Roo’s Who

Friday, 21st Jun, 2013 by Krilling for Company in Binary numbers, Digit-sums, Mathematics, Prime numbers and tagged addition, Arithmetic, boxing prime, boxing primes, digit arrangement, digit arrangements, digit sum, digit sums, digit-multiplication, kangaroo prime, kangaroo primes, math, mathematics, maths, multiplication, prime, prime number, prime numbers, prime sequence, prime sequences, primes, recreational math, recreational mathematics, recreational maths, subtraction | Leave a comment

11 is a prime number, divisible by only itself and 1. If you add its digits, 1 + 1, you get 2. 11 + 2 = 13, another prime number. And 13 + (1 + 3) = 17, a third prime number. And there it ends, because 17 + (1 + 7) = 25 = 5 x 5. I call (11, 13, 17) kangaroo primes, because one jumps to another. In base 10, the record for numbers below 1,000,000 is this:

6 primes: 516493 + 28 = 516521 + 20 = 516541 + 22 = 516563 + 26 = 516589 + 34 = 516623.

In base 16, the record is this:

8 primes: 97397 = 17,C75_[b=16] + 32 = 97429 = 17,C95_[b=16] + 34 = 97463 = 17,CB7_[b=16] + 38 = 97501 = 17,CDD_[b=16] + 46 = 97547 = 17,D0B_[b=16] + 32 = 97579 = 17,D2B_[b=16] + 34 = 97613 = 17,D4D_[b=16] + 38 = 97651 = 17,D73_[b=16].

Another kind of kangaroo prime is found not by adding the sum of digits, but by adding their product, i.e., the result of multiplying the digits (except 0). 23 + (2 x 3) = 29. 29 + (2 x 9) = 47. But 47 + (4 x 7) = 75 = 3 x 5 x 5. So (23, 29, 47) are kangaroo primes too. In base 10, the record for numbers below 1,000,000 is this:

9 primes: 524219 + 720 = 524939 + 9720 = 534659 + 16200 = 550859 + 9000 = 559859 + 81000 = 640859 + 8640 = 649499 + 69984 = 719483 + 6048 = 725531.

But what about subtraction? For a reason I don’t understand, subtracting the digit-sum doesn’t seem to create any kangaroo-primes in base 10. But 11 in base 8 is 13 = 1 x 8^1 + 3 x 8^0 and 13_[b=8] – (1 + 3) = 7. In base 2, this sequence appears:

1619 = 11,001,010,011_[b=2] – 6 = 1613 = 11,001,001,101_[b=2] – 6 = 1607 = 11,001,000,111_[b=2] – 6 = 1601 = 11,001,000,001_[b=2] – 4 = 1597.

However, subtracting the digit-product creates kangaroo-primes in base 10. For example, 23 – (2 x 3) = 17. The record below 1,000,000 is this (when 0 is found in the digits of a number, it is not included in the multiplication):

7 primes: 64037 – 504 = 63533 – 810 = 62723 – 504 = 62219 – 216 = 62003 – 36 = 61967 – 2268 = 59699.

Base 2 also provides examples of addition/subtraction pairs of kangaroo-primes, like this:

3 = 11_[b=2] + 2 = 5 = 101_[b=2] | 5 = 101_[b=2] – 2 = 3

277 = 100,010,101_[b=2] + 4 = 281 = 100,011,001_[b=2] | 281 – 4 = 277

311 = 100,110,111_[b=2] + 6 = 317 = 100,111,101_[b=2] | 317 – 6 = 311

In base 10, addition/subtraction pairs are created by the digit-product, like this:

239 + 54 = 293 | 293 – 54 = 239
563 + 90 = 653 | 653 – 90 = 563
613 + 18 = 631 | 631 – 18 = 613
2791 + 126 = 2917 | 2917 – 126 = 2791
3259 + 270 = 3529 | 3529 – 270 = 3259
5233 + 90 = 5323 | 5323 – 90 = 5233
5297 + 630 = 5927 | 5927 – 630 = 5297
6113 + 18 = 6131 | 6131 – 18 = 6113
10613 + 18 = 10631 | 10631 – 18 = 10613
12791 + 126 = 12917 | 12917 – 126 = 12791

You could call these boxing primes, like boxing kangaroos. The two primes in the pair usually have the same digits in different arrangements, but there are also pairs like these:

24527 + 560 = 25087 | 25087 – 560 = 24527
25183 + 240 = 25423 | 25423 – 240 = 25183
50849 + 1440 = 52289 | 52289 – 1440 = 50849

Clock around the Rock

Tuesday, 16th Apr, 2013 by Krilling for Company in Art, Binary numbers, Mathematics, Palindromic numbers, Prime numbers and tagged base 10, base 2, binary, binary arithmetic, binary digits, binary numbers, binary primes, bit, bits, brime, brimes, clockface, deciminimalism, deciminimalist, deciminimalist prime, deciminimalist primes, dragon biting its own tail, elegance, hexadecimal, integer, integers, math, mathematics, maths, minimalism, minimalist, number-ring, octal, Ouroboprime, Ouroboprimes, Ouroboros, palindrome, palindromes, palindromic binary primes, palindromic numbers, palindromic prime, palindromic primes, prime, prime numbers, primes, primes in binary, recreational math, recreational mathematics, recreational maths, ring of numbers, serpent biting its own tail, simplicity | Leave a comment

If you like minimalism, you should like binary. There is unsurpassable simplicity and elegance in the idea that any number can be reduced to a series of 1’s and 0’s. It’s unsurpassable because you can’t get any simpler: unless you use finger-counting, two symbols are the minimum possible. But with those two – a stark 1 and 0, true and false, yin and yang, sun and moon, black and white – you can conquer any number you please. 2 = 10_[2]. 5 = 101. 100 = 1100100. 666 = 1010011010. 2013 = 11111011101. 9^9 = 387420489 = 10111000101111001000101001001. You can also perform any mathematics you please, from counting sheep to modelling the evolution of the universe.

