# Puzzler of 01/21/2012: What is X in the series 4, 6, 12, 18, 30, 42, 60, X, 102, ...?

Hint: The numbers following 102 are (ROLL OVER➔) [ 108, 138, … ].

Feel like I’m taking my SAT’s…

78

MikeInNH" Feel like I'm taking my SAT's...."

78

You just missed! Off by [ 6 ].

If 102 were 108, 78 would fit into the pattern. As it is, I cannot pick up on a pattern. I look forward to the explanation.

I guess 72 would work, but that would be a bizarre pattern.

mark9207"If 102 were 108, 78 would fit into the pattern. As it is, I cannot pick up on a pattern. I look forward to the explanation."
Slide cursor for another hint ➔ The numbers 4, 6, 12, 18, 30, 42, 60, 72, 102, 108, 138, ... are unique in the "immediate company they keep." This problem is more difficult than the average puzzler.

I didn’t realize that there were clues in the brackets that would show up if you highlighted them with the cursor (I have no idea how to do that, or any other formatting for that matter since the instructions on how to format these things got discarded with the old system). It appears that I did figure it out, and there does seem to be a pattern that makes some sense, mathematically. However, I don’t understand the “immediate company they keep” hint. I never was too great at math. Perhaps I will ask my pastor about this in the morning since he’s also the professor of mathematics at the local community college.

78 since pattern becomes add multiple of 6 increasing every two additions. 4 seems to be an outlier however. After 4 add 6 to 6, add 6 to 12 then add 12 to 18 then 12 to 30, add 18 to 42 then 18 to 60 to get 78. Finally 102 comes from adding 24 to 78.

mark9207"I didn't realize that there were clues in the brackets that would show up if you highlighted them with the cursor (I have no idea how to do that ..."
To print a white font on a white background, use ☟ this is a white font on a white background ☚ highlight to make visible

The initial post to every thread is embedded in a light gray or silver background.
To print a light silver font on a light silver background, use ☟
this is a silver font on a white background
The hex number #F0F0F0 represents three two-digit hex numbers #F0, #F0, #F0, each with values in the range 0-255, that gives the intensity, respectively, of the red, green, and blue pixels on the screen.

The word “white” (in quotes) may be substituted for the six-byte hex number #FFFFFF; the silver color used in the car talk forum is non-standard and hence has no officially sanctioned color name. A list of standard HTML colors and color names is at HTML Color Names

mark9207"I don't understand the "immediate company they keep" hint."
That was my poor attempt at being subtle.

WARNING! SOLUTION ALERT! Highlight to see give-away hint below ☟

Each of the numbers in the sequence 4, 6, 12, …, 72, 102, … is bounded above and below by a pair of numbers that share a particular mathematical quality. For example, 12 is bounded by 11 and 13. Also, 72 is bounded by 71 and 73. The numbers 11, 13, 71, and 73 all share a unique feature. On the other hand, 78 is bounded by 77 and 79, one of which does not have this quality. WARNING OVER!

This numerical sequence is rather difficult. The sequence is covered in number theory, but that is usually given as a graduate course in pure mathematics. I don’t see how the average person could come up with a solution, as THE NUMBERS ARE NOT A FUNCTION OF PREVIOUS NUMBERS IN THE SERIES.

Finally, I have one bone to pick with the tappet brothers. As purportedly graduates of MIT, they should know that their list of numbers form a sequence, not a series. A series is the summation of a sequence, as in a power series, a Fourier series, or a Taylor’s series and so on. Perhaps an MIT degree should come with an expiration date.

KenCD"78 since pattern becomes add multiple of 6 increasing every two additions. 4 seems to be an outlier, however. After 4 add 6 to 6, add 6 to 12 then add 12 to 18 then 12 to 30, add 18 to 42 then 18 to 60 to get 78. Finally 102 comes from adding 24 to 78."
No, the next number after 60 in the sequence 4, 6, 12, 18, 30, 42, 60, ... is 72, not 78.

As I mentioned in my post above, the numbers in the sequence are not formed from previous numbers in the sequence.

