Pete's Blog

The following are the first twenty-five terms of a sequence:

0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, 40, 41, 45, 49, 50, . . .

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.

I'm not sure how tough this one will be. I'm running out of ideas for good ones to post. If it hasn't been solved by Thursday, I'll give a hint then. 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 (4, 6, 9, 10, 14, 15, 21, 25, 35, 49, . . .) was the list of composite numbers with two prime factors of equal digit length. E.g., a(7)=21, since 21=(3)(7), and each of 3 and 7 are only one digit long. That's why there's the gap between a(10)=49 and a(11)=121. KWTsui pretty-much solved it.

Update: Sorry for being so late in bringing you the hint. Hint: You'd come back again and again to this list when dealing with the Pythagorean Theorem.

Recently the Southern Baptist Convention's "Executive Committee voted to end cooperative relationship with Broadway [Baptist Church] over church's failure to clarify or correct stand." Dr. Mohler tweeted that he was in the meeting at 2:20 this afternoon, tweeted the announcement at 3:20.

Searching the Baptist Press (the Southern Baptists' newspaper), the earliest mention I could find of the issue was from December 4^th, 2007. Apparently their pastor (who'd started in April 2001) wanted to allow homosexual members to have a single picture of their couples to appear in the members' directory.

Reading a little later, another article came out March 6^th, 2008, stating that there was a vote on the issue (it is a Baptist church), and the vote passed (with less than a third of the membership voting—boo-urns). (Update: I misunderstood the vote. They took out any idea of having couples together. My current understanding is that they put pictures of the members doing ministry together, which might have the couple together, but not necessarily.) But this vote led to a petition for the pastor to be removed from the church. The vote was on March 9^th and the congregation voted to keep their pastor (although, you can hardly say that the congregation voted to keep their pastor if only half the membership voted).

The pastor ultimately resigned anyway to take a teaching post at a theological school in Georgia.

The Southern Baptist Convention's Executive Committee a year ago started considering what to do with Broadway since the denomination is confessionally opposed to homosexuality: "In the spirit of Christ, Christians should oppose . . . homosexuality." Finally they've decided. I think they made the right call.

You might find this boring since it doesn't really impact any of us. Still, I wanted to post it because the SBC's (Southern Baptist Convention) always had a special place in my heart. They started off strong, swung liberal, and then in the late 70s and 80s regained the orthodoxy the denomination was founded on. Also, they're real Baptists, honouring both credobaptism and congregationalism, plus they're complementarian. A few other things in the mix are the presence of Drs. Albert Mohler and Mark Dever, both are Southern Baptists, and the International Mission Board which works really hard to send anyone who has a desire and is qualified to go into the mission field. They really want to tell the world about Jesus. So I get pretty excited when I hear about them. Especially when it's news of them doing the right thing.

In a talk last year Dr. Mohler said that he was uncertain what direction the denomination was going to take when it came to congregations that were walking in a direction opposed to their confession. I wonder whether he had this issue in mind when he mentioned it. I'm glad to know that they had the courage to separate from people who were in clear opposition, not only to the confession, but to the Scriptures. It gives me hopes that this cooperative organization (when I called the SBC a denomination I say it in the very loosest sense of the word, since the SBC is committed to congregationalism and local church autonomy) might stand for some good years ahead.

Update: Read the Baptist Press'
article.

--

Another thing that interested me in the course of this process, was the number of members who participated in the votes, compared with the total membership. Membership inflation has been a problem in the SBC. For example, officially, there are 16.5 million members on their rolls. But Dr. Tom Ascol, pastor of an SBC church in Cape Coral Florida, said that of the 16.5 million members, only 6.1 million of them regularly attend a Sunday worship service. And we Baptists complain about the paedobaptists having unsaved members.

Sorry Dan. I haven't posted anything of substance because I don't have too much to say. I still don't.

• Did you know that the same country where Calvinism's TULIP acronym was thought up has the tulip as its national flower?
  
• I appreciate that Dr. MacArthur's Question and Answer page has a feature that allows you to report typos. I would like it if more websites had it.
  
• Saturday Dr. Mohler Tweeted that he was picking out a card for his wife, even though her birthday wasn't until Sunday. Seems strange; I wouldn't want to alert my wife (who might be reading my online material) to what I was doing in advance for her birthday.
  
• I have never heard Piper use so much humour before in a message. Very interesting and engaging autobiographical talk.
  
• I wonder whether people who enjoy speaking on the phone for endless hours have difficulty reading nonverbal communication.
  
• If I were a postmodernist I'd never say the self-defeating statement, "There is no absolute truth." I'd say, "There is only one absolute truth."
  
• I'm often surprised at how few people know this worship song. I think it's great.

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.

The following are the first ten terms of a sequence:

0, 1, 0, 2, 0, 1, 0, 3, 0, 1, . . .