1 + 0 = ∞

But one disadvantage of binary, from the human point of view, is that numbers get long quickly: every doubling in size adds an extra digit. You can overcome that disadvantage using octal or hexadecimal, which compress blocks of binary into single digits, but those number systems need more symbols: eight and sixteen, as their names suggest. There’s an elegance there too, but binary goes masked, hiding its minimalist appeal beneath apparent complexity. It doesn’t need to wear a mask for computers, but human beings can appreciate bare binary too, even with our weak memories and easily tiring nervous systems. I especially like minimalist binary when it’s put to work on those most maximalist of numbers: the primes. You can compare integers, or whole numbers, to minerals. Some are like mica or shale, breaking readily into smaller parts, but primes are like granite or some other ultra-hard, resistant rock. In other words, some integers are easy to divide by other integers and some, like the primes, are not. Compare 256 with 257. 256 = 2^8, so it’s divisible by 128, 64, 32, 16, 8, 4, 2 and 1. 257 is a prime, so it’s divisible by nothing but itself and 1. Powers of two are easy to calculate and, in binary, very easy to represent:

2^0 = 1 = 1
2^1 = 2 = 10_[2]
2^2 = 4 = 100
2^3 = 8 = 1000
2^4 = 16 = 10000
2^5 = 32 = 100000
2^6 = 64 = 1000000
2^7 = 128 = 10000000
2^8 = 256 = 100000000

Primes are the opposite: hard to calculate and usually hard to represent, whatever the base:

02 = 000010_[2]
03 = 000011
05 = 000101
07 = 000111
11 = 001011
13 = 001101
17 = 010001
19 = 010011
23 = 010111
29 = 011101
31 = 011111
37 = 100101
41 = 101001
43 = 101011

Maximalist numbers, minimalist base: it’s a potent combination. But “brimes”, or binary primes, nearly all have one thing in common. Apart from 2, a special case, each brime must begin and end with 1. For the digits in-between, the God of Mathematics seems to be tossing a coin, putting 1 for heads, 0 for tails. But sometimes the coin will come up all heads or all tails: 127 = 1111111_[2] and 257 = 100000001, for example. Brimes like that have a stark simplicity amid the jumble of 83 = 1010011_[2], 113 = 1110001, 239 = 11101111, 251 = 11111011, 277 = 100010101, and so on. Brimes like 127 and 257 are also palindromes, or the same reading in both directions. But less simple brimes can be palindromes too:

73 = 1001001
107 = 1101011
313 = 100111001
443 = 110111011
1193 = 10010101001
1453 = 10110101101
1571 = 11000100011
1619 = 11001010011
1787 = 11011111011
1831 = 11100100111
1879 = 11101010111

But, whether they’re palindromes or not, all brimes except 2 begin and end with 1, so they can be represented as rings, like this:

Those twelve bits, or binary digits, actually represent the thirteen bits of 5227 = 1,010,001,101,011. Start at twelve o’clock (digit 1 of the prime) and count clockwise, adding 1’s and 0’s till you reach 12 o’clock again and add the final 1. Then you’ve clocked around the rock and created the granite of 5227, which can’t be divided by any integers but itself and 1. Another way to see the brime-ring is as an Ouroboros (pronounced “or-ROB-or-us”), a serpent or dragon biting its own tail, like this:

Alchemical Ouroboros (1478)

Another alchemical Ouroboros (1599)

But you don’t have to start clocking around the rock at midday or midnight. Take the Ouroboprime of 5227 and start at eleven o’clock (digit 12 of the prime), adding 1’s and 0’s as you move clockwise. When you’ve clocked around the rock, you’ll have created the granite of 6709, another prime:

Other Ouroboprimes produce brimes both clockwise and anti-clockwise, like 47 = 101,111.

Clockwise

101,111 = 47
111,011 = 59
111,101 = 61

Anti-Clockwise

111,101 = 61
111,011 = 59
101,111 = 47

If you demand the clock-rocked brime produce distinct primes, you sometimes get more in one direction than the other. Here is 151 = 10,010,111:

Clockwise

10,010,111 = 151
11,100,101 = 229

Anti-Clockwise

11,101,001 = 233
11,010,011 = 211
10,100,111 = 167
10,011,101 = 157

The most productive brime I’ve discovered so far is 2,326,439 = 1,000,110,111,111,110,100,111_[2], which produces fifteen distinct primes:

Clockwise (7 brimes)

1,000,110,111,111,110,100,111 = 2326439
1,100,011,011,111,111,010,011 = 3260371
1,110,100,111,000,110,111,111 = 3830207
1,111,101,001,110,001,101,111 = 4103279
1,111,110,100,111,000,110,111 = 4148791
1,111,111,010,011,100,011,011 = 4171547
1,101,111,111,101,001,110,001 = 3668593

Anti-Clockwise (8 brimes)

1,110,010,111,111,110,110,001 = 3768241
1,100,101,111,111,101,100,011 = 3342179
1,111,111,011,000,111,001,011 = 4174283
1,111,110,110,001,110,010,111 = 4154263
1,111,101,100,011,100,101,111 = 4114223
1,111,011,000,111,001,011,111 = 4034143
1,110,110,001,110,010,111,111 = 3873983
1,000,111,001,011,111,111,011 = 2332667

Appendix: Deciminimalist Primes

Some primes in base ten use only the two most basic symbols too. That is, primes like 11_[10], 101_[10], 10111_[10] and 1011001_[10] are composed of only 1’s and 0’s. Furthermore, when these numbers are read as binary instead, they are still prime: 11_[2] = 3, 101_[2] = 5, 10111_[2] = 23 and 1011001_[2] = 89. Here is an incomplete list of these deciminimalist primes:

11_[10] = 1,011_[2]; 11_[2] = 3_[10] is also prime.

