magicdragon2 ([info]magicdragon2) wrote,
@ 2004-08-12 14:25:00
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Tomorrow is Friday the 13th...
Tomorrow is Friday the 13th... But I don't believe in Bad Luck. (knock wood!) As it says on How Stuff Works:
"Like many human beliefs, the fear of Friday the 13th (known as paraskevidekatriaphobia) isn't exactly grounded in scientific logic. But the really strange thing is that most of the people who believe the day is unlucky offer no explanation at all, logical or illogical. As with most superstitions, people fear Friday the 13th for its own sake, without any need for background information...."

I've had some good things and bad things happen since my last posting on LiveJournal.

Bad: I was offline for a couple of weeks due to an allegedly offensive anonymous posting. My wife and I guessed (early on) the twist ending of "The Village." My car was illegally towed from a Ralph's Grocery parking lot where I had the Assistant Manager's permission to park -- he forgot to inform his boss, or Security. Cost me $760 in towing and impound fees; plus $188 to repair damage from the towing. My wife and son had bad chest colds. I'm dogsitting a canine who is trying to break the Guiness Book of Records for loudest bark.

Good: I've continued to be incredibly productive in mathematical insights. As recounted in greater detail on Math Pages of Jonathan Vos Post:

Inventory Status Summary:
35 Math papers written in past year;
9 presented and accepted in proceedings of International Conferences;
4 still in editorial hands at Mathematics Magazine;
3 rejected by Mathematics Magazine, being resubmitted elsewhere;
1 still in editorial hands at Fibonacci Quarterly;
1 rejected by Fibonacci Quarterly, being resubmitted elsewhere;
2 rejected by American Mathematical Monthly, being resubmitted elsewhere;
14 still being completed/polished for first submission.

My latest theorem:

"Primes, Semiprimes, and Factorizations of
N-dimensional Centered Tetrahedral and Centered Cube
Numbers"

ABSTRACT:

The following theorem is introduced: for nonnegative
integers X and positive integer N, the equation
NdCC(X) = X^N + (X+1)^N, which gives the
N-dimensional Centered Cube Numbers, has NO prime nor
semiprime values iff the binary representation of N
does NOT contain as substrings the binary
representations of ALL factors of N.

This paper consists of a table of prime factorizations
of N-dimensional Centered Cube Numbers for N from 1 to
12, which have the form X^N + X^(n+1); and N-dimensional Centered Tetrahedral Numbers for N from 4 to 10.

This table has purposes including a search for primes
and, due to related papers by this author [Post 2004
a,b,c...], a search for semiprimes. In some cases,
relating the the key theorem, we gave a polynomial
irreducibility argument for why there can be no primes
of a given form. A table of polynomial factorization
is an appendix. The paper's results are summarized in
an Extended Abstract....

(20 comments) - (Post a new comment)

How common are Friday the 13ths?
[info]magicdragon2
2004-08-12 03:07 pm UTC (link)
How common are Friday the 13ths?
[From the Dr. Math archives]:

You can use a perpetual calendar to work out how many Friday the 13ths there are for every possible pattern of weekdays in a year. It turns out that every year has at least one Friday the 13th. The largest possible number of Friday the 13ths in one year is three. Surprisingly, it turns out that a 13th is slightly more likely to fall on a Friday than any other day. This happens because the leap year rules make some patterns of weekdays happen more often than others. The winning patterns include many Friday the 13ths. Here's a [slightly restructured] table from Eric Weisstein's World of Astronomy that shows how often the 13th falls on each day of the week, over four hundred years.