I gave the solution in my previous post as well, Highlight with your cursor the blank space following the warning if you want to read it…

A difficult puzzler by those perfidious old farts from MIT.

KenCD"4 seems to be an outlier, however."
There are no "outliers" in mathematics. In physics, maybe. In engineering, all the time.

I switched to Firefox from chrome…I was able to see the numbers hidden. I just thought they were missing.

Interesting puzzle…

So, are all the numbers in this sequence supposed to have prime numbers before and after them??? I feel kind of dumb for having to ask, but that is what I deduce from these hints and what not. If that is the case, then 4 does fit in, and you would pretty much have to look at a prime number chart to see the series. 78 also works as a solution, but for a different reason, and the pattern would have to jump around and make little sense, as with 72 as an answer with a sequential pattern. Mike mentioned something about SAT’s. I could see this being a question on one of those tests, of course with only one correct answer and only one correct explanation.

A couple of years ago, I stumbled across a nice little site called The On-Line Encyclopedia of Integer Sequences at http://oeis.org/Seis.html. If a sequence has turned up anywhere in the mathematical literature, you can find it here, along with its definition, additional terms, relationship to other sequences, and even an option to listen to the sequence as a musical tune (check out the Fibonacci sequence A000045).

If you feed it the numbers up through 60 in the puzzler, it gives you back four sequences, A014574 (Average of twin-prime pairs), A129297 (Nonnegative integers m such that m^2-1 has no divisors d with 1<d<m-1, which also has the numbers 0,1,2,3 at the beginning), A072570 (Even interprimes i = (p+q)/2 (where p, q are consecutive primes) such that (q-p)/2 is not divisible by 3, which is further described as a superset of A014574) and A167777 (Even single (or even isolated) numbers, which also includes the number 2 at the beginning).

All four of these, including the ones with extra terms at the beginning, continue with the same numbers for some time, so whichever one is being used the missing number in the Puzzler is the same.

(The sequence A133377 is one I came up with, but someone else actually submitted it.)

I think that the sequence is the number between twin primes:

3 and 5 are twin primes and the number between them is 4
5 and 7 are twin primes and the number between them is 6
11 and 13 are twin primes and the number between them is 12
17 and 19 are twin primes and the number between them is 18
29 and 31 are primes and the number between them is 30
41 and 43 are twin primes and the number between them is 42
59 and 61 are twin primes and the number between them is 60
_____ and _____are twin primes and the number between them is ______
101 and 103 are twin primes and the number between them is 102

This should make it easy to figure out what number is in the sequence between 60 and 102.
Prime numbers: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61, 67, , ,

Twin Primes

Twin primes are pairs of primes of the form (p, p+2). The term “twin prime” was coined by Paul Stäckel (1862-1919; Tietze 1965, p. 19). The first few twin primes are n+/-1 for n=4, 6, 12, 18, 30, 42, 60, 72, 102, 108, 138, 150, 180, 192, 198, 228, 240, 270, 282, … (Sloane’s A014574). Explicitly, these are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), … (Sloane’s A001359 and A006512).

All twin primes except (3, 5) are of the form 6n+/-1.

It is conjectured that there are an infinite number of twin primes (this is one form of the twin prime conjecture), but proving this remains one of the most elusive open problems in number theory. An important result for twin primes is Brun’s theorem, which states that the number obtained by adding the reciprocals of the odd twin primes,
B=(1/3+1/5)+(1/5+1/7)+(1/(11)+1/(13))+(1/(17)+1/(19))+…,
(1)

converges to a definite number (“Brun’s constant”), which expresses the scarcity of twin primes, even if there are infinitely many of them (Ribenboim 1996, p. 201). By contrast, the series of all prime reciprocals diverges to infinity, as follows from the Mertens second theorem by letting x->infty.