101_[10] = 1,100,101_[2]; 101_[2] = 5_[10] is also prime.

10,111_[10] = 10,011,101,111,111_[2]; 10,111_[2] = 23_[10] is also prime.

101,111_[10] = 11,000,101,011,110,111_[2]; 101,111_[2] = 47_[10] is also prime.

1,011,001_[10] = 11,110,110,110,100,111,001_[2]; 1,011,001_[2] = 89_[10] is also prime.

1,100,101_[10] = 100,001,100,100,101,000,101_[2]; 1,100,101_[2] = 101_[10] is also prime.

10,010,101_[10] = 100,110,001,011,110,111,110,101_[2]; 10,010,101_[2] = 149_[10] is also prime.

10,011,101_[10] = 100,110,001,100,000,111,011,101_[2]; 10,011,101_[2] = 157_[10] is also prime.

10,100,011_[10] = 100,110,100,001,110,100,101,011_[2]; 10,100,011_[2] = 163_[10] is also prime.

10,101,101_[10] = 100,110,100,010,000,101,101,101_[2]; 10,101,101_[2] = 173_[10] is also prime.

10,110,011_[10] = 100,110,100,100,010,000,111,011_[2]; 10,110,011_[2] = 179_[10] is also prime.

10,111,001_[10] = 100,110,100,100,100,000,011,001_[2].

11,000,111_[10] = 101,001,111,101,100,100,101,111_[2]; 11,000,111_[2] = 199_[10] is also prime.

11,100,101_[10] = 101,010,010,101,111,111,000,101_[2]; 11,100,101_[2] = 229_[10] is also prime.

11,110,111_[10] = 101,010,011,000,011,011,011,111_[2].

11,111,101_[10] = 101,010,011,000,101,010,111,101_[2].

100,011,001_[10] = 101,111,101,100,000,101,111,111,001_[2]; 100,011,001_[2] = 281_[10] is also prime.

100,100,111_[10] = 101,111,101,110,110,100,000,001,111_[2].

100,111,001_[10] = 101,111,101,111,001,001,010,011,001_[2]; 100,111,001_[2] = 313_[10] is also prime.

101,001,001_[10] = 110,000,001,010,010,011,100,101,001_[2].

101,001,011_[10] = 110,000,001,010,010,011,100,110,011_[2]; 101,001,011_[2] = 331_[10] is also prime.

101,001,101_[10] = 110,000,001,010,010,011,110,001,101_[2].

101,100,011_[10] = 110,000,001,101,010,100,111,101,011_[2].

101,101,001_[10] = 110,000,001,101,010,110,111,001,001_[2].

101,101,111_[10] = 110,000,001,101,010,111,000,110,111_[2]; 101,101,111_[2] = 367_[10] is also prime.

101,110,111_[10] = 110,000,001,101,101,000,101,011,111_[2].

101,111,011_[10] = 110,000,001,101,101,010,011,100,011_[2]; 101,111,011_[2] = 379_[10] is also prime.

101,111,111_[10] = 110,000,001,101,101,010,101,000,111_[2]; 101,111,111_[2] = 383_[10] is also prime.

110,010,101_[10] = 110,100,011,101,001,111,011,110,101_[2].

110,100,101_[10] = 110,100,011,111,111,111,010,000,101_[2]; 110,100,101_[2] = 421_[10] is also prime.

110,101,001_[10] = 110,100,100,000,000,001,000,001,001_[2].

110,110,001_[10] = 110,100,100,000,010,010,100,110,001_[2]; 110,110,001_[2] = 433_[10] is also prime.

110,111,011_[10] = 110,100,100,000,010,100,100,100,011_[2]; 110,111,011_[2] = 443_[10] is also prime.

Sumbertime Views

Friday, 5th Apr, 2013 by Krilling for Company in 666, Digit-sums, Mathematics, Number of the Beast and tagged 153, 666, Arithmetic, base 10, base 11, base 47, base 9, Biblical numbers, Book of Revelation, cube, cubes, Gospel of John, integer bases, math, mathematics, maths, New Testament numbers, number sums, numbers in the Bible, numbers in the New Testament, powers, recreational math, recreational mathematics, recreational maths, square, Squares, sum of integers, sum of successive integers, sumber, sumbers, sums of cubes, sums of powers, sums of squares | Leave a comment

Like 666 (see Revelation 13:18), 153 (see John 21:11) appears in the Bible. And perhaps for the same reason: because it is the sum of successive integers. 153 = 1+2+3+…+17 = Σ(17), just as 666 = Σ(36). So both numbers are sum-numbers or sumbers. But 153 has other interesting properties, including one that can’t have been known in Biblical times, because numbers weren’t represented in the right way. It’s also the sum of the cubes of its digits: 153 = 1^3 + 5^3 + 3^3 = 1 + 125 + 27. So 153 is a cube-sumber or 3-sumber. The other 3-sumbers are 370, 371 and 407. There are 4-sumbers too, like 1,634 = 1^4 + 6^4 + 3^4 + 4^4, and 5-sumbers, like 194,979 = 1^5 + 9^5 + 4^5 + 9^5 + 7^5 + 9^5, and 6-sumbers, like 548,834 = 5^6 + 4^6 + 8^6 + 8^6 + 3^6 + 4^6.