Number of 13ths Percent Day
--------------- ------- ---
687 14.31% Sunday
685 14.27% Monday
685 14.27% Tuesday
687 14.31% Wednesday
684 14.25% Thursday
688 14.33% Friday
684 14.25% Saturday

(Reply to this) (Thread)
Re: How common are Friday the 13ths?
(Anonymous)
2004-08-12 05:20 pm UTC (link)
I see that N-dimensional Centered Cube Numbers and N-dimensional Centered Tetrahedron Numbers are listed in Sloane's Online Encyclopedia of Integer Sequences.

centered cube numbers, higher-dimensional (1): A008514, A008515, A008516, A036085, A036086, A036087, A036088 A036089, A036090, A036091

centered cube numbers, higher-dimensional (2): A036092, A036093, A036094, A036095, A036096, A036097, A036098, A036099, A036100, A036101,

A036102
centered cube numbers: A005898*

A005894 Centered tetrahedral numbers.
A008498 4-dimensional centered tetrahedral numbers.
A008499 Number of 5-dimensional centered tetrahedral numbers.
A008500 6-dimensional centered tetrahedral numbers.
A008501 7-dimensional centered tetrahedral numbers.
A008502 8-dimensional centered tetrahedral numbers.
A008503 9-dimensional centered tetrahedral numbers.
A008504 10-dimensional centered tetrahedral numbers.
A008505 11-dimensional centered tetrahedral numbers.
A008506 12-dimensional centered tetrahedral numbers.

But why would anyone want to factor them into primes, let alone care if they are semiprimes?

-- Math Skeptic

(Reply to this) (Parent)(Thread)
Re: How common are Friday the 13ths?
[info]magicdragon2
2004-08-12 06:08 pm UTC (link)
This is actually a comment on the parent of this thread, not on the Friday the 13ths comment, but be that as it may...

Factoring numbers into primes is a practice that goes back to prehistory. It's roughly 2000 years since Euclid proved that there are an infinite number of primes. Primes are, after the integers themselves, the most important numbers in the branch of Mathematics called Number Theory.

The use of computers has changed the approach to factoring (or factorizing). D. H. Lehmer used mechanical computers for factoring, and this is enormously faster using techniques such as Elliptic Curve Factorization.

Semiprimes are less-studied than primes... until recently. Semiprimes with large numbers of digits are the pairs of keys in the Rivest Shamir Adelman (RSA) Public Key Cryptosystems, which are very important in internet security.

This is one a number of related papers I've written recently which focus on unusual properties of semiprimes, or on finding semiprimes in unusual places.

(Reply to this) (Parent)
Semirpimes & Semiprimes
[info]magicdragon2
2004-08-12 07:40 pm UTC (link)
"Semirpimes"

Jonathan Vos Post
12 August 2004, 7 p.m.
Copyright (c) 2004 by Magic Dragon Multimedia

A Semiprime, to review, is a whole number which is the
product of exactly two (not necessarily distinct)
prime numbers.

Define "emirpimes" as a semiprime whose reverse (base
10) is a different semiprime. The first is 15,
because: 15 reversed is 51, and both are semiprimes
(15 = 3 x 5), (51 = 3 x 17).

Emirpimes begin: 15, 26, 39, 49, 51, 62, 93, 94, 115,
122, 123, 129, 143, 155, 158, 159, 169, 177, 178, 183,
185, 187, 203, 205, 221, 265, 289, 302, 314, 319, 321,
326, 327, 329, 335, 339, 341, 355, 381, 394, 398, 413,
415, 437, 493, 497, 502, 511, 514, 533, 538, 551, 553,
559, 581, 586, 589, 623, 629, 667, 685, 718, 723, 734,
771, 781, 794, 817, 835, 851, 871, 893, 899, 913, 921,
923, 926, 933, 951, 955, 961, 982, 985, 998, 1006,
1011, 1027, 1041, 1043, 1046, 1047, 1057, 1059, 1067,
1079, 1101, 1115, 1119, 1121, 1135, 1137, 1141, 1145,
1149, 1154, 1165, 1169, 1174, 1177, 1195, 1202,...

Note that Emirpimes ends with an "s" but is singular.
The plural is Semirpimes. Oddly, Semirpimes is an
anagram of Semiprimes (as one need only switch the
central pr / rp).

Although it's trivial that there are
an infinite number of Semiprimes: here's the
First Puzzle: prove that there are an infinite number of
Semirpimes.