The following table gives the first few p for the twin primes (p, p+2), cousin primes (p, p+4), sexy primes (p, p+6), etc.
pair Sloane first member
(p, p+2) A001359 3, 5, 11, 17, 29, 41, 59, 71, …
(p, p+4) A023200 3, 7, 13, 19, 37, 43, 67, 79, …
(p, p+6) A023201 5, 7, 11, 13, 17, 23, 31, 37, …
(p, p+8) A023202 3, 5, 11, 23, 29, 53, 59, 71, …
(p, p+10) A023203 3, 7, 13, 19, 31, 37, 43, 61, …
(p, p+12) A046133 5, 7, 11, 17, 19, 29, 31, 41, …

Let pi_2(n) be the number of twin primes p and p+2 such that p<=n. It is not known if there are an infinite number of such primes (Wells 1986, p. 41; Shanks 1993), but it seems almost certain to be true (Hardy and Wright 1979, p. 5).

J. R. Chen has shown there exists an infinite number of primes p such that p+2 has at most two factors (Le Lionnais 1979, p. 49). Brun proved that there exists a computable integer x_0 such that if x>=x_0, then
pi_2(x)<(100x)/((lnx)^2)
(2)

(Ribenboim 1996, p. 261). It has been shown that
pi_2(x)<=cproduct_(p>2)[1-1/((p-1)^2)]x/((lnx)^2)[1+O((lnlnx)/(lnx))],
(3)

written more concisely as
pi_2(x)<=cPi_2x/((lnx)^2)[1+O((lnlnx)/(lnx))],
(4)

where Pi_2 is known as the twin primes constant and c is another constant. The constant c has been reduced to 68/9 approx 7.5556 (Fouvry and Iwaniec 1983), 128/17 approx 7.5294 (Fouvry 1984), 7 (Bombieri et al. 1986), 6.9075 (Fouvry and Grupp 1986), 6.8354 (Wu 1990), and 6.8325 (Haugland 1999). The latter calculation involved evaluation of 7-fold integrals and fitting of three different parameters.

Hardy and Littlewood (1923) conjectured that c=2 (Ribenboim 1996, p. 262), and that pi_2(x) is asymptotically equal to
pi_2(x)∼2Pi_2int_2^x(dx)/((lnx)^2).
(5)

This result is sometimes called the strong twin prime conjecture and is a special case of the k-tuple conjecture. A necessary (but not sufficient) condition for the twin prime conjecture to hold is that the prime gaps constant, defined by
Delta=limsup_(n->infty)(p_(n+1)-p_n)/(p_n),
(6)

where p_n is the nth prime and d_n=p_(n+1)-p_n is the prime difference function, satisfies Delta=0.

Wolf notes that the formula
pi_2(x)∼2Pi_2([pi(x)]^2)/x,
(7)

(which has asymptotic growth ∼Pi_2x/(lnx)^2) agrees with numerical data much better than does Pi_2x/(lnx)^2, although not as well as Pi_2Li_2(x).

Extending the search done by Brent in 1974 or 1975, Wolf has searched for the analog of the Skewes number for twins, i.e., an x such that pi_2(x)-Pi_2Li_2(x) changes sign. Wolf checked numbers up to 2^(42) and found more than 90000 sign changes. From this data, Wolf conjectured that the number of sign changes nu(n) for x<n of pi_2(x)-Pi_2Li_2(x) is given by
nu(n)∼(sqrt(n))/(lnn).
(8)

Proof of this conjecture would also imply the existence an infinite number of twin primes.

The largest known twin primes as of Jan. 2012 are
q=3756801695685·2^(666669)-1
(9)

and q+2, each having 200700 decimal digits, and were found by PrimeGrid on Dec. 25, 2011 (http://primes.utm.edu/top20/page.php?id=1#records).

In 1995, Nicely discovered a flaw in the Intel® PentiumTM microprocessor by computing the reciprocals of 824633702441 and 824633702443, which should have been accurate to 19 decimal places but were incorrect from the tenth decimal place on (Cipra 1995, 1996; Nicely 1996).