But there are no 2-sumbers, or numbers that are the sum of the squares of their digits. It doesn’t take long to confirm this, because numbers above a certain size can’t be 2-sumbers. 9^2 + 9^2 = 162, but 9^2 + 9^2 + 9^2 = 243. So 2-sumbers can’t exist above 99 and if you search that high you’ll find that they don’t exist at all. At least not in this house, but they do exist in the houses next door. Base 10 yields nothing, so what about base 9?

4^2 + 5^2 = 45_[9] = 41_[10]
5^2 + 5^2 = 55_[9] = 50

And base 11?

5^2 + 6^2 = 56_[11] = 61_[10]
6^2 + 6^2 = 66_[11] = 72

This happens because odd bases always yield a pair of 2-sumbers whose second digit is one more than half the base and whose first digit is the same or one less. See above (and the appendix). Such a pair is found among the 14 sumbers of base 47, which is the best total till base 157 and its 22 sumbers. Here are the 2-sumbers for base 47:

2^2 + 10^2 = 104
3^2 + 12^2 = 153
5^2 + 15^2 = 250
9^2 + 19^2 = 442
12^2 + 21^2 = 585
14^2 + 22^2 = 680
23^2 + 24^2 = 1,105
24^2 + 24^2 = 1,152
33^2 + 22^2 = 1,573
35^2 + 21^2 = 1,666
38^2 + 19^2 = 1,805
42^2 + 15^2 = 1,989
44^2 + 12^2 = 2,080
45^2 + 10^2 = 2,125

As the progressive records for 2-sumber-totals are set, subsequent bases seem to either match or surpass them, except in three cases below base 450:

2 in base 5
4 in base 7
6 in base 13
10 in base 43
14 in base 47
22 in base 157
8 in base 182*
16 in base 268*
30 in base 307
18 in base 443*

Totals for sums-of–squares in bases 4 to 450 (click for larger image)

Appendix: Odd Bases and 2-sumbers

Take an even number and half of that even number: say 12 and 6. 12 x 6 = 11 x 6 + 6. Further, 12 x 6 = 2 x 6 x 6 = 2 x 6^2 = 6^2 + 6^2. Accordingly, 66_[11] = 6 x 11 + 6 = 12 x 6 = 6^2 + 6^2. So 66 in base 11 is a 2-sumber. Similar reasoning applies to every other odd base except base-3 [update: wrong!]. Now, take 12 x 5 = 2 x 6 x 5 = 2 x (5×5 + 5) = 5^2+5 + 5^5+5 = 5^5 + 5^5+2×5. Further, 5^5+2×5 = (5+1)(5+1) – 1 = 6^2 – 1. Accordingly, 56_[11] = 11×5 + 6 = 12×5 + 1 = 5^2 + 6^2. Again, similar reasoning applies to every other odd base except base-3 [update: no — 1^2 + 2^2 = 12_[3] = 5; 2^2 + 2^2 = 22_[3] = 8]. This means that every odd base b, except base-3, will supply a pair of 2-sumbers with digits [d-1][d] and [d][d], where d = (b + 1) / 2.

Three Is The Key

Saturday, 23rd Feb, 2013 by Krilling for Company in Art, Magic squares, Mathematics, Post-Weird, Prime numbers and tagged 142857, 222, 223, 3, 37, David Wells, magic square, magic squares, math, mathematics, maths, Penguin Dictionary of Curious and Interesting Numbers, prime numbers, prime reciprocal, prime reciprocal magic square, prime reciprocal magic squares, prime reciprocals, primes, reciprocal, reciprocals, recreational math, recreational mathematics, recreational maths, Roses of Heliogabalus, Sir Lawrence Alma-Tadema, two hundred and twenty-two | Leave a comment

If The Roses of Heliogabalus (1888) is any guide, Sir Lawrence Alma-Tadema (1836-1912) thought that 222 is a special number. But his painting doesn’t exhaust its secrets. To get to another curiosity of 222, start with 142857. As David Wells puts it in his Penguin Dictionary of Curious and Interesting Numbers (1986), 142857 is a “number beloved of all recreational mathematicians”. He then describes some of its properties, including this:

142857 x 1 = 142857
142857 x 2 = 285714
142857 x 3 = 428571
142857 x 4 = 571428
142857 x 5 = 714285
142857 x 6 = 857142

The multiples are cyclic permutations: the order of the six numbers doesn’t change, only their starting point. Because each row contains the same numbers, it sums to the same total: 1 + 4 + 2 + 8 + 5 + 7 = 27. And because each row begins with a different number, each column contains the same six numbers and also sums to 27, like this:

1 4 2 8 5 7
+ + + + + +
2 8 5 7 1 4
+ + + + + +
4 2 8 5 7 1
+ + + + + +
5 7 1 4 2 8
+ + + + + +
7 1 4 2 8 5
+ + + + + +
8 5 7 1 4 2

= = = = = =

2 2 2 2 2 2
7 7 7 7 7 7

If the diagonals of the square also summed to the same total, the multiples of 142857 would create a full magic square. But the diagonals don’t have the same total: the left-right diagonal sums to 31 and the right-left to 23 (note that 31 + 23 = 54 = 27 x 2).