Second Puzzle: For N digit positive integers, there
are obviously an even number of Semirpimes. For N>1,
is that even number ever 0?

Third Puzzle:

There's a nice palindrome from the semirpimes 169 and
961:

13 x 13 = 169 sides reversed is 961 = 31 x 31

Are there any more of this form for N>3 digits?

(Reply to this) (Thread)
Re: Semirpimes & Semiprimes
[info]kappa_l
2004-08-13 01:22 am UTC (link)
I don't see that proving the infiniteness of the set of all semirpimes is trivial. Euklid's classic doesn't seem to make sense while working semirpimes. I'm not a math prof, though.
How much number theory do I have to solve the puzzles?

(Reply to this) (Parent)(Thread)
Re: Semirpimes & Semiprimes
[info]magicdragon2
2004-08-13 08:54 am UTC (link)
I don't think that it is trivial to prove that there are an infinite number of semirpimes. How much Number Theory is needed? We'll see. I haven't solved that puzzle myself. Either I will, soon, and show the answer on the blog; or someone else will solve it, and I'll show their solution, with acknowledgment; or we won't know anytime soon. Let me give some more information, in the next posting.

(Reply to this) (Parent)(Thread)
Re: Semirpimes & Semiprimes
[info]kappa_l
2004-08-13 09:01 am UTC (link)
Oh, I thought you already solved all three of them and posted them as a challenge for the readers.
Mostly sets of numbers that have some 'primality' feature are of infinite cardinality, I experienced. But since there is no solution to question no.1 should the question read "prove or disprove?".

Nevertheless, I doubt there will be a largest emirpime.

(Reply to this) (Parent)(Thread)
Re: Semirpimes & Semiprimes
[info]magicdragon2
2004-08-13 09:32 am UTC (link)
I'm "thinking out loud" here, in posting incomplete results in real time.

As to your statement: "Mostly sets of numbers that have some 'primality' feature are of infinite cardinality" -- would that this were true!

It has not yet been proved that there are an infinite number of twin primes. Twin Primes are a pair of primes P and P+2 with a single even number between them, such as 3 and 5, 11 and 13, 101 and 103. Except for 3 and 5, the twin primes are always of the form 6n-1 and 6n+1.

It has not yet been proved (as of 1998) that there are an infinite number of Sophie Germain primes. Those are primes P such that the "companion" prime 2p+1 is also prime. We know examples up to 34547 digits in length.

We typically develop an analytic formula for how many members of a set of integers there are in some given range, such as from 1 through a maximum N.

For instance, for primes, Gauss made a first cut at the Prime Number Theorem in 1792, when only 15 years old. He suggested that the number of primes less than or equal to N converged to n / ln n, the ratio of n to the natural logarithm of n. He then refined the estimate to Li(n), the logarithmic integral. See the above-linked Mathworld page for details, and Prime Counting Function.

Also, we typically investigate infinite series based on the set in question. For example, as we see in Prime Sums:

"In 1737, Euler showed that the harmonic series of primes, (i.e., sum of the reciprocals of the primes) diverges."

However, the sum of the reciprocals of the squares of primes converges. The sum of the reciprocals of the twin primes converges. And, in a forthcoming paper by myself and Dr. Geoffrey Landis, we show that sum of the reciprocals of the semiprimes converges, and we give an exact analytical expression.

But we don't yet know the equivalents for semirpimes!

(Reply to this) (Parent)(Thread)
Re: Semirpimes & Semiprimes
[info]kappa_l
2004-08-13 10:01 am UTC (link)
Well, I do not know very much about number theory. I was aware of Twin Primes Problem, though.

Maybe it's possible to solve 1 by making a general statement for 3.
Perhaps one could formulate a equation system by the Cauchy product of the decimal series of the involved primes.
I'm stuck there, however, and so I'll stop falling on your nerves. Thanks for your answer, though.

(Reply to this) (Parent)(Thread)
Re: Semirpimes & Semiprimes
[info]magicdragon2
2004-08-13 10:43 am UTC (link)
Excellent suggestion. Well done!

