 1 are both odd, so 2 is excluded. E.g., 5=1²+2². Considering 1 and 2, they share no prime factors, 2 is even so they're not both odd, and neither number is 0, so there's no reason to exclude 5. Thus, 5 is the first candidate to go into last week's sequence. E.g., 13=2²+3². Considering 2 and 3, they share no prime factors, 2 is even so they're not both odd, and neither number is 0, so there's no reason to exclude 13. It is the second candidate to go into last week's sequence.

It was not solved.