But where does 142857 come from? It’s actually the first six digits of the reciprocal of 7, i.e. 1/7 = 0·142857… Those six numbers repeat for ever, because 1/7 is a prime reciprocal with maximum period: when you calculate 1/7, all integers below 7 are represented in the remainders. The square of multiples above is simply another way of representing this:

1/7 = 0·142857…
2/7 = 0·285714…
3/7 = 0·428571…
4/7 = 0·571428…
5/7 = 0·714285…
6/7 = 0·857142…
7/7 = 0·999999…

The prime reciprocals 1/17 and 1/19 also have maximum period, so the squares created by their multiples have the same property: each row and each column sums to the same total, 72 and 81, respectively. But the 1/19 square has an additional property: both diagonals sum to 81, so it is fully magic:

01/19 = 0·0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1…
02/19 = 0·1 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2…
03/19 = 0·1 5 7 8 9 4 7 3 6 8 4 2 1 0 5 2 6 3…
04/19 = 0·2 1 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4…
05/19 = 0·2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 0 5…
06/19 = 0·3 1 5 7 8 9 4 7 3 6 8 4 2 1 0 5 2 6…
07/19 = 0·3 6 8 4 2 1 0 5 2 6 3 1 5 7 8 9 4 7…
08/19 = 0·4 2 1 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8…
09/19 = 0·4 7 3 6 8 4 2 1 0 5 2 6 3 1 5 7 8 9…
10/19 = 0·5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 0…
11/19 = 0·5 7 8 9 4 7 3 6 8 4 2 1 0 5 2 6 3 1…
12/19 = 0·6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 0 5 2…
13/19 = 0·6 8 4 2 1 0 5 2 6 3 1 5 7 8 9 4 7 3…
14/19 = 0·7 3 6 8 4 2 1 0 5 2 6 3 1 5 7 8 9 4…
15/19 = 0·7 8 9 4 7 3 6 8 4 2 1 0 5 2 6 3 1 5…
16/19 = 0·8 4 2 1 0 5 2 6 3 1 5 7 8 9 4 7 3 6…
17/19 = 0·8 9 4 7 3 6 8 4 2 1 0 5 2 6 3 1 5 7…
18/19 = 0·9 4 7 3 6 8 4 2 1 0 5 2 6 3 1 5 7 8…

First line = 0 + 5 + 2 + 6 + 3 + 1 + 5 + 7 + 8 + 9 + 4 + 7 + 3 + 6 + 8 + 4 + 2 + 1 = 81

Left-right diagonal = 0 + 0 + 7 + 5 + 5 + 9 + 0 + 3 + 0 + 4 + 2 + 8 + 7 + 5 + 6 + 7 + 5 + 8 = 81

Right-left diagonal = 9 + 9 + 2 + 4 + 4 + 0 + 9 + 6 + 9 + 5 + 7 + 1 + 2 + 4 + 3 + 2 + 4 + 1 = 81

In base 10, this doesn’t happen again until the 1/383 square, whose magic total is 1719 (= 383-1 x 10-1 / 2). But recreational maths isn’t restricted to base 10 and lots more magic squares are created by lots more primes in lots more bases. The prime 223 in base 3 is one of them. Here the first line is

1/223 = 1/22021₃ = 0·

0000100210210102121211101202221112202
2110211112001012200122102202002122220
2110110201020210001211000222011010010
2222122012012120101011121020001110020
0112011110221210022100120020220100002
0112112021202012221011222000211212212…

The digits sum to 222, so 222 is the magic total for all rows and columns of the 1/223 square. It is also the total for both diagonals, so the square is fully magic. I doubt that Alma-Tadema knew this, because he lived before computers made calculations like that fast and easy. But he was probably a Freemason and, if so, would have been pleased to learn that 222 had a link with squares.

Summer-Climb Views

Thursday, 31st Jan, 2013 by Krilling for Company in Mathematics, Palindromic numbers and tagged algorithm, algorithms, amateur math, amateur mathematics, amateur maths, digit sum, digit sums, lychrel number, lychrel numbers, math, mathematical computing, mathematics, maths, palindrome, palindromes, palindromic, recreational math, recreational mathematics, recreational maths | Leave a comment

Simple things can sometimes baffle advanced minds. If you take a number, reverse its digits, add the result to the original number, then repeat all that, will you eventually get a palindrome? (I.e., a number, like 343 or 27172, that reads the same in both directions.) Many numbers do seem to produce palindromes sooner or later. Here are 195 and 197:

195 + 591 = 786 + 687 = 1473 + 3741 = 5214 + 4125 = 9339 (4 steps)

197 + 791 = 988 + 889 = 1877 + 7781 = 9658 + 8569 = 18227 + 72281 = 90508 + 80509 = 171017 + 710171 = 881188 (7 steps)

But what about 196? Well, it starts like this:

196 + 691 = 887 + 788 = 1675 + 5761 = 7436 + 6347 = 13783 + 38731 = 52514 + 41525 = 94039 + 93049 = 187088 + 880781 = 1067869 + 9687601 = 10755470 + 7455701 = 18211171 + 17111281 = 35322452 + 25422353 = 60744805 + 50844706 = 111589511 + 115985111 = 227574622 + 226475722 = 454050344 + 443050454 = 897100798 + 897001798 = 1794102596 + 6952014971 = 8746117567 + 7657116478 = 16403234045 + 54043230461 = 70446464506 + 60546464407 = 130992928913 + 319829299031 = 450822227944 + 449722228054 = 900544455998…

And so far, after literally years of computing by mathematicians, it hasn’t produced a palindrome. It seems very unlikely it ever will, but no-one can prove this and say that 196 is, in base 10, a Lychrel number, or a number that never produces a palindrome. In other words, a simple thing has baffled advanced minds.