Cauchy, or someone in his style, is dealing with the problem of the emirp. As Eric W. Weisstein puts it, in his MathWorld page on Emirp:

"An emirp ('prime' spelled backwards) is a prime whose reversal is also prime, but which is not a palindromic prime. The first few are 13, 17, 31, 37, 71, 73, 79, 97, 107, 113, 149, 157, ... (Sloane's A006567)."

The formula gives the digits of the product of an emirp with the companion reversed prime, in terms of the digits of the two emirps.

Eric W. Weisstein also (personal correspondence) made some little corrections to what I've posted so far on semirpimes, and gives a nice short mathematica program for calculating them.

(Reply to this) (Parent)

(Reply from suspended user)

[info]carriewamix
2008-07-17 01:14 am UTC (link)
I'm not sure though. I'm still going over the book and thinking as I go. Centaurs are the seventh creature listed in the book and come in several colours.

(Reply to this) (Parent)
More About Semirpimes
[info]magicdragon2
2004-08-13 09:01 am UTC (link)
Here's a list of the semiprimes under 2000:

Semirpimes begin: 15, 26, 39, 49, 51, 62, 93, 94, 115, 122, 123, 129, 143, 155, 158, 159, 169, 177, 178, 183, 185, 187, 203, 205, 221, 265, 289, 302, 314, 319, 321, 326, 327, 329, 335, 339, 341, 355, 381, 394, 398, 413, 415, 437, 493, 497, 502, 511, 514, 533, 538, 551, 553, 559, 581, 586, 589, 623, 629, 667, 685, 718, 723, 734, 771, 781, 794, 817, 835, 851, 871, 893, 899, 913, 921, 923, 926, 933, 951, 955, 961, 982, 985, 998, 1006, 1011, 1027, 1041, 1043, 1046, 1047, 1057, 1059, 1067, 1079, 1101, 1115, 1119, 1121, 1135, 1137, 1141, 1145, 1149, 1154, 1165, 1169, 1174, 1177, 1195, 1202, 1211, 1214, 1219, 1234, 1238, 1243, 1247, 1253, 1273, 1286, 1293, 1299, 1306, 1313, 1315, 1317, 1318, 1322, 1343, 1346, 1347, 1354, 1357, 1366, 1371, 1379, 1382, 1383, 1387, 1401, 1402, 1405, 1411, 1415, 1417, 1418, 1437, 1438, 1454, 1461, 1466, 1469, 1497, 1502, 1507, 1513, 1514, 1527, 1538, 1561, 1563, 1569, 1577, 1623, 1641, 1642, 1651, 1655, 1658, 1671, 1679, 1687, 1691, 1703, 1706, 1707, 1714, 1717, 1727, 1731, 1735, 1754, 1757, 1761, 1765, 1766, 1769, 1774, 1779, 1795, 1797,1807, 1814, 1817, 1843, 1849, 1874, 1882, 1891, 1893, 1894, 1897, 1903, 1915, 1923, 1927, 1937, 1942, 1961, 1969, 1981, 1983, 1985,...

How are the semirpimes distributed? Sometimes they form “clumps.” For instance, 1043, 1046, 1047 is three semirpimes in one decade. Here are 7 semirpimes which are consecutive in the list of semiprimes: 1299, 1306, 1313, 1315, 1317, 1318, 1322. Again, here are 7 semirpimes which are consecutive in the list of semiprimes: 1405, 1411, 1415, 1417, 1418, 1437, 1438. It is clear that there cannot be four consecutive integers which are semiprimes, let alone semirpimes. That's because one of those four integers must be a multiple of 4, which means that it has factors 2, 2 again, and X for some X>1, hence having at least three factors means that it is not, by definition, a semiprime.

Yet sometimes there are gaps. From semirpime 62 to semirpime 93 is a difference of 31, yet the next integer, 94, is also semirpime. From semirpime 1577 to semirpime 1623 is a difference of 46. One expects to see, on the average, longer and longer such gaps.