If n>=2, the integers n and n+2 form a pair of twin primes iff
4[(n-1)!+1]+n=0 (mod n(n+2)).
(10)

n=pp^’ where (p,p^’) is a pair of twin primes iff
phi(n)sigma(n)=(n-3)(n+1)
(11)

(Ribenboim 1996, p. 259). S. M. Ruiz has found the unexpected result that (n,n+2) are twin primes iff
sum_(i=1)^ni^a(|(n+2)/i|+|n/i|)=2+n^a+sum_(i=1)^ni^a(|(n+1)/i|+|(n-1)/i|)
(12)

for a>=0, where |x| is the floor function.

The values of pi_2(n) were found by Brent (1976) up to n=10^(11). T. Nicely calculated them up to 10^(14) in his calculation of Brun’s constant. Fry et al. (2001) and Sebah (2002) independently obtained pi_2(10^(16)) using distributed computation. The following table gives known values of pi_2(10^n) (Sloane’s A007508; Ribenboim 1996, p. 263; Nicely 1999; Sebah 2002).
n pi_2(n)
10^3 35
10^4 205
10^5 1224
10^6 8169
10^7 58980
10^8 440312
10^9 3424506
10^(10) 27412679
10^(11) 224376048
10^(12) 1870585220
10^(13) 15834664872
10^(14) 135780321665
10^(15) 1177209242304
10^(16) 10304195697298

It is conjectured that every even number is a sum of a pair of twin primes except a finite number of

After this long explanation of TwinPrimes the answer should be 72.
Can’t wait to hear the explanation from Tom next week.

Thanks,
Ted
exceptions whose first few terms are 2, 4, 94, 96, 98, 400, 402, 404, 514, 516, 518, … (Sloane’s A007534; Wells 1986, p. 132).

mark9207'"I could see this being a question on one of those tests, of course with only one correct answer and only one correct explanation."
If I remember correctly, my BlueBook exam at the University of Chicago, Spring 1975, Mathematics Dept. asked the questions:
• why is each number a product of 6?
• what is it a product of in a hexidecimal system?
All the entries after yours were copied from Google searches and the copiers don't know crap about what they are posting. They are just a bunch of math wannabes.

Disgusting. It’s the last time I submit an answer to a puzzler. I’ll let the pseudo math-gurus answer first.

“All the entries after yours were copied from Google searches and the copiers don’t know crap about what they are posting”.
Mechaniker–I didn’t do any Google searching. After trying and failing to find a formula to generate the values in the sequence, I just happened to think of the integer values on each side of the integer in the sequence: 3,4,5 5,6,7 11,12,13 17,18,19 29,30,31 41,42,43 59,60,61 etc,
Each number of the sequence 4,6,12,18,30, 42, 60, has a prime number that is one less than the value in the sequence and a prime number that is one greater than the value in the sequence. Any prime integer except 2 must be odd since even numbers can be divided by 2. I recalled a discussion some years ago in a college mathematics class about twin primes (I’m 70 years old, so it has been a while back) which I found fascinating. I know that number theory covers topics such as twin primes, but I elected to take numerical analysis as opposed to theory of numbers because my schedule would’t allow both courses.

@Mechaniker

why is each number a product of 6?

The middle number is always even (between two odd numbers), so it’s divisible by 2. It’s also always divisible by 3, since for any 3 consecutive numbers, one must be divisible by three. The two primes are not divisible by 3, so the middle one must be. So the middle number is always divisible by 6. The only exception is 4, since the lower prime (3) is divisible by three.

what is it a product of in a hexadecimal system?

trick question. 6.

"All the entries after yours were copied from Google searches and the copiers don’t know crap about what they are posting. They are just a bunch of math wannabes.

Disgusting. It’s the last time I submit an answer to a puzzler. I’ll let the pseudo math-gurus answer first."

Given your tone, perhaps that’s for the best.

After reading the solution (not here, but where Cartalk gives their official solution) I filed this under “I wouldn’t have gotten this in a million years”.

Not only does this require a bizarre mathematical insight the average person wouldn’t get, they explain it in a way that doesn’t make it obvious at all how you would go about finding the solution.

I came up with 78, not 72.

Now I wonder how many correct answers to this they would have gotten. They never tally that up for us.