I don’t know whether it can baffle advanced minds, but here’s another simple mathematical technique: sum all the digits of a number, then add the result to the original number and repeat. How long before a palindrome appears in this case? Sum it and see:

10 + 1 = 11

12 + 3 = 15 + 6 = 21 + 3 = 24 + 6 = 30 + 3 = 33 (5 steps)

13 + 4 = 17 + 8 = 25 + 7 = 32 + 5 = 37 + 10 = 47 + 11 = 58 + 13 = 71 + 8 = 79 + 16 = 95 + 14 = 109 + 10 = 119 + 11 = 130 + 4 = 134 + 8 = 142 + 7 = 149 + 14 = 163 + 10 = 173 + 11 = 184 + 13 = 197 + 17 = 214 + 7 = 221 + 5 = 226 + 10 = 236 + 11 = 247 + 13 = 260 + 8 = 268 + 16 = 284 + 14 = 298 + 19 = 317 + 11 = 328 + 13 = 341 + 8 = 349 + 16 = 365 + 14 = 379 + 19 = 398 + 20 = 418 + 13 = 431 + 8 = 439 + 16 = 455 + 14 = 469 + 19 = 488 + 20 = 508 + 13 = 521 + 8 = 529 + 16 = 545 (45 steps)

14 + 5 = 19 + 10 = 29 + 11 = 40 + 4 = 44 (4 steps)

15 + 6 = 21 + 3 = 24 + 6 = 30 + 3 = 33 (4 steps)

16 + 7 = 23 + 5 = 28 + 10 = 38 + 11 = 49 + 13 = 62 + 8 = 70 + 7 = 77 (7 steps)

17 + 8 = 25 + 7 = 32 + 5 = 37 + 10 = 47 + 11 = 58 + 13 = 71 + 8 = 79 + 16 = 95 + 14 = 109 + 10 = 119 + 11 = 130 + 4 = 134 + 8 = 142 + 7 = 149 + 14 = 163 + 10 = 173 + 11 = 184 + 13 = 197 + 17 = 214 + 7 = 221 + 5 = 226 + 10 = 236 + 11 = 247 + 13 = 260 + 8 = 268 + 16 = 284 + 14 = 298 + 19 = 317 + 11 = 328 + 13 = 341 + 8 = 349 + 16 = 365 + 14 = 379 + 19 = 398 + 20 = 418 + 13 = 431 + 8 = 439 + 16 = 455 + 14 = 469 + 19 = 488 + 20 = 508 + 13 = 521 + 8 = 529 + 16 = 545 (44 steps)

18 + 9 = 27 + 9 = 36 + 9 = 45 + 9 = 54 + 9 = 63 + 9 = 72 + 9 = 81 + 9 = 90 + 9 = 99 (9 steps)

19 + 10 = 29 + 11 = 40 + 4 = 44 (3 steps)

20 + 2 = 22

I haven’t looked very thoroughly at this technique, so I don’t know whether it throws up a seemingly unpalindromizable number. If it does, I don’t have an advanced mind, so I won’t be able to prove that it is unpalindromizable. But an adaptation of the technique produces something interesting when it is represented on a graph. This time, if s > 9, where s = digit-sum(n), let s = digit-sum(s) until s <= 9 (i.e, s < 10, the base). I call this the condensed digit-sum:

140 + 5 = 145 + 1 = 146 + 2 = 148 + 4 = 152 + 8 = 160 + 7 = 167 + 5 = 172 + 1 = 173 + 2 = 175 + 4 = 179 + 8 = 187 + 7 = 194 + 5 = 199 + 1 = 200 + 2 = 202 (15 steps)

Here, for comparison, is the sequence for 140 using uncondensed digit-sums:

140 + 5 = 145 + 10 = 155 + 11 = 166 + 13 = 179 + 17 = 196 + 16 = 212 (6 steps)

When all the numbers (including palindromes) created using condensed digit-sums are shown on a graph, they create an interesting pattern in base 10 (the x-axis represents n, the y-axis represents n, n1 = n + digit-sum(n), n2 = n1 + digit-sum(n1), etc):

(Please open images in a new window if they fail to animate.)

And here, for comparison, are the patterns created by uncondensed digit-sums in base 2 to 10:

Watch this Sbase

Friday, 28th Dec, 2012 by Krilling for Company in Mathematics and tagged Arithmetic, base, bases, binary, digit sum, digit sums, math, mathematics, maths, recreational math, recreational mathematics, recreational maths | Leave a comment

In standard notation, there are two ways to represent 2: 10, in base 2, and 2 in every other base. Accordingly, there are three ways to represent 3: 11 in base 2, 10 in base 3, and 3 in every other base. There are four ways to represent 4, five ways to represent 5, and so on. Now, suppose you sum all the digits of all the representations of n in the bases 2 to n, like this:

Σ(2) = 1+0₂ = 1
Σ(3) = 1+1₂ + 1+0₃ = 3 (+2)
Σ(4) = 1+0+0₂ + 1+1₃ + 1+0₄ = 4 (+1)
Σ(5) = 1+0+1₂ + 1+2₃ + 1+1₄ + 1+0₅ = 8 (+4)
Σ(6) = 1+1+0₂ + 2+0₃ + 1+2₄ + 1+1₅ + 1+0₆ = 10 (+2)
Σ(7) = 1+1+1₂ + 2+1₃ + 1+3₄ + 1+2₅ + 1+1₆ + 1+0₇ = 16 (+6)
Σ(8) = 1+0+0+0₂ + 2+2₃ + 2+0₄ + 1+3₅ + 1+2₆ + 1+1₇ + 1+0₈ = 17 (+1)
Σ(9) = 1+0+0+1₂ + 1+0+0₃ + 2+1₄ + 1+4₅ + 1+3₆ + 1+2₇ + 1+1₈ + 1+0₉ = 21 (+4)
Σ(10) = 1+0+1+0₂ + 1+0+1₃ + 2+2₄ + 2+0₅ + 1+4₆ + 1+3₇ + 1+2₈ + 1+1₉ + 1+0₁₀ = 25 (+4)