-- ----------------------------------------------
N Number of Semirpimes with N Digits
-- ----------------------------------------------
0 0
1 0
2 8 of 90 = 8.8888%
3 76 of 900 = 8.4444%
4 over 1000
-- ----------------------------------------------

Let's look at the distribution "century by century" in the range 1000 to 2000, where semiprimes are 27.8% of all integers, semirpimes are 13% of all integers, and semirpimes are 46.76% of the semiprimes..

------- -------------- --------------
Range Semiprimes Semirpimes
------- -------------- --------------
1000-1100 23 11
1100-1200 31 15
1200-1300 29 13
1300-1400 32 17
1400-1500 22 14
1500-1600 26 10
1600-1700 26 10
1700-1800 30 18
1800-1900 28 11
1100-1200 31 11
------- -------------- --------------
1000-2000 278 130
------- -------------- --------------


One can pretend that all is chance, although the semiprimes and semipimes are completely predictable. What is the chance that an integer N is a semiprime, based on the size of N? What is the chance that a semprime N, when reversed, is another semiprimes, thus making N a emirpime?

Hugo Pfoertner has established that there are 1,237,088,048,653 semiprimes under 10 trillion, which means that, up to 13 digits, semiprimes are roughly 1/8 of all integers. Is it true, then, that the chance of an N up to 10 trillion has 1/8 x 1/8 = 1/64 chance of being a semipime?

Or does that probability shrink towards zero, making it increasingly unlikely for larger numbers to be semirpimes? We know in great detail what the answers are to the corresponding questions about primes. Primes matter. But semiprimes are almost as interesting as primes, and semirpimes are a quirky subset of semiprimes. Let's keep working on this...

(Reply to this)

Cauchy Product formula and emirps
[info]magicdragon2
2004-08-13 12:58 pm UTC (link)
The Cauchy Product formula gives the digits of the product of an emirp with the companion reversed prime, in terms of the digits of the two emirps. Adapting the cited page to text:

Let P(1) and P(2) be prime numbers with representation in decimal digits:

P(1) = SUM[from I = 0 to s] (10^I) a(I)
P(2) = SUM[from I = 0 to r] (10^I) b(I)

Taking the Cauchy product, we obtain:

P(1)^2 = SUM[from I = 0 to 2s] SUM[for c+d = I] (10^I) a(c) a(d)
P(2)^2 = SUM[from I = 0 to 2r] SUM[for c+d = I] (10^I) b(c) b(d)

With a(k),b(k) = 0 if k>s or k>r respectively. We have to show that there exist a(I), b(I), such that P(1)^2 = Mirror[P(2)^2] where Mirror is the obvious mirror-reversed sequence of digits.

Let's look at the Jth digit of Mirror[P(2)^2].For convenience, let J be even [of course, we deduce that r = s in the above representations of P(1), P(2)].

SUM [for c + d = 2r – J] b(c) b(d) = SUM [for c + d = J] a(c) a(d)

Hence:

b(0) b(2r-J) + ... + b(r – J/2) b(r – J/2) = a(0) a(J) + ... + a(J/2) a(J/2)

(Reply to this)

Dr. Eric Weisstein calculates deeper into semirpimes
[info]magicdragon2
2004-08-13 01:08 pm UTC (link)
The remarkable mathematician/astronomer/encyclopedist Dr. Eric W. Weisstein has made some corrections to my data, and extended the count of semirpimes with a given number of digits.

The proper count, thanks to Eric, is:

0 semirpimes with 0 digit;
0 semirpimes with 1 digit;
10 semirpimes with 2 digits;
80 semirpimes with 3 digits;
898 semirpimes with 4 digits;
7200 semirpimes with 5 digits;
60732 semirpimes with 6 digits;
506724 semirpimes with 7 digits;
...

I asked: "to what does count/N converge?"