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 Wednesday, I'll give a hint. It's pretty easy, so you probably won't need 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.

Update: Hint: It's related to powers of 2.

I'll post the solution on Thursday. (Since I forgot to put the hint up for last week's sequence until today, I won't put the solution up for it until Wednesday of this week.)

Update: The sequence is the highest exponent for which 2 to the exponent of the entry divides the index. E.g., 2¹=2 divides 3 but 2²=4 does not, so a(3)=1. E.g., 2³=8 divides 8 but 2⁴=16 does not, so a(8)=3. E.g., 2⁰=1 divides 7 but 2 does not, so a(7)=0. E.g., 2¹=2 divides 10 but 2²=4 does not, so a(10)=1. It was not solved.

Update: Last week's second sequence (1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, . . .) was the list of smallest integers with exactly the index's number of proper divisors, that is, a(n) was the smallest positive integer with exactly n divisors. E.g., 1, 2, 3, 4, 6, 9, 12, 18, 36 all divide 36, and they are the only divisors of 36. Furthermore, no positive integer less than 36 has exactly nine divisors. So a(9)=36. It was not solved. As an extra exercise, prove that for each i=1, 2, 3, . . . there exists an integer with exactly i divisors.

The following is a list of sequences (of varying difficulty, where the difficulty is reported based on the number of terms listed, on my subjective assessment) for the following seven weeks, each with a varying number of terms listed:

1. 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, . . .
   
2. 5, 7, 13, 19, 31, 43, 61, 73, 103, 109, 139, 151, 181, 193, 199, 229, 241, 271, 283, 313, 349, 421, 433, 463, 523, 571, 601, 619, 643, 661, 811, 823, 829, 859, 883, 1021, 1033, 1051, 1063, 1093, 1153, 1231, 1279, 1291, 1303, 1321, 1429, 1453, 1483, 1489, . . . (Update: Hint: Suppose p and q are primes. Which for which values of p and q is p + q in the sequence?)
   
3. 4, 8, 9, 12, 16, 18, 20, 24, 25, 27, . . . (Update: Hint: Look at the prime factors.)
   
4. 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, . . . (Update: Hint: (Upon further thinking, I should have given more terms to indicate that this sequence was tough.) It's something to do with 2^a(n).)
   
5. 1, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, 20, 22, 11, . . . (Update: Hint: First, notice that the greatest common divisor for successive terms beyond the first (i.e., gcd(a(i), a(i + 1))>1 for each i є {2, 3, 4, . . .}. Second, this sequence contains every positive integer.)
   
6. 1, 4, 6, 7, 8, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 23, 25, 27, 29, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 70, 71, 72, 73, 74, . . . (Update: Hint: Notice that i is in the sequence if and only if a(i) is odd.)
   
7. 5, 53, 157, 173, 211, 257, 263, 373, 563, 593, 607, 653, 733, 947, 977, 1103, 1123, 1187, 1223, 1367, . . . (Update: Hint: This sequence is also a subset of the primes. It's somewhat similar to the sequence five above. For each positive integer i, a(i) is related to two primes, p and q, and more specifically the sum p + q. But, this sequence is dissimilar from the one above in that this one's terms, while all are directly related to p + q they're not necessarily of the form p + q. The terms are of the form p + q with something more done to them.)
   

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.

Within eight weeks from now, for any of the sequences that are still left unsolved I'll post a hint, and then on June 15^th, with that week's sequence, post the solutions.

I think that it's easy to think that the healings in the Bible are always related to physical ailments. For example, Jesus healed blindness (Jn. 9:1-7), a withered hand (Lk. 6:6-10), a leper (Mk. 1:40-42), paralysis (Mt. 9:2-7), and many other conditions. As a result, it's easy (at least for me) to think that prayers for healing should be limited to physical infirmities. But consider Psalm 107:17-20:

If I hadn't read it, I would have never guessed that the Bible had anything to say about anorexia. I'm happy that I read it, because it reminds me once again that God does care. Not only does God care when a person is poor, hungry, fatherless, physically broken, persecuted, etc., but He also cares when our minds aren't working properly. It gives me hope that God is willing to heal struggles with gluttony or depression or paranoia, or other psychological problems.

As a note of criticism of much of Christian counseling, this verse clearly says that it's sin that's responsible for this malady, not a deficient childhood where you didn't learn the lessons you needed to. (I'm very pro-biblical counseling, where Christian counseling seems semi-Pelagian at best.)

This week's was solved really quickly, so I'm posting another. The following are the first twenty terms of a sequence:

1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144, 120, 65536, 180, 262144, 240, . . .

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.