It seems reasonable to suppose that as n increases, so the all-digit-sum of n increases. But that isn’t always the case: occasionally it decreases. Here are the sums for n=11..100 (with prime factors when the sum is composite):

Σ(11) = 35 = 5·7 (+10)
Σ(12) = 34 = 2·17 (-1)
Σ(13) = 46 = 2·23 (+12)
Σ(14) = 52 = 2²·13 (+6)
Σ(15) = 60 = 2²·3·5 (+8)
Σ(16) = 58 = 2·29 (-2)
Σ(17) = 74 = 2·37 (+16)
Σ(18) = 73 (-1)
Σ(19) = 91 = 7·13 (+18)
Σ(20) = 92 = 2²·23 (+1)
Σ(21) = 104 = 2³·13 (+12)
Σ(22) = 114 = 2·3·19 (+10)
Σ(23) = 136 = 2³·17 (+22)
Σ(24) = 128 = 2⁷ (-8)
Σ(25) = 144 = 2⁴·3² (+16)
Σ(26) = 156 = 2²·3·13 (+12)
Σ(27) = 168 = 2³·3·7 (+12)
Σ(28) = 171 = 3²·19 (+3)
Σ(29) = 199 (+28)
Σ(30) = 193 (-6)
Σ(31) = 223 (+30)
Σ(32) = 221 = 13·17 (-2)
Σ(33) = 241 (+20)
Σ(34) = 257 (+16)
Σ(35) = 281 (+24)
Σ(36) = 261 = 3²·29 (-20)
Σ(37) = 297 = 3³·11 (+36)
Σ(38) = 315 = 3²·5·7 (+18)
Σ(39) = 339 = 3·113 (+24)
Σ(40) = 333 = 3²·37 (-6)
Σ(41) = 373 (+40)
Σ(42) = 367 (-6)
Σ(43) = 409 (+42)
Σ(44) = 416 = 2⁵·13 (+7)
Σ(45) = 430 = 2·5·43 (+14)
Σ(46) = 452 = 2²·113 (+22)
Σ(47) = 498 = 2·3·83 (+46)
Σ(48) = 472 = 2³·59 (-26)
Σ(49) = 508 = 2²·127 (+36)
Σ(50) = 515 = 5·103 (+7)
Σ(51) = 547 (+32)
Σ(52) = 556 = 2²·139 (+9)
Σ(53) = 608 = 2⁵·19 (+52)
Σ(54) = 598 = 2·13·23 (-10)
Σ(55) = 638 = 2·11·29 (+40)
Σ(56) = 634 = 2·317 (-4)
Σ(57) = 670 = 2·5·67 (+36)
Σ(58) = 698 = 2·349 (+28)
Σ(59) = 756 = 2²·3³·7 (+58)
Σ(60) = 717 = 3·239 (-39)
Σ(61) = 777 = 3·7·37 (+60)
Σ(62) = 807 = 3·269 (+30)
Σ(63) = 831 = 3·277 (+24)
Σ(64) = 819 = 3²·7·13 (-12)
Σ(65) = 867 = 3·17² (+48)
Σ(66) = 861 = 3·7·41 (-6)
Σ(67) = 927 = 3²·103 (+66)
Σ(68) = 940 = 2²·5·47 (+13)
Σ(69) = 984 = 2³·3·41 (+44)
Σ(70) = 986 = 2·17·29 (+2)
Σ(71) = 1056 = 2⁵·3·11 (+70)
Σ(72) = 1006 = 2·503 (-50)
Σ(73) = 1078 = 2·7²·11 (+72)
Σ(74) = 1114 = 2·557 (+36)
Σ(75) = 1140 = 2²·3·5·19 (+26)
Σ(76) = 1155 = 3·5·7·11 (+15)
Σ(77) = 1215 = 3⁵·5 (+60)
Σ(78) = 1209 = 3·13·31 (-6)
Σ(79) = 1287 = 3²·11·13 (+78)
Σ(80) = 1263 = 3·421 (-24)
Σ(81) = 1293 = 3·431 (+30)
Σ(82) = 1333 = 31·43 (+40)
Σ(83) = 1415 = 5·283 (+82)
Σ(84) = 1368 = 2³·3²·19 (-47)
Σ(85) = 1432 = 2³·179 (+64)
Σ(86) = 1474 = 2·11·67 (+42)
Σ(87) = 1530 = 2·3²·5·17 (+56)
Σ(88) = 1530 = 2·3²·5·17 (=)
Σ(89) = 1618 = 2·809 (+88)
Σ(90) = 1572 = 2²·3·131 (-46)
Σ(91) = 1644 = 2²·3·137 (+72)
Σ(92) = 1663 (+19)
Σ(93) = 1723 (+60)
Σ(94) = 1769 = 29·61 (+46)
Σ(95) = 1841 = 7·263 (+72)
Σ(96) = 1784 = 2³·223 (-57)
Σ(97) = 1880 = 2³·5·47 (+96)
Σ(98) = 1903 = 11·173 (+23)
Σ(99) = 1947 = 3·11·59 (+44)
Σ(100) = 1923 = 3·641 (-24)

The sum usually increases, occasionally decreases. In one case, when 87 = n = 88, it stays the same. This also happens when 463 = n = 464, where Σ(463) = Σ(464) = 39,375. Does it happen again? I don’t know. The ratio of sum-ups to sum-downs seems to tend towards 3:1. Is that the exact ratio at infinity? I don’t know. Watch this sbase.