Eric answers:

"We need a few more terms to be able to say. There are probably more
clever ways than brute force, but brute force usually works reasonably
well :) "

(Reply to this)

Correction: proper running count of semipirmes
[info]magicdragon2
2004-08-13 01:49 pm UTC (link)
Through no fault of Eric Weisstein's, I incorrectly tabulated in the above posting. It should read:

0 semirpimes with 0 digit;
0 semirpimes with 1 digit;
10 semirpimes with 2 digits;
90 semirpimes with 3 or fewer digits;
898 semirpimes with 4 or fewer digits;
7200 semirpimes with 5 or fewer digits;
60732 semirpimes with 6 or fewer digits;
506724 semirpimes with 7 or fewer digits.

According to Patrick De Geest, as posted on Sloane's A066265
N Number of semiprimes < 10^n
--- ---------------------------
0 0
1 3
2 34
3 299
4 2625
5 23378
6 210035
7 1904324
8 17427258
9 160788536
10 1493776443
11 13959990342
12 131126017178
13 1237088048653

Combining these tables, we get the following:

N Percentage of N-digit semiprimes which are semiprimes
-- -----------------------------------------------------
0 0/0
1 0% = 0/3
2 29.41% = 10/34
3 30.10% = 90/299
4 34.21% = 898/2625
5 30.798% = 7200/23378
6 28.9% = 60732/210035
7 26.6% = 506724/1904324

It is not clear to me why that percentage rises, peaks at 4 digits, and then gradually diminishes. It is not clear where that percentage is going, or the corresponding percentage of integers up to N which are semiprimes.

(Reply to this)

Andrew Numbers: 2nd-order Semirpimes
[info]magicdragon2
2004-08-14 03:28 pm UTC (link)
Here's another twist, suggested explicitly by my 15-year-old son Andrew Carmichael Post (who has co-authored Math and Physics papers before): we mirror-reverse the prime factors of semiprimes or semirpimes, and sometimes get a semiprime again. Perhaps we should call these "Andrew Numbers."

Semiprimes whose product of mirror-reversed factors is again a Semiprime

1837 = 11 x 167; 761 x 11 = 8371 which is again a Semiprime
[1837 is a semiprime, not an emirpime] [ this a sort of 2nd degree semirpime]

Semirpime whose product of mirror-reversed factors is again a Semiprime

1795 = 5 x 359; 953 x 5 = 4765 which is again a Semiprime

1814 = 2 x 907; 709 x 2 = 1418 which is again a Semiprime

1874 = 2 x 937; 739 x 2 = 1478 which is again a Semiprime

1882 = 2 x 941; 149 x 2 = 298 which is again a Semiprime

1915 = 5 x 383 ; which is a product of two palindromes hence, when the factors are mirror reversed, they multiply to be the original 1915 [a fixed point of the process]

1937 = 13 x 149; 941 x 31 = 29171 which is again a Semiprime

1942 = 2 x 971; 179 x 2 = 358 which is again a Semiprime

1969 = 11 x 179; 791 x 11 = 10681 which is again a Semiprime

The process referred to for 1915 is as follows:

Input integer N. Factor N into primes. Mirror-reverse each prime, and multiply them all together. This is output, M. Repeat process, starting with M.

Conjecture: this process always reaches a fixed point such as 1915, where all prime factors are palindromes.

4th Puzzle: What is the first 3rd order Andrew Number?

(Reply to this)

More Andrew Numbers
[info]magicdragon2
2004-08-14 04:14 pm UTC (link)
143 = 11 x 13; 31 x 11 = 341 which is again a Semiprime: AND 341 is 143 backwards!

155 = 31 x 5; 5 x 13 = 65 which is again a Semiprime

158 = 2 x 79; 97 x 2 = 194 which is again a Semiprime

169 = 13 x 13; 31 x 31 = 961 which is again a Semiprime: AND 961 is 169 backwards! I pointed this out before and asked for more such examples.

185 = 5 x 37; 73 x 5 = 365 which is again a Semiprime

187 = 11 x 17; 71 x 11 = 781 AND 781 is 187 backwards!