This afternoon's sequence (1, 4, 8, 9, 16, 25, 27, 32, 36, 49, . . .) was the sequence of Powerful Numbers. That is to say, the list of positive numbers where if the number has a prime divisor, the square of that prime divides it. E.g., 2 divides 8, so, since 8 is in the list it follows that 2²=4 divides 8, which is obviously true. T solved it.

Update: I've added another ten terms to the sequence so that there are now twenty.

Update: Hint: There's a relationship between n and the divisors of a(n), but probably not the category of relationship that first comes to mind. Since I put the hint up late, I'm not going to put the solution on next week's sequence (that of April 20^th) until Wednesday, April 22^nd.

I love John Piper's ministry. I've referred to it here before, and I have the sidebar on the right of my blog devoted to the resources of his that I've read or listened to lately. Yesterday I listened to a talk that he gave at Driscoll's pastors conference last year. It was the first time in a long time (a few months) that I listened to something from Piper. It filled me with conviction of sin, and a joy in the glory and holiness of God. As with most of the audio I link to in my entries, I highly recommend it.

The talk reminded me of his book The Supremacy of God in Preaching (2003). The book was probably one of the three most important books that I've read in my Christian walk (alongside Desiring God [Piper] and Evangelism and the Sovereignty of God [Packer]).

Here's a few paragraphs from the first chapter (pp. 26-27):

At 109 pages, it's a short read with fairly large print and fairly small pages. It's Piper at his best. Adam said that it's the most readable of Piper's books. I completely agree. Get it. Read it.

One of the blurbs on its back reads, "Here's a book that every preacher should read at least once a year. This book is a powerful antidote to the unbalance, self-centered preaching of today."

The following are the first twenty terms of a sequence:

1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, . . .

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.

Last week's sequence (1, 2, 3, 4, 5, 3, 7, 4, 6, 5, . . .) was the smallest number, say m, such that n divided m! (m-factorial). E.g., 10 divides 5!=120, but not 4!=24, so a(10)=5. E.g., 16 divides 6!=720, but not 5!=120, so a(16)=6. It was not solved.

The following are the first twenty terms of a sequence:

1, 2, 3, 4, 5, 3, 7, 4, 6, 5, 11, 4, 13, 7, 5, 6, 17, 6, 19, 5, 7, 11, 23, 4, 10, 13, 9, 7, 29, 5, 31, 8, 11, 17, 7, 6, 37, 19, 13, 5, . . .

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 then. 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, 2, 3, 1, 5, 6, 7, 2, 1, 10, . . .) was what's called the square-free part of n. That is to say, a(n) is the smallest positive (i.e., nonzero) divisor of n such that na(n) is a square. E.g., 18=2∙3² so (2)(18)= 2²∙3²=(2∙3)², yielding a(18)=2. KWTsui solved it.

Update: I've added another 20 terms to the sequence. Furthermore, here's a very strong hint: The sequence describes a certain relationship between n and a(n)! (i.e., a(n) factorial).

Blogger's note: The following contrast is extreme. I did not write it myself. I would like to give credit where it's due, but it was sent to me in an email. If you know to whom credit is due, let me know. Follows is the story of The Ant and the Grasshopper:

CLASSIC VERSION:

The ant works hard in the withering heat all summer long, building his house and laying up supplies for the winter. The grasshopper thinks he's a fool, and laughs and dances and plays the summer away. Come winter, the ant is warm and well fed. The shivering grasshopper has no food or shelter, so he dies out in the cold.

THE CANADIAN VERSION:

The ant works hard in the withering heat all summer long, building his house and laying up supplies for the winter. The grasshopper thinks he's a fool, and laughs and dances and plays the summer away. Come winter, the ant is warm and well fed.

So far, so good, eh?

The shivering grasshopper calls a press conference and demands to know why the ant should be allowed to be warm and well fed while others less fortunate, like him, are cold and starving. The CBC shows up to provide live coverage of the shivering grasshopper, with cuts to a video of the ant in his comfortable warm home with a table laden with food. Canadians are stunned that in a ountry of such wealth, this poor grasshopper is allowed to suffer so while others have plenty.

The NDP, the CAW and the Coalition Against Poverty demonstrate in front of the ant's house. The CBC, interrupting an Inuit cultural festival special from Nunavut with breaking news, broadcasts them singing "We Shall Overcome." Jack Layton grants in an interview with Mike Duffy that the ant has gotten rich off the backs of grasshoppers, and calls for an immediate tax hike on the ant to make him pay his "fair share".

In response to polls, the Liberal Government drafts the Economic Equity and Grasshopper Anti-Discrimination Act, retroactive to the beginning of the summer. The ant's taxes are reassessed, and he is also fined for failing to hire grasshoppers as helpers. Without enough money to pay both the fine and his newly imposed retroactive taxes, his home is confiscated by the government. The ant moves to the US and starts a successful agribiz company.