Back to Bases

Thursday, 22nd Nov, 2012 by Krilling for Company in Mathematics and tagged animated gif, animated gifs, Arithmetic, factors, math, mathematics, maths, palindrome, palindromes, palindromic number, prime factors, recreational math, recreational mathematics, recreational maths | Leave a comment

(N.B. I am not a mathematician and often make stupid mistakes in my recreational maths. Caveat lector.)

101 isn’t a number, it’s a label for a number. In fact, it’s a label for infinitely many numbers. In base 2, 101₂ = 5; in base 3, 101₃ = 10; 101₄ = 17; 101₅ = 26; and so on, for ever. In some bases, like 2 and 4, the number labelled 101 is prime. In other bases, it isn’t. But it is always a palindrome: that is, it’s the same read forward and back. But 101, the number itself, is a palindrome in only two bases: base 10 and base 100.¹ Note that 100 = 101-1: with the exception of 2, all integers, or whole numbers, are palindromic in at least one base, the base that is one less than the integer itself. So 3 = 11₂; 4 = 11₃; 5 = 11₄; 101 = 11₁₀₀; and so on.

Less trivial is the question of which integers set progressive records for palindromicity, or for the number of palindromes they create in bases less than the integers themselves. You might guess that the bigger the integer, the more palindromes it will create, but it isn’t as simple as that. Here is 10 represented in bases 2 through 9:

1010₂ | 101₃* | 22₄* | 20₅ | 14₆ | 13₇ | 12₈ | 11₉*

10 is a palindrome in bases 3, 4, and 9. Now here is 30 represented in bases 2 through 29 (note that a number between square brackets represents a single digit in that base):²

11110₂ | 1010₃ | 132₄ | 110₅ | 50₆ | 42₇ | 36₈ | 33₉* | 30 | 28₁₁ | 26₁₂ | 24₁₃ | 22₁₄* | 20₁₅ | 1[14]₁₆ | 1[13]₁₇ | 1[12]₁₈ | 1[11]₁₉ | 1[10]₂₀ | 19₂₁ | 18₂₂ | 17₂₃ | 16₂₄ | 15₂₅ | 14₂₆ | 13₂₇ | 12₂₈| 11₂₉*

30, despite being three times bigger than 10, creates only three palindromes too: in bases 9, 14, and 29. Here is a graph showing the number of palindromes for each number from 3 to 100 (prime numbers are in red):

The number of palindromes a number has is related to the number of factors, or divisors, it has. A prime number has only one factor, itself (and 1), so primes tend to be less palindromic than composite numbers. Even large primes can have only one palindrome, in the base b=n-1 (55,440 has 119 factors and 61 palindromes; 65,381 has one factor and one palindrome, 11₆₅₃₈₀). Here is a graph showing the number of factors for each number from 3 to 100:

And here is an animated gif combining the two previous images:

Here is a graph indicating where palindromes appear when n, along the x-axis, is represented in the bases b=2 to n-1, along the y-axis:

The red line are the palindromes in base b=n-1, which is “11” for every n>2. The lines below it arise because every sufficiently large n with divisor d can be represented in the form d·n₁ + d. For example, 8 = 2·3 + 2, so 8 in base 3 = 22₃; 18 = 3·5 + 3, so 18 = 33₅; 32 = 4.7 + 4, so 32 = 44₇; 391 = 17·22 + 17, so 391 = [17][17]₂₂.

And here, finally, is a table showing integers that set progressive records for palindromicity (p = number of palindromes, f = total number of factors, prime and non-prime):

n	Prime Factors	p	f	n	Prime Factors	p	f
3	3	1	1	2,520	2³·3²·5·7	25	47
5	5	2	1	3,600	2⁴·3²·5²	26	44
10	2·5	3	3	5,040	2⁴·3²·5·7	30	59
21	3·7	4	3	7,560	2³·3³·5·7	32	63
36	2²·3²	5	8	9,240	2³·3·5·7·11	35	63
60	2²·3·5	6	11	10,080	2⁵·3²·5·7	36	71
80	2⁴·5	7	9	12,600	2³·3²·5²·7	38	71
120	2³·3·5	8	15	15,120	2⁴·3³·5·7	40	79
180	2²·3²·5	9	17	18,480	2⁴·3·5·7·11	43	79
252	2²·3²·7	11	17	25,200	2⁴·3²·5²·7	47	89
300	2²·3·5²	13	17	27,720	2³·3²·5·7·11	49	95
720	2⁴·3²·5	16	29	36,960	2⁵·3·5·7·11	50	95
1,080	2³·3³·5	17	31	41,580	2²·3³·5·7·11	51	95
1,440	2⁵·3²·5	18	35	45,360	2⁴·3⁴·5·7	52	99
1,680	2⁴·3·5·7	20	39	50,400	2⁵·3²·5²·7	54	107
2,160	2⁴·3³·5	21	39	55,440	2⁴·3²·5·7·11	61	119

Notes

1. That is, it’s only a palindrome in two bases less than 101. In higher bases, “101” is a single digit, so is trivially a palindrome (as the numbers 1 through 9 are in base 10).

2. In base b, there are b digits, including 0. So base 2 has two digits, 0 and 1; base 10 has ten digits, 0-9; base 16 has sixteen digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F.

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