221 = 13 x 17; 71 x 31 = 2201 which is again a Semiprime

289 = 17 x 17; 71 x 71 = 5041 which is again a Semiprime; AND 1405 is again a semirpime; 1405 = 5 x 281 . That is, we have found the first semirpime which, upon applying the process, yields not just a semiprime but another semirpime.

302 = 2 x 151 and, because these 2 prime factors are both palindromes, we have a fixed point of the process.

314 = 2 x 157; 751 x 2 = 1502 which is again a Semiprime. AND, reversing 1502, we get 2051 = 7 x 293. That is, we have found the second semirpime which, upon applying the process, yields not just a semiprime but another semirpime.

(Reply to this)

Even More Andrew Numbers
[info]magicdragon2
2004-08-14 06:34 pm UTC (link)
718 = 2 x 359, 953 x 2 = 1906 which is again a Semiprime, AND the reverse of 1906 is 6091, a Prime

781 = 11 x 71 sides reversed is 17 x 11 = 187

835 = 5 x 167, 761 x 5 = 3805 which is again a Semiprime

923 = 13 x 71, 17 x 31 = 527 which is again a Semiprime

933 = 3 x 311 sides reversed is 113 x 3 = 339

955 = 5 x 191, a palindromic-factored fixed point

961 = 31 x 31 sides reversed is 13 x 13 = 169

1011 = 3 x 337, 733 x 3 = 2199 which is again a Semiprime

1027 = 13 x 79, 97 x 31 = 3007, which is again a Semiprime
AND 3007 reversed is 7003= 47 x 149

1041 = 3 x 347, 743 x 3 = 2229 which is again a Semiprime

1043 = 7 x 149, 941 x 7 = 6587

note that this makes 1041 and 1043 "Twin Andrew Numbers"

1057 = 7 x 151, palindromic-factor fixed point

1059 = 3 x 353, palindromic-factor fixed point

note that this makes 1057 and 1059 "Twin Andrew palindromic-factor fixed point Numbers"

1067 = 11 x 97, 79 x 11 = 869

1119 = 3 x 373, palindromic-factor fixed point

1149 = 3 x 383, palindromic-factor fixed point

1169 = 7 x 167, 761 x 7 = 5327, which is again a Semiprime
AND 5327 reversed is 7235 = 5 x 1447 which is again a Semiprime
AND 1447 reversed is 7441 = 7 x 1063 which is again a Semiprime
AND 1063 reversed is 3601 = 13 x 277 which is again a Semiprime

1177 = 11 x 107 sides reversed is 701 x 11 = 7711

1243 = 11 x 113 sides reversed is 311 x 11 = 3421

1253 = 7 x 179, 971 x 7 = 6797 which is again a Semiprime

1313 = 13 x 101, palindromic-factor fixed point

1343 = 17 x 79, 97 x 71 = 3421 = 6887 which is again a Semiprime
AND 6887 reversed is 7886 = 2 x 3943 which is again a Semiprime




(Reply to this)

10/13/2006 is double-13 Friday
[info]magicdragon2
2006-10-14 07:42 pm UTC (link)
For the fearful, this Friday has their number
By Jennifer Harper
THE WASHINGTON TIMES
October 13, 2006

This is not a good day for paraskevidekatriaphobics -- those who fear Friday the 13th. It's double-13 Friday. All the numbers in the numerical notation -- 10/13/2006 -- add up to 13 as well, giving great pause to the superstitious.

The phenomenon hasn't happened in 476 years, said Heinrich Hemme, a physicist at Germany's University of Aachen who crunched the numbers to find that the double-whammy last occurred Jan. 13, 1520.

"Pure chance," the good professor told the press yesterday.

But it's not exactly TGIF for the 21 million Americans who fear the day. Some may not travel or even get out of bed, said Donald Dossey, a North Carolina psychologist who coined the term "paraskevidekatriaphobia" 20 years ago. He estimates that the nation is out $900 million in lost productivity because of Friday the 13th sick-outs...

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