The CBC later shows the now fat grasshopper finishing up the last of the ant's food, though spring is still months away, while the government house he is in, which just happens to be the ant's old house, crumbles around him because he hasn't bothered to maintain it. Inadequate government funding is blamed, Bob Rae is appointed to head a commission of enquiry that will cost $10,000,000.

The grasshopper is soon dead of a drug overdose, the Toronto Star blames it on the obvious failure of government to address the root causes of despair arising from social inequity. The abandoned house is taken over by a gang of immigrant spiders, praised by the government for enriching Canada's multicultural diversity, who promptly set up a marijuana grow op and terrorize the community.

THE END

Blogger's note: I don't agree that laziness should be punished with death, but I think this parable does a good job of showing that circumstances are often self inflicted (though, entirely within the sovereign providence of God).

The following are the first twenty terms of a sequence:

1, 2, 3, 1, 5, 6, 7, 2, 1, 10, 11, 3, 13, 14, 15, 1, 17, 2, 19, 5, . . .

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.

I think this is the toughest one I've posted so far, so you'll probably need a hint. If it hasn't been solved by Thursday, I'll give a hint then. 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, 1, 2, 1, 2, 2, 3, 1, 2, 2, . . .) was the number of 1s in the binary expression of the decimal number. E.g., a(7)=3, since 7₁₀=111₂. T solved it.

Update: I've added five more terms to the sequence. It originally only had the first ten terms.

Update: I've added another five more terms to the sequence.

If you're a fan of Driscoll, or if you're an avid inspector of Piper's resource site (Desiring God), you might be aware of a talk Piper gave at one of Driscoll's pastors' conferences. It was titled "Why I Trust the Scriptures." In it Piper gave an amazing illustration of why Christians should be concerned about know why they believe. It goes something like this:

Suppose a man walks up to, hands you a briefcase, telling you there's a million dollars in it, and asks you to take it to the bank. You ask him why he chose you, and he responds that he picked you at random. Do you feel trusted? No.

Now, suppose the same situation happens—a man walks up to, hands you a briefcase, telling you there's a million dollars in it, and asks you to take it to the bank. This time, though, after you ask him why he's given you the money, he says that it's because he knows you. He's studied your behaviour over the course of months. "I know that you treat your boss fairly, that when you say you're going to do something, you'll do it, that your friends have complete trust in you. I know your character. I have every confidence that if you take this briefcase from me it will get to the bank." You feel trusted from this person.

Likewise, is it glorifying to God is you only trust Him because it's better than nothing? Maybe. But if you trust Him because He's clearly the only choice, because it makes a difference, because He has made you confident in Him, He is very glorified.

I'm not saying every Christian needs to be able to respond to Dawkins, Dennett, Harris and Hitchens (today's four horsemen of atheism), but it's good to know why you trust God. I think that God will be most glorified simply at our words to Him, "Why did I trust your Son? I knew Him." But don't insinuate that confidence in the Truth of the Gospel is about as important as a coin toss by spending as much time thinking about arguments for the faith as you would watching the fate of the flip fall with the coin.

The following are the first ten terms of a sequence:

1, 1, 2, 1, 2, 2, 3, 1, 2, 2, . . .

It may be of interest, though probably unhelpful in determining the pattern, that the limit at infinity of the sequence does not exist, though max_{nє{1,2,3,4, . . .}}{a(n)}=∞. Therefore, it follows that the derivative of the mystical curve of best fit is also non-existent at infinity.

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.

This one's easy, but, if it hasn't been solved by Thursday, I'll give a hint then. 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, 2, 3, 2, 5, 3, 7, 2, 3, 5, . . .) was the greatest prime that divided the placeholder, with the first term defined as 1, that is, a(1)=1. E.g., a(30)=5, since the primes that divide 30 are 2, 3 and 5, and 5 is the highest of them. T solved it. No one remarked that the first term needed to be defined apart from the pattern.

The following are the first forty terms of a sequence:

1, 2, 3, 2, 5, 3, 7, 2, 3, 5, 11, 3, 13, 7, 5, 2, 17, 3, 19, 5, 7, 11, 23, 3, 5, 13, 3, 7, 29, 5, 31, 2, 11, 17, 7, 3, 37, 19, 13, 5, . . .

It may be of interest, though probably unhelpful in determining the pattern, that the limit at infinity of the sequence does not exist, though max_{nє{1,2,3,4, . . .}}{a(n)}=∞. Therefore, it follows that the derivative of the mystical curve of best fit is also non-existent at infinity.

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.

This one's easy, but, if it hasn't been solved by Thursday, I'll give a hint then. 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 (0, 1, 1, 1, 1, 2, 1, 1, 1, 2, . . .) was the number of distinct primes that divided the placeholder. E.g., a(30)=3, since the primes that divide 30 are 2, 3 and 5. KWTsui